# Conservation of Momentum

The law of conservation of momentum states that total momentum doesn't change over time, even if two objects collide. When a property doesn't change over time we say that the property is conserved.

$$m_1v_1 + m_2v_2 = \text{total momentum}$$ Click the simulation below a few times and watch the sum of the total horizontal momentum before and after the collision.

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)

derivation of conservation of momentum

First we are going to rewrite force in terms of a change in velocity.

$$F = ma$$ $$a = \frac{\Delta v}{\Delta t}$$ $$F = m \frac{\Delta v}{\Delta t}$$ $$F = m \frac{(v-u)}{\Delta t}$$

When two bodies collide they each experience an equal but opposite force for an equal period of time as explained by Newton's third law.

$$F_1 \Delta t = -F_2 \Delta t$$ $$m_1 \frac{(v_1-u_1)}{\Delta t} \Delta t= -m_2 \frac{(v_2-u_2)}{\Delta t}\Delta t$$ $$m_1 (v_1-u_1) = -m_2 (v_2-u_2)$$ $$m_1 v_1- m_1 u_1 = -m_2 v_2 + m_2 u_2$$ $$m_1 u_1 + m_2 u_2= m_2 v_2 + m_1 v_1$$

symmetry and conservation laws

Conservation of momentum is a law. It's a pattern we see when we collect data. We didn't have a explanation for why until 1915 when Emmy Noether published a mathematical proof that conservation can be understood as a consequence of symmetry. So, momentum is conserved because space is symmetrical.

Symmetry is a property that doesn't change after a transformation. If you translate (move) your location in space the total momentum of a system of particles doesn't change. So we say that momentum is conserved.

There are a few different conservation laws in classical physics:

• Conservation of energy occurs because of time symmetry. The laws of physics work the same at any time.
• Conservation of momentum occurs because of translational symmetry. The laws of physics work the same anywhere in space.
• Conservation of angular momentum occurs because of rotational symmetry. The laws of physics work the same at any angle.
• Conservation of electric charge occurs because of gauge invariance. This is a quantum mechanical principle related to magnitude and phase of a wave function.
• There are also a few more conservation laws in quantum mechanics: parity, lepton number, baryon number
• You might have heard of conservation of mass, but mass isn't always conserved. You can destroy or produce mass because mass is a type of energy.

$$\sum p_i = \sum p_f$$ $$m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$p$$ = momentum [kg m/s] vector
$$u$$ = initial velocity [m/s] vector
$$v$$ = final velocity [m/s] vector
$$m$$ = mass [kg]
Example: A green box and a  white  box collide. The green box has a mass of 2 kg and is moving to the right at 3 m/s. The white box is moving to the left at 1 m/s and has a mass of 10 kg. If the white box has stopped after the collision what is the velocity of the green box?
solution $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{green + white = green + white}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(2)(3)+(10)(-1)=(2)v_{1}+(10)(0)$$ $$6-10=(2)v_{1}$$ $$-4=(2)v_{1}$$ $$-2 \mathrm{\tfrac{m}{s}} =v_{1}$$
Question: When does the total momentum change in this simulation?

The total momentum changes when one of the objects collides with the walls.

Question: Why does the total momentum change? Is this a violation of conservation of momentum?

The total momentum is conserved, but we aren't including the walls in our calculations. It can be very difficult to measure the change in velocity for very massive objects, like the ground or a wall.

green box blue box orange box
m = 1.60 kg m = 4.90 kg m = 3.60 kg
u = 3.0 m/s u = 2.0 m/s u = -1.0 m/s
v = ??? v = 1.18 m/s v = 1.58 m/s
Example: Three boxes collide. Use the table to find the final velocity of the green box.
solution $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $${\color{Lime} \blacksquare} \quad +\quad {\color{Blue} \blacksquare} \quad+\quad {\color{orange} \blacksquare} \quad = \quad {\color{Lime} \blacksquare} \quad+\quad {\color{Blue} \blacksquare} \quad+\quad {\color{orange} \blacksquare}$$ $$m_{1}u_{1} + m_{2}u_{2} + m_{3}u_{3} = m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}$$ $$(1.6)(3)+(4.9)(2)+(3.6)(-1)=(1.6)v_{1}+(4.9)(1.18)+(3.6)(1.58)$$ $$11=1.6v_{1}+11.47$$ $$-0.29 \mathrm{\tfrac{m}{s}} = v_{1}$$
Example: A photon of red light has a momentum of 9.45 × 10-28 kg m/s. How many photons have to collide with a 100 kg spaceship to get it to speed up from rest to 5 m/s? Assume the photons reflect off the ship after they collide.
solution

