We need to find when the velocity will be equal to zero, given a constant frictional force
Declare known variables and coordinate system:
First, we convert mass into kilograms
[math]\displaystyle{ \mathbf{m} = \frac{1000} {1000} = 1 kg }[/math]
Next, we restate the velocity and coefficient of friction
[math]\displaystyle{ \vec{\mathbf{v}} = (3,4,0) \; \frac{m} {s} }[/math]
[math]\displaystyle{ \mu = .3 }[/math]
Next, from the context of the problem, we take x and y to be perpendicular to the force of gravity (that is, the floor is flat), so gravity will be directed in the negative z direction.
Find the initial momentum:
[math]\displaystyle{ \vec{p}_{i} = \mathbf{m} \vec{\mathbf{v}} = (1 \; kg)((3,4,0) \frac{m}{s}) = (3,4,0) \; \frac{kg \; m}{s} }[/math]
Find time passed:
We have the momentum principle:
[math]\displaystyle{ \Delta{\vec{p}} = \vec{\mathbf{F}}_{net} \Delta t }[/math]
Next, we compute the force due to friction (see Friction):
[math]\displaystyle{ \vec{\mathbf{F}}_{frict} = \vec{\mathbf{F}}_{N}*\mu }[/math]
Expanding in the coordinate system we defined earlier:
[math]\displaystyle{ \vec{\mathbf{F}}_{normal} = (0,0,\mathbf{m}|\mathbf{g}|) = (0,0,9.8) N }[/math]
Now we compute friction. Friction will always be directed in the opposite direction of the velocity, so just as velocity has vector for [math]\displaystyle{ \vec{\mathbf{v}} = (3,4,0) \; m/s }[/math], friction will have vector form [math]\displaystyle{ \vec{\mathbf{F}}_{frict} = |\vec{\mathbf{F}}_{frict}|(-3/5,-4/5,0) }[/math], where [math]\displaystyle{ |\vec{\mathbf{F}}_{frict}| }[/math] is the magnitude of the frictional force, and the vector has been normalized, so that the its magnitude is equal to one (you should recognize 3,4, and 5 as a pythagorean triple). Therefore, we calculate
[math]\displaystyle{ |\vec{\mathbf{F}}_{frict}| = \mu |\vec{\mathbf{F}}_{normal}| = 2.94 \;N }[/math]
[math]\displaystyle{ \vec{\mathbf{F}}_{frict} = (2.94 \; N)(-3/5,-4/5,0) = (-1.764,-2.352,0) \; N }[/math]
Now, we compute the time it takes to stop based on the change in momentum
[math]\displaystyle{ \Delta \vec{p} = \vec{p}_{i} = (\Delta t)\vec{\mathbf{f}}_{frict} }[/math]
Since the final momentum will be 0, the change in momentum is simply [math]\displaystyle{ \Delta \vec{\mathbf{p}} = - \vec{\mathbf{p}}_i }[/math], so we have
[math]\displaystyle{ (-3,-4,0) = (-1.764,-2.352,0) N * \Delta {t} }[/math]
[math]\displaystyle{ \Delta {t} = \frac {(3,-4.,0) \; kg \; m/s} {(-1.764,-2.352,0) \; N} }[/math]
[math]\displaystyle{ \Delta {t} = 1.7 s }[/math]
Find displacement
We have two options. We may use average velocity:
[math]\displaystyle{ \Delta {d} = \vec{v}_{avg} * \Delta {t} = \frac {(3,4,0)} {2} \frac{m} {s} * 1.7 s = (2.55,3.4,0) m }[/math]
and then find the final position:
[math]\displaystyle{ \mathbf{r}_{f} = \mathbf{r}_{i} + \Delta {d} = (0,0,0) m + (2.55,3.4,0) m = (2.55, 3.4,0) m }[/math]
Alternatively, we may use the kinematics equation
[math]\displaystyle{ \vec{\mathbf{r}}_f = \frac{\vec{\mathbf{a}} t^2}{2} + \vec{\mathbf{v}}_i t + \vec{\mathbf{r}}_i }[/math]
where acceleration [math]\displaystyle{ \vec{a} = \vec{\mathbf{F}}/\mathbf{m} = (-1.764, -2.352, 0) m/s^2 }[/math]. Thus this yields
[math]\displaystyle{ \vec{\mathbf{r}}_f = \frac{(-1.764, -2.352, 0)(1.7 \; s)^2}{2} + (3,4,0)(1.7 \; s) + (0,0,0) = (2.55, 3.4, 0) \; m
'''Find final position'''
\lt /div\gt \lt /div\gt
===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'''
\lt math\gt \mathbf{m} = 5 kg }[/math]
[math]\displaystyle{ \mathbf{h} = 100 m }[/math]
[math]\displaystyle{ \overrightarrow{\mathbf{v}}_{initial} = (10,15,10) \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 * (10,15,10) \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]\displaystyle{ \overrightarrow{\mathbf{p}}_{final} = (50,0,50) 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 in X
[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_x}} = \Delta{t} * \mathbf{v}_{avg x} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m }[/math]
Find Displacement in Z
[math]\displaystyle{ \mathbf{v}_{avg z} = \frac{\mathbf{v}_{final z} + \mathbf{v}_{initial z}} {2} = 50 \frac{m} {s} }[/math]
[math]\displaystyle{ \Delta {\mathbf{d_z}} = \Delta{t} * \mathbf{v}_{avg z} = 1.42857 s * 50 \frac{m} {s} = 71.4285 m }[/math]
[math]\displaystyle{ \Delta {\overrightarrow{\mathbf{d}}} = (71.4285,0,71.4285) m }[/math]
Find final position
[math]\displaystyle{ \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]\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]
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