# Momentum

Momentum is defined as the product of mass and velocity. For example, a heavy truck moving rapidly has a large momentum—it takes a large or prolonged force to get the truck up to this speed, and it takes a large or prolonged force to bring it to a stop afterwards. If the truck were lighter, or moving more slowly, then it would have less momentum.

# $$p = mv$$

$$p$$ = momentum [kg m/s] vector
$$m$$ = mass [kg]
$$v$$ = velocity[m/s] vector

# Impulse

An impulse is defined as both a force multiplied by time (FΔt) and a change in momentum (Δp). The longer a force is applied the more the momentum will change.

Examples of impulse:
cars speeding up
cars crashing
rockets accelerating
kicking a soccer ball
punching
jumping

derivation of impulse

The equation for impulse can be produce by replacing acceleration in F = ma with Δv/Δt.

$$F = ma$$ $$a = \frac{\Delta v}{\Delta t}$$ $$F = m \frac{\Delta v}{\Delta t}$$ $$F \Delta t = m \Delta v$$
$$F \Delta t = m(v_f - v_i)$$ $$F \Delta t = mv_f - mv_i$$ $$F \Delta t = p_f - p_i$$ $$F \Delta t = \Delta p$$

# $$J = F \Delta t = \Delta p$$ $$F \Delta t = m\Delta v$$

$$J$$ = impulse [Ns, kg m/s] vector
$$F$$ = force [N, kg m/s²] vector
$$\Delta t$$ = time period [s]
$$\Delta p$$ = change in momentum [kg m/s] vector
Example: A medicine ball is hurtling towards you at 15 m/s. You weakly try to stop it by applying a 100N force for 0.1s. You don't stop the ball, but it slows to 14 m/s. Calculate the mass of the ball.
solution $$F \Delta t = m\Delta v$$ $$\frac{F \Delta t}{\Delta v} = m$$ $$\frac{(100)(0.1)}{15-14} = m$$ $$\frac{(100)(0.1)}{1} = m$$ $$10kg = m$$
Example: The Tesla Model S car has a mass of 2,250kg. It holds the record for fastest 0 to 60 miles/hour acceleration of a production car with 2.8 seconds. Convert the miles/hour into m/s, and find the force produced by the car.
solution $$60\left(\frac{\color{red}{mile}}{\color{blue}{hour}}\right)\left(\frac{1609 m}{1 \color{red}{mile}}\right)\left(\frac{1 \color{blue}{hour}}{3600s}\right) = 26.8\small\frac{m}{s}$$
$$F \Delta t = m\Delta v$$ $$F = \frac{m\Delta v}{\Delta t}$$ $$F = \frac{(2250)(26.8)}{2.8}$$ $$F = \frac{60300}{2.8}$$ $$F = 21536 N$$
Example: How does the 0 to 60 miles/hour acceleration of the Tesla Model S from the previous example compare to the acceleration of gravity?
solution $$F = 21536 N$$ $$F = ma$$ $$\frac{F}{m}=a$$ $$\frac{21536}{2250}=a$$ $$a = 9.57 \small \frac{m}{s^2}$$ $$g = 9.81 \small \frac{m}{s^2}$$

Wow, how is that going to feel?

# Conservation of Momentum

Momentum is useful because the total momentum for a system of objects is conserved. This means the sum of all momentums won't change even if the objects collide.

Click the simulation above a few times and note that the total momentum is the same before and after a collision.

Conservation of momentum isn't true if one of the objects in the system interacts with an object outside the system. For example if an object hits the ground, and the ground isn't part of our system.

Conservation of momentum means we can set the total momentums at two points in time equal to each other. (as long as there are no outside forces)

## $$p_{initial} = p_{final}$$ $$p_{1i}+p_{2i}+...=p_{1f}+p_{2f}+...$$

### $$m_{1}u_{1}+m_{2}u_{2}+...=m_{1}v_{1}+m_{2}v_{2}+...$$

$$u$$ = initial velocity [m/s] vector
$$v$$ = final velocity [m/s] vector
$$m$$ = mass [kg]

Example: A green and white box collide. The green box has a mass of 2kg and is moving to the right at 3m/s. The white box is moving to the left at 1m/s and has a mass of 10kg. If the white box has stopped after the collision what is the velocity of the green box?
solution

initial momentum = final momentum

green + white = green + white

$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(2)(3)+(10)(-1)=(2)v_{1}+(1)(0)$$ $$6-10=(2)v_{1}$$ $$-4=(2)v_{1}$$ $$-2 \tfrac{m}{s}= v_{1}$$

green box blue box orange box
m = 1.60kg m = 4.90kg m = 3.60kg
u = 3.0m/s u = 2.0m/s u = -1.0m/s
v = ??? v = 1.18m/s v = 1.58m/s
Example: Three boxes collide. Use the table to find the final velocity of the green box
solution

initial momentum = final momentum

$$m_{1}u_{1}+m_{2}u_{2}+m_{3}u_{3}=m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}$$ $$(1.6)(3)+(4.9)(2)+(3.6)(-1)=(1.6)v_{1}+(4.9)(1.18)+(3.6)(1.58)$$ $$11=1.6v_{1}+11.47$$ $$-0.29\tfrac{m}{s}= v_{1}$$

# 2-D Conservation of Momentum

Momentum is a vector. It has a direction and a magnitude. We can solve for the horizontal and vertical components separately just like how we solved 2-d Motion problems. Although in this case there isn't a time variable to link up the vertical and horizontal equations.

Horizontal Example: A pink hexagon and a cyan triangle collide. Before the collision the hexagon has a mass of 5.20kg and is moving to the right at 2m/s. The cyan triangle is moving to the left at 1m/s and has a mass of 1.17kg. If the triangle bounces off the hexagon at 2.92 m/s to the right what is the horizontal velocity of the hexagon? (ignore the vertical information)
solution

initial momentum = final momentum

hexagon + triangle = hexagon + triangle

$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(2)+(1.17)(-1)=(5.20)v_{1}+(1.17)(2.92)$$ $$10.40-1.17=(5.20)v_{1}+3.42$$ $$5.81=(5.20)v_{1}$$ $$1.12 \small\frac{m}{s}=\normalsize v_{1}$$

Vertical Example: Lets take a look at the vertical aspect of the pink hexagon and cyan triangle collide. Before the collision the hexagon has a mass of 5.20kg and is not moving up or down. The final vertical velocity of the hexagon is 0.23 down. The final vertical velocity of the triangle is 0.67 up. What is the triangle's initial vertical velocity.
solution

initial momentum = final momentum

hexagon + triangle = hexagon + triangle

$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(0)+(1.17)(u_{2})=(5.20)(-0.23)+(1.17)(0.67)$$ $$(1.17)(u_{2})=-0.41$$ $$u_{2}=-0.35\small\frac{m}{s}$$