Momentum is defined as the product of mass and velocity.

A rapidly moving truck has a large momentum. It takes a large or prolonged force to get the truck up to this speed, or to bring it to a stop afterwards.

A slow moving snail has a small momentum. It would be easy to slow down or speed up a little snail.

$$p = mv$$

\(p\) = momentum [kg m/s] vector
\(m\) = mass [kg]
\(v\) = velocity[m/s] vector
Example: What is the momentum of a bowling ball?
Google a typical mass and velocity.
solution $$\text{convert to m/s}$$ $$ 18 \left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 8.0 \mathrm{\tfrac{m}{s}} $$
$$p=mv$$ $$p=(5 \, \mathrm{kg})(7.6 \, \mathrm{\tfrac{m}{s}})$$ $$p=38 \,\mathrm{kg \tfrac{m}{s}}$$


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

derivation of impulse $$F = ma$$ $$a = \frac{\Delta v}{\Delta t}$$ $$\text{substitute acceleration with Δv/Δt}$$ $$F = m \frac{\Delta v}{\Delta t}$$ $$F \Delta t = m \Delta v$$
$$F \Delta t = m(v_f - v_i)$$ $$F \Delta t = mv_f - mv_i$$ $$F \Delta t = p_f - p_i$$ $$F \Delta t = \Delta p$$

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

\(J\) = 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
\(m\) = mass [kg]
\(v\) = velocity[m/s] vector
Example: A medicine ball is hurtling towards you at 15 m/s. You weakly try to stop it by applying a 100 N force for 0.1 s. You don't stop the ball, but it slows to 14 m/s. Calculate the mass of the ball.
solution $$F \Delta t = m\Delta v$$ $$\frac{F \Delta t}{\Delta v} = m$$ $$\frac{(-100)(0.1)}{14-15} = m$$ $$\frac{(-100)(0.1)}{-1} = m$$ $$10 \, \mathrm{kg} = m$$
Example: The Tesla Model S car has a mass of 2,250 kg. 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 $$ 60 \left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}} $$
$$F \Delta t = m\Delta v$$ $$F = \frac{m\Delta v}{\Delta t}$$ $$F = \frac{(2250)(26.8)}{2.8}$$ $$F = \frac{60300}{2.8}$$ $$F = 21536 \, \mathrm{N}$$

How does the 0 to 60 miles/hour acceleration compare to the acceleration of gravity?
solution $$F = 21536 N$$ $$F = ma$$ $$\frac{F}{m}=a$$ $$\frac{21536}{2250}=a$$ $$a = 9.57 \,\mathrm{ \tfrac{m}{s^2} }$$ $$ g = 9.81 \, \mathrm{\tfrac{m}{s^2}} $$

It's just a bit under the acceleration of gravity. How would that 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: conservation doesn't hold 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)

$$ \sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$m_1 u_1 + m_2u_2 + \cdots = m_1v_1 + m_2v_2 + \cdots$$

\(p\) = momentum [kg m/s] vector
\(u\) = 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 $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{green + white = green + white}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(2)(3)+(10)(-1)=(2)v_{1}+(1)(10)$$ $$6-10=(2)v_{1}$$ $$-4=(2)v_{1}$$ $$-2 \mathrm{\tfrac{m}{s}} = v_{1}$$

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 $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$ {\color{#0f0} \blacksquare} \quad +\quad {\color{#419eff} \blacksquare} \quad+\quad {\color{orange} \blacksquare} = {\color{#0f0} \blacksquare} \quad+\quad {\color{#419eff} \blacksquare} \quad+\quad {\color{orange} \blacksquare}$$ $$m_{1}u_{1}+m_{2}u_{2}+m_{3}u_{3}=m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}$$ $$(1.6)(3)+(4.9)(2)+(3.6)(-1)=(1.6)v_{1}+(4.9)(1.18)+(3.6)(1.58)$$ $$11=1.6v_{1}+11.47$$ $$-0.29 \mathrm{\tfrac{m}{s}} = v_{1}$$