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:
Below is a particle that has no net force and therefore moves at a constant velocity:
A object with no net force on it
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
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).
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
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 (1,0,4) m/s. What is the location of the ball after 10s, if it starts at the origin. Assume the coefficient of friction is 0.3.
Declare known variables:
Change mass to kilograms:
[math]\displaystyle{ \mathbf{m} = \frac{1000} {1000} = 1 kg }[/math]
[math]\displaystyle{ \overrightarrow{\mathbf{v}} = (1,0,4) \frac{m} {s} }[/math]
[math]\displaystyle{ \Delta{t} = 10 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 * (1,0,4) }[/math] = [math]\displaystyle{ (1,0,4) kg * \frac{m} {s} }[/math]
Find change in momentum:
[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}} = (0,2.94,0) N * 10 s = (0,29.4,0) kg * \frac{m} {s} }[/math]
Find the final momentum
[math]\displaystyle{ \overrightarrow{p}_{final} = \overrightarrow{p}_{initial} + \Delta{\overrightarrow{p}} = (1,0,4) + (0,29.4,0) = (1,29.4,4) kg * \frac{m} {s} }[/math]
Find final velocity
[math]\displaystyle{ \overrightarrow{v}_{final} = \frac{\overrightarrow{p}_{final}} {\mathbf{m}} = \frac{(1,29.4,4)} {1} = (1,29.4,4) kg * \frac{m} {s} }[/math]
Find displacement
[math]\displaystyle{ \Delta {d} = \overrightarrow{v}_{final} * \Delta {t} = (1,29.4,4) \frac{m} {s} * 10 s = (10,294,40) m }[/math]
Find final position
[math]\displaystyle{ \mathbf{r}_{final} = \mathbf{r}_{initial} + \Delta {d} = (0,0,0) m + (10,294,40) m = (10,294,40) m }[/math]
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