Physics Book - User contributions [en]

Predicting Change in multiple dimensions

2015-12-05T21:43:58Z

Rishavbose365: /* A Computational Model */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

(If it does not work take the '?outputOnly=true' out of the url and try again)

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

You kick a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Before Newton, French scientist and philosopher Descartes introduced the concept of momentum. He used the concept of momentum to describe how people moved when objects were thrown at them. He focused generally on the conservation of momentum when dealing with collisions. Newton's laws further expanded on the idea of conservation of momentum. The ideas that F = ma and the idea that for every action there is an equal and opposite reaction are the basis for many problems and concepts explained in this section. More information here:

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

https://en.wikipedia.org/wiki/Momentum#History_of_the_concept

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

==References==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

[[Category:Momentum]]

Predicting Change in multiple dimensions

2015-12-03T01:13:16Z

Rishavbose365: /* Difficult */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

You kick a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Before Newton, French scientist and philosopher Descartes introduced the concept of momentum. He used the concept of momentum to describe how people moved when objects were thrown at them. He focused generally on the conservation of momentum when dealing with collisions. Newton's laws further expanded on the idea of conservation of momentum. The ideas that F = ma and the idea that for every action there is an equal and opposite reaction are the basis for many problems and concepts explained in this section. More information here:

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

https://en.wikipedia.org/wiki/Momentum#History_of_the_concept

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

==References==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

[[Category:Momentum]]

Predicting Change in multiple dimensions

2015-12-03T00:40:28Z

Rishavbose365: /* See also */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Before Newton, French scientist and philosopher Descartes introduced the concept of momentum. He used the concept of momentum to describe how people moved when objects were thrown at them. He focused generally on the conservation of momentum when dealing with collisions. Newton's laws further expanded on the idea of conservation of momentum. The ideas that F = ma and the idea that for every action there is an equal and opposite reaction are the basis for many problems and concepts explained in this section. More information here:

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

https://en.wikipedia.org/wiki/Momentum#History_of_the_concept

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

==References==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

[[Category:Momentum]]

Predicting Change in multiple dimensions

2015-12-03T00:40:19Z

Rishavbose365: /* References */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Before Newton, French scientist and philosopher Descartes introduced the concept of momentum. He used the concept of momentum to describe how people moved when objects were thrown at them. He focused generally on the conservation of momentum when dealing with collisions. Newton's laws further expanded on the idea of conservation of momentum. The ideas that F = ma and the idea that for every action there is an equal and opposite reaction are the basis for many problems and concepts explained in this section. More information here:

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

https://en.wikipedia.org/wiki/Momentum#History_of_the_concept

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

==References==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

[[Category:Momentum]]

Predicting Change in multiple dimensions

2015-12-03T00:39:56Z

Rishavbose365: /* History */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Before Newton, French scientist and philosopher Descartes introduced the concept of momentum. He used the concept of momentum to describe how people moved when objects were thrown at them. He focused generally on the conservation of momentum when dealing with collisions. Newton's laws further expanded on the idea of conservation of momentum. The ideas that F = ma and the idea that for every action there is an equal and opposite reaction are the basis for many problems and concepts explained in this section. More information here:

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

https://en.wikipedia.org/wiki/Momentum#History_of_the_concept

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

==References==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

[[Category:Momentum]]

Predicting Change in multiple dimensions

2015-12-03T00:27:02Z

Rishavbose365: /* Connectedness */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

==References==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

[[Category:Momentum]]

Predicting Change in multiple dimensions

2015-12-03T00:18:31Z

Rishavbose365: /* References */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

==References==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

[[Category:Momentum]]

Predicting Change in multiple dimensions

2015-12-03T00:17:50Z

Rishavbose365: /* See also */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

https://en.wikipedia.org/wiki/Momentum

https://en.wikibooks.org/wiki/General_Mechanics/Momentum

http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-03T00:12:13Z

Rishavbose365: /* A Computational Model */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

====Others====

[https://phet.colorado.edu/en/simulation/legacy/collision-lab Collision Lab]

[https://phet.colorado.edu/en/simulation/legacy/projectile-motion Projectile Motion]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-03T00:03:35Z

Rishavbose365: /* A Computational Model */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move (Click run to start simulation):