I'll start by solving for the total momentum needed, so I will not use the momentum per red photon yet.

$$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$p_{\mathrm{photons}} + p_\mathrm{{ship}} = p_\mathrm{{photons}} + p_\mathrm{{ship}}$$ $$p + (100)(0) = -p + (100)(5)$$ $$p = -p + 500$$ $$2p = 500$$ $$p = 250 \space \mathrm{ kg \tfrac{m}{s}}$$

Light with 250 kg m/s of momentum will accelerate the ship to 5 m/s. In order to count the number of red photons that make up the light we need to divide that number by the momentum of one red photon.

$$n = \frac{250 \space \mathrm{ kg \tfrac{m}{s}}}{9.45 \times 10^{-28} \space \mathrm{ kg \tfrac{m}{s}}}$$ $$n = 2.77 \times 10^{29} \space \mathrm{photons}$$

# Perfectly Inelastic Collisions

When objects collide and bounce off without any energy loss we say the collision is elastic. We will learn about how to solve those situations at the end of the energy notes.

When objects collide and stick together we say the collision is perfectly inelastic. In those situations the equations for conservation of momentum simplify a bit.

$$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$$ $$m_1u_1 + m_2u_2 = \left(m_1+m_2\right)v$$
Click to Randomize Problem: Click above to generate a random problem.
solution

Objects can also start as one mass and separate, like in an explosion.

$$\left(m_1+m_2\right)u = m_1v_1 + m_2v_2$$
Click to Randomize Problem: Click above to generate a random problem.
solution
Example: A 100.0 kg goalie throws a 1.1 kg soccer ball while jumping forward at 3.2 m/s. If the ball flies out of the goalie's hands at 10.0 m/s what speed is the goalie?
solution $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{goalie holding ball = goalie + ball}$$ $$(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}$$ $$(100.0+1.1)(3.2)=(100.0)v_{1}+(1.1)(10.0)$$ $$323.52=(100.0)v_{1}+11$$ $$3.13 \, \mathrm{\tfrac{m}{s}} = v_{1}$$

# 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: A pink hexagon and a cyan triangle collide. Before the collision the hexagon has a mass of 5.20 kg and is moving to the right at 2 m/s. The cyan triangle is moving to the left at 1 m/s and has a mass of 1.17 kg. If the triangle bounces off the hexagon at 2.72 m/s to the right what is the horizontal velocity of the hexagon? (ignore the vertical information)
solution $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{hexagon + triangle = hexagon + triangle}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(2)+(1.17)(-1)=(5.20)v_{1}+(1.17)(2.72)$$ $$10.40-1.17=(5.20)v_{1}+3.18$$ $$6.05=(5.20)v_{1}$$ $$1.16 \, \mathrm{\tfrac{m}{s}}= v_{1}$$

Vertical: Lets take a look at the vertical aspect of the pink hexagon and cyan triangle collision. Before the collision the hexagon has a mass of 5.20 kg and is not moving up or down. The final vertical velocity of the hexagon is 0.23 m/s down. The final vertical velocity of the triangle is 0.67 m/s up. What is the triangle's initial vertical velocity?
solution $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{hexagon + triangle = hexagon + triangle}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(0)+(1.17)(u_{2})=(5.20)(-0.23)+(1.17)(0.67)$$ $$(1.17)(u_{2})=-0.41$$ $$u_{2}=-0.35 \, \mathrm{\tfrac{m}{s}}$$