Students will learn what momentum is and how to calculate momentum of objects. In addition, students will learn how to use conservation of momentum to solve basic problems.

### Key Equations

\begin{align*} p = m v \; \; \text{Momentum is equal to the objects mass multiplied by its velocity}\end{align*}

\begin{align*} \sum p_{\text{initial}} = \sum p_{\text{final}} \; \; \text{The total momentum does not change in closed systems}\end{align*}

- Momentum is a vector that points in the direction of the velocity vector. The magnitude of this vector is the product of mass and speed.
- The total momentum of the universe is always the same and is equal to zero. The total momentum of an isolated system never changes.
- Momentum can be transferred from one body to another. In an isolated system in which momentum is transferred internally, the total initial momentum is the same as the total final momentum.