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T21:04:04Z

Rishavbose365: /* Examples */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

==Examples==

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T20:50:51Z

Rishavbose365: /* Elastic Collision */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T20:50:34Z

Rishavbose365: /* A Computational Model */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Elastic Collision====

Two objects undergo perfectly elastic collision

[https://trinket.io/embed/glowscript/765aa96eac?outputOnly=true Two objects collide perfectly elastically]

====Inelastic Collision====

Two objects undergo inelastic collision

[https://trinket.io/embed/glowscript/8bc7a1d125?outputOnly=true Two inelastic objects collide]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T05:33:00Z

Rishavbose365: /* Relate by Force */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

or:

<math> \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,.</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Collision====

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T05:31:09Z

Rishavbose365: /* A Computational Model */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

====Collision====

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T05:30:29Z

Rishavbose365: /* The Main Idea */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is the vector quantity equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T05:24:35Z

Rishavbose365: /* Relate by Force */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

This can also be expressed as:

<math> \overrightarrow{F}_{net} = \frac {d\overrightarrow{p}} {dt}</math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T05:03:24Z

Rishavbose365: /* Connectedness */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

Momentum is what some physicists call the fundamental principle. Almost everything in physics is based off this principle. This is why it is one of the most interesting parts of physics. From getting a few measurements, we have shown how we can predict the final destination of any particle. We can create models for real life examples, like finding how hard we have to kick a ball in a sport, where a ball will stop rolling on the ground and for many instances of projectile motion. It basically allows us to create mathematical models and predict the final destination of an object for almost any situation.

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T02:20:18Z

Rishavbose365: /* Middling */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,10). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,10) \frac{m} {s} = (50,75,50) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,50) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement in X'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find Displacement in Z'''

<math> \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

<math> \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,71.4285) m = (71.4285,0,71.4285) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T02:16:45Z

Rishavbose365: /* Part E */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,0). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,0) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,0) \frac{m} {s} = (50,75,0) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,0) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,0) m = (71.4285,0,0) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

'''State Known Variables'''

<math> \mathbf{m} = 0.76 kg </math>

<math> \Delta{t} = .034 s </math>

'''Find Change in Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = 0 </math>

<math> \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} </math>

Use V Initial from Part A:

<math> \Delta{\overrightarrow{\mathbf{p}}} = \overrightarrow{\mathbf{p}}_{final} - \overrightarrow{\mathbf{p}}_{initial} = m * \overrightarrow{v}_{initial} - 0 = 0.76 kg * (24,14.25,0) \frac{m} {s} = (18.24,10.83,0) kg \frac{m} {s}</math>

<math> |\Delta{\mathbf{p}}| = \sqrt{18.24^2 + 10.83^2} = 21.213 kg \frac{m} {s}</math>

'''Find Net Force'''

<math> |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} </math>

<math> |\mathbf{F}_{net}| = \frac{|\Delta{\mathbf{p}}|} {\Delta{t}} = \frac{21.213} {.034} = 623.908 N</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T02:15:45Z

Rishavbose365: /* Difficult */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,0). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,0) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,0) \frac{m} {s} = (50,75,0) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,0) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,0) m = (71.4285,0,0) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 34 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T02:00:00Z

Rishavbose365: /* Difficult */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,0). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,0) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,0) \frac{m} {s} = (50,75,0) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,0) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,0) m = (71.4285,0,0) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 340 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T01:59:35Z

Rishavbose365: /* Difficult */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,0). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,0) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,0) \frac{m} {s} = (50,75,0) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,0) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,0) m = (71.4285,0,0) m</math>

===Difficult===

Using a cannon you shoot a 0.5 kg ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 340 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T01:58:07Z

Rishavbose365: /* Part D */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,0). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,0) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,0) \frac{m} {s} = (50,75,0) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,0) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,0) m = (71.4285,0,0) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 340 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

'''Find Max Time'''

The velocity at the max point is 0. Use equation from part A:

<math> v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0</math>

<math>v_{initial y} = g \Delta{t_{max}}</math>

<math>\Delta{t_{max}} = \frac{v_{initial y}} {g}</math>

'''Find Max Height'''

<math> \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}}</math>

<math> y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2</math>

<math> y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})}</math>

<math> y_{max} = \frac{1}{2} {(\frac{v_{initial y}^2} {g})} = \frac{1}{2} {(\frac{14.25^2} {9.8})} = 10.3603 m</math>

====Part E====

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T01:43:21Z

Rishavbose365: /* Part C */

This page discusses the use of momentum to predict change in multi-dimensions and examples of how it is used.

Claimed by rbose7

==The Main Idea==

The '''linear momentum''', or '''translational momentum''' of an object is equal to the product of the mass and velocity of an object. A change in any of these properties is reflected in the momentum.

If the object(s) are in a closed system not affected by external forces the total momentum of the system cannot change.

We can apply these properties to all three dimensions and use momentum to predict the path an object will follow over time by observing the change in momentum in the same way we did in one-dimension.

===A Mathematical Model===

This change in momentum is shown by the formula:

<math>\Delta \overrightarrow{p}</math> = <math>\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}</math> = <math>m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial}</math>

Or by relating it to force:

<math>\Delta p = F \Delta t\,</math>

====Relate by Velocity====

Given the velocity:

<math>\overrightarrow{v} = \left(v_x,v_y,v_z \right) </math>

For an object with mass <math> \mathbf{m} </math>

The object has a momentum of :

<math>\overrightarrow{p} </math> = <math>\overrightarrow{v} * \mathbf{m} </math>
= <math> \left(v_x,v_y,v_z \right) * \mathbf{m} </math>
= <math> \left(\mathbf{m} v_x,\mathbf{m} v_y,\mathbf{m} v_z \right) </math>

====Relate by Force====

Given the force:

<math>\overrightarrow{F} = \left(F_x,F_y,F_z \right) </math>

And change in time:

<math>\Delta t</math>

<math>\Delta p = \overrightarrow{F} \Delta t\,</math>
= <math> \left(F_x,F_y,F_z \right) * \Delta t </math>
= <math> \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

<math>\overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p </math>
= <math> \overrightarrow{p}_{initial} + \left(\Delta tF_x,\Delta tF_y,\Delta tF_z \right) </math>

====Multiple Particles====

If we have multiple particles with a force acting on it, we can use the same process to predict its path. The only difference is that we pretend the particles are just on large particle with its center at the '''center of mass'''.

Center of Mass:

<math> \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}.</math>

<math> \overrightarrow{r}_{cm} = \frac{m_1 (r_{x1},r_{y1},r_{z1}) + m_2 (r_{x2},r_{y2},r_{z2}) + \cdots}{m_1 + m_2 + \cdots}.</math>

Using this we carry out the same calculations, but use the mass:

<math> m_{total} = m_1 + m_2 + \cdots </math>

and use the velocity:

<math> \overrightarrow{v}_{cm} </math>

and only move the mass as a whole from the center:

<math> \overrightarrow{r}_{cm} </math>

===A Computational Model===

Below are models that use change in momentum to predict how particles move:

====A object with no net force on it====

Below is a particle that has no net force and therefore moves at a constant velocity:

[https://trinket.io/glowscript/b40327b7c7?outputOnly=true A object with no net force on it]

====A object with the force of gravity====

Below is an object moving with gravity acting on it. Because gravity acts in the 'y' direction, the object's y component for velocity decreases:

[https://trinket.io/embed/glowscript/7065ab6e93?outputOnly=true A object with the force of gravity]

====Many Particles====

Below are several objects moving with gravity acting on it, using calculations from center of mass (it is usually more accurate to apply calculations on each particle individually, but this is good for a big picture).

[https://trinket.io/glowscript/80c86eca7b?outputOnly=true Many Particles]

====A object launched from a cliff====

Below is an object launched with an initial velocity that has gravity acting on it. It loses velocity in the y direction due to gravity until it hits the ground:

[https://trinket.io/embed/glowscript/4507df1ea2?outputOnly=true A object launched from a cliff]

====An electron and proton====

We can also use momentum to model the path of more complex models, like a proton and electron near each other:

[https://trinket.io/embed/glowscript/bc77968855?outputOnly=true An electron and proton with non-zero velocities with electric force included]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A ball of mass 1000 g rolls across the floor with a velocity of (0,10,0) m/s. After how much time does the ball stop? Where does it stop if it starts at the origin? Assume the coefficient of friction is 0.3.

