Predicting Change in multiple dimensions

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]\displaystyle{ \Delta \overrightarrow{p} }[/math] = [math]\displaystyle{ \overrightarrow{p}_{final}-\overrightarrow{p}_{initial} }[/math] = [math]\displaystyle{ m\overrightarrow{v}_{final}-m\overrightarrow{v}_{initial} }[/math]

Or by relating it to force:

[math]\displaystyle{ \Delta p = F \Delta t\, }[/math]

Relate by Velocity

Given the velocity:

[math]\displaystyle{ \overrightarrow{v} = \left(v_x,v_y,v_z \right) }[/math]

For an object with mass [math]\displaystyle{ \mathbf{m} }[/math]

The object has a momentum of :

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

Relate by Force

Given the force:

[math]\displaystyle{ \overrightarrow{F} = \left(F_x,F_y,F_z \right) }[/math]

And change in time:

[math]\displaystyle{ \Delta t }[/math]

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

[math]\displaystyle{ \overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta p }[/math] = [math]\displaystyle{ \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]\displaystyle{ \overrightarrow{r}_{cm} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + \cdots}{m_1 + m_2 + \cdots}. }[/math]

[math]\displaystyle{ \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]\displaystyle{ m_{total} = m_1 + m_2 + \cdots }[/math]

and use the velocity:

[math]\displaystyle{ \overrightarrow{v}_{cm} }[/math]

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

[math]\displaystyle{ \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:

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:

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).

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:

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:

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]\displaystyle{ \mathbf{m} = \frac{1000} {1000} = 1 kg }[/math]

[math]\displaystyle{ \overrightarrow{\mathbf{v}} = (0,10,0) \frac{m} {s} }[/math]

[math]\displaystyle{ \mu = .3 }[/math]

Find the initial momentum:

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

Find time passed:

[math]\displaystyle{ \Delta{\overrightarrow{p}} = \overrightarrow{\mathbf{F}}_{net} * \Delta{t} }[/math]

[math]\displaystyle{ \mathbf{\overrightarrow{F}}_{net} = \overrightarrow{\mathbf{F}}_{normal}*\mu }[/math]

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

[math]\displaystyle{ \Delta{\overrightarrow{p}} = \overrightarrow{p}_{initial} = (0,2.94,0) N * \Delta {t} }[/math]

[math]\displaystyle{ (0,10,0) = (0,2.94,0) N * \Delta {t} }[/math]

[math]\displaystyle{ \Delta {t} = \frac {(0,2.94,0)} {(0,10,0)} }[/math]

[math]\displaystyle{ \Delta {t} = .294 s }[/math]

Find displacement

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \mathbf{m} = 5 kg }[/math]

[math]\displaystyle{ \mathbf{h} = 100 m }[/math]

[math]\displaystyle{ \overrightarrow{\mathbf{v}}_{initial} = (10,15,0) \frac{m} {s} }[/math]

Find Initial Momentum

[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{initial} = \mathbf{m} * \overrightarrow{\mathbf{v}}_{initial} }[/math]

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

Find Net Force

[math]\displaystyle{ \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]\displaystyle{ \Delta{\overrightarrow{\mathbf{p}}} = \Delta {t} * \overrightarrow{\mathbf{F}}_{net} }[/math]

[math]\displaystyle{ (0,70,0) = \Delta {t} * (0,49,0) }[/math]

[math]\displaystyle{ \Delta {t} = 1.42857 s }[/math]

Find Displacement

[math]\displaystyle{ \mathbf{v}_{avg x} = \frac{\mathbf{v}_{final x} + \mathbf{v}_{initial x}} {2} = 50 \frac{m} {s} }[/math]

[math]\displaystyle{ \Delta {\mathbf{d}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m }[/math]

Find final position

[math]\displaystyle{ \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]\displaystyle{ \frac{\Delta {p}_x} {\Delta {t}} = \mathbf{F}_{net} = 0 }[/math]

[math]\displaystyle{ \frac{m *\Delta {v}_x} {\Delta {t}} = 0 }[/math]

Or the x component of the velocity is constant.