We need to find when velocity is 0 or when final momentum is 0.

'''Declare known variables:'''

Change mass to kilograms:

<math> \mathbf{m} = \frac{1000} {1000} = 1 kg </math>

<math> \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} </math>

<math> \mu = .3 </math>

'''Find the initial momentum:'''

<math> \overrightarrow{p}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}} </math>
= <math> 1 * (0,10,0) </math>
= <math> (0,10,0) kg * \frac{m} {s} </math>

'''Find time passed:'''

<math> \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} </math>

<math> \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu </math>

<math> \overrightarrow{\mathbf{F}}_{normal} = (0,\mathbf{m} * \mathbf{g} = 1 kg * 9.8 \frac{m} {s^2},0) = (0,9.8,0) N </math>

<math> \mathbf{\overrightarrow{F}}_{net} = (0,9.8,0) N * 0.3 = (0,2.94,0) N</math>

<math> \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} </math>

<math> (0,10,0) = (0,2.94,0) N * \Delta {t} </math>

<math> \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} </math>

<math> \Delta {t} = .294 s </math>

'''Find displacement'''

<math> \Delta {d} = \overrightarrow{v}_{avg} * \Delta {t} = \frac {(0,10,0)} {2} \frac{m} {s} * 0.294 s = (0,1.47,0) m </math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (0,1.47,0) m = (0,1.47,0) m</math>

===Middling===

You kick a 5kg ball off a 100m cliff at a velocity of (10,15,0). How long does it take for the ball to reach the ground? How far away does it land?

'''Declare Known Variables'''

<math> \mathbf{m} = 5 kg </math>

<math> \mathbf{h} = 100 m </math>

<math> \overrightarrow{\mathbf{v}}_{initial} = (10,15,0) \frac{m} {s} </math>

'''Find Initial Momentum'''

<math> \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} </math>

<math> \overrightarrow{\mathbf{p}}_{initial} = 5kg * (10,15,0) \frac{m} {s} = (50,75,0) kg * \frac{m} {s}</math>

'''Find Final Momentum'''

Because the ball will hit the ground the final y component of the velocity and final momentum will be 0. Because gravity only affects the y component, the x and z components are unchanged.

<math> \overrightarrow{\mathbf{p}}_{final} = (50,0,0) kg * \frac{m} {s}</math>

'''Find Net Force'''

<math> \overrightarrow{\mathbf{F}}_{net} = \overrightarrow{\mathbf{F}}_{g} = (0,\mathbf{m} * \mathbf{g},0) = (0, 5kg * 9.8 \frac{m} {s^2},0) = (0,49,0) N</math>

'''Find time passed'''

<math> \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} </math>

<math> (0,70,0) = \Delta {t} * (0,49,0) </math>

<math> \Delta {t} = 1.42857 s </math>

'''Find Displacement'''

<math> \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s}</math>

<math> \Delta {\mathbf{d}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m</math>

'''Find final position'''

<math> \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,100,0) m + (71.4285,-100,0) m = (71.4285,0,0) m</math>

===Difficult===

Using a cannon you shoot a ball through 5 meter high posts from 60 meters away. It takes 2.5s to travel through the posts. Answer the following questions:

a)What is the initial velocity?

b)What angle did you shoot it from?

c)What is the velocity of the ball as it crosses the posts?

d)What was the balls maximum height?

e)What was the force on the ball if the ball is .76 kg and the impact lasts for 340 ms?

====Part A====

'''Find Initial Velocity in X direction'''

<math> \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 </math>

<math> \frac{m *\Delta {v}_x} {\Delta {t}} = 0 </math>

Or the x component of the velocity is constant.

<math> \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} </math>

'''Find Initial Velocity in Y direction'''

<math> \frac{\Delta {p}_y} {\Delta {t}} = \mathbf{F}_{net} = - \mathbf{m}\mathbf{g} = \frac{m *\Delta {v}_y} {\Delta {t}} </math>

<math> \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g}</math>

<math> \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y}</math>

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} </math>

<math> v_{final y} = 2v_{avg y} - v_{initial y} </math>

<math>- \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} </math>

<math> v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} </math>

<math> \frac{\Delta {y}} {\Delta {t}} = v_{avg y} </math>

<math> \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2</math>

<math> v_{initial y} = \frac{\Delta {y}} {\Delta {t}} + \frac{1} {2} \mathbf{g} \Delta {t} = \frac{5} {2.5} + \frac{1} {2} * 9.8 * 2.5 = 14.25 \frac {m} {s} </math>

'''Find Initial Velocity in Z direction'''

No change in Z direction, therefore:

<math> v_{initial z} = 0 \frac{m} {s} </math>

'''Find Initial Velocity'''

<math> \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s}</math>

====Part B====

'''Find the Angle'''

<math> \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375</math>

<math> \theta = \tan^{-1} 0.59375 = 0.53358 radians</math> or 30.6997 degrees

====Part C====

'''Find Final Velocities'''

Final velocity is equal to the velocity as it goes through the poles.

No change in x velocity:

<math> v_{final x} = 24 \frac{m} {s} </math>

From part A:

<math> v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y}</math>

<math> v_{final y} = - 9.8 * 2.5 + 14.25</math>

<math> v_{final y} = - 10.25 \frac{m} {s}</math>

<math>\overrightarrow{v}_{final} = (24, -10.25, 0) \frac{m} {s}</math>

====Part D====

====Part E====

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

==References==

This section contains the the references you used while writing this page

[[Category:Momentum]]

https://en.wikipedia.org/wiki/Momentum

Predicting Change in multiple dimensions

2015-12-02T00:57:12Z

Rishavbose365: /* Part B */

Predicting Change in multiple dimensions

2015-12-01T22:59:25Z

Rishavbose365: /* Difficult */

Predicting Change in multiple dimensions

2015-12-01T22:58:14Z

Rishavbose365: /* Part A */

Predicting Change in multiple dimensions

2015-12-01T22:33:36Z

Rishavbose365: /* Part A */

Predicting Change in multiple dimensions

2015-12-01T22:31:18Z

Rishavbose365: /* Difficult */

Predicting Change in multiple dimensions

2015-12-01T22:09:33Z

Rishavbose365: /* Middling */

Predicting Change in multiple dimensions

2015-12-01T22:00:15Z

Rishavbose365: /* A Computational Model */

Predicting Change in multiple dimensions

2015-12-01T21:57:57Z

Rishavbose365: /* Examples */

Predicting Change in multiple dimensions

2015-12-01T21:55:01Z

Rishavbose365: /* Middling */

Predicting Change in multiple dimensions

2015-12-01T21:18:47Z

Rishavbose365: /* Examples */

Predicting Change in multiple dimensions

2015-12-01T21:17:42Z

Rishavbose365: /* Middling */

Predicting Change in multiple dimensions

2015-12-01T20:57:29Z

Rishavbose365: /* Simple */

Predicting Change in multiple dimensions

2015-11-30T21:56:49Z

Rishavbose365: /* Simple */

Predicting Change in multiple dimensions

2015-11-30T21:05:54Z

Rishavbose365: /* Simple */

Predicting Change in multiple dimensions

2015-11-30T21:00:20Z

Rishavbose365: /* Simple */

Predicting Change in multiple dimensions

2015-11-30T20:53:50Z

Rishavbose365: /* Simple */

Predicting Change in multiple dimensions

2015-11-30T20:52:21Z

Rishavbose365: /* Examples */

Predicting Change in multiple dimensions

2015-11-30T20:48:56Z

Rishavbose365: /* A Mathematical Model */

Predicting Change in multiple dimensions

2015-11-30T20:45:59Z

Rishavbose365:

Predicting Change in multiple dimensions

2015-11-30T15:51:43Z

Rishavbose365: /* Simple */

Predicting Change in multiple dimensions

2015-11-30T15:42:40Z

Rishavbose365: /* A Computational Model */

Predicting Change in multiple dimensions

2015-11-30T15:27:46Z

Rishavbose365: /* Multiple Particles */

Predicting Change in multiple dimensions

2015-11-30T14:49:02Z

Rishavbose365: /* A Mathematical Model */

Predicting Change in multiple dimensions

2015-11-30T14:48:14Z

Rishavbose365: /* Multiple Particles */

Predicting Change in multiple dimensions

2015-11-30T14:47:44Z

Rishavbose365: /* Multiple Particles */