[math]\displaystyle{ \overrightarrow{v}_x = \frac{\Delta {x}} {\Delta {t}} = \frac{60} {2.5} = 24 \frac{m}{s} }[/math]

Find Initial Velocity in Y direction

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

[math]\displaystyle{ \frac{\Delta {v}_y} {\Delta {t}} = -\mathbf{g} }[/math]

[math]\displaystyle{ \Delta {v}_y = - \mathbf{g} * \Delta {t} = v_{final y} - v_{initial y} }[/math]

[math]\displaystyle{ v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y} }[/math]

[math]\displaystyle{ v_{avg y} = \frac{v_{final y} + v_{initial y}} {2} }[/math]

[math]\displaystyle{ v_{final y} = 2v_{avg y} - v_{initial y} }[/math]

[math]\displaystyle{ - \mathbf{g} * \Delta {t} + v_{initial y} = 2v_{avg y} - v_{initial y} }[/math]

[math]\displaystyle{ v_{avg y} = v_{initial y} - \frac{1} {2} \mathbf{g} \Delta {t} }[/math]

[math]\displaystyle{ \frac{\Delta {y}} {\Delta {t}} = v_{avg y} }[/math]

[math]\displaystyle{ \Delta {y} = v_{initial y} \Delta {t} - \frac{1} {2} \mathbf{g} \Delta {t}^2 }[/math]

[math]\displaystyle{ 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]\displaystyle{ v_{initial z} = 0 \frac{m} {s} }[/math]

Find Initial Velocity

[math]\displaystyle{ \overrightarrow{v}_{initial} = (24,14.25,0) \frac {m} {s} }[/math]

Part B

Find the Angle

[math]\displaystyle{ \tan {\theta} = \frac{\mathbf{v_{initial y}}} {\mathbf{v_{initial x}}} = \frac{14.25} {24} = 0.59375 }[/math]

[math]\displaystyle{ \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]\displaystyle{ v_{final x} = 24 \frac{m} {s} }[/math]

From part A:

[math]\displaystyle{ v_{final y} = - \mathbf{g} * \Delta {t} + v_{initial y} }[/math]

[math]\displaystyle{ v_{final y} = - 9.8 * 2.5 + 14.25 }[/math]

[math]\displaystyle{ v_{final y} = - 10.25 \frac{m} {s} }[/math]

[math]\displaystyle{ \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]\displaystyle{ v_{max y} = v_{initial y} - g \Delta{t_{max}} = 0 }[/math]

[math]\displaystyle{ v_{initial y} = g \Delta{t_{max}} }[/math]

[math]\displaystyle{ \Delta{t_{max}} = \frac{v_{initial y}} {g} }[/math]

Find Max Height

[math]\displaystyle{ \frac{\Delta {y}} {\Delta{t}} = v_{initial y} - \frac{1}{2} g \Delta{t_{max}} }[/math]

[math]\displaystyle{ y_{max} - y_{initial} = v_{initial y} \Delta{t_{max}} - \frac{1}{2} g \Delta{t_{max}}^2 }[/math]

[math]\displaystyle{ y_{max} = \frac{v_{initial y}^2} {g} - \frac{1}{2} {(\frac{v_{initial y}^2} {g})} }[/math]

[math]\displaystyle{ 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]\displaystyle{ \mathbf{m} = 0.76 kg }[/math]

[math]\displaystyle{ \Delta{t} = .034 s }[/math]

Find Change in Momentum

[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{initial} = 0 }[/math]

[math]\displaystyle{ \overrightarrow{\mathbf{p}}_{final} = m * \overrightarrow{v}_{initial} }[/math]

Use V Initial from Part A:

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

Find Net Force

[math]\displaystyle{ |\Delta{\mathbf{p}}| = |\mathbf{F}_{net}| * \Delta{t} }[/math]

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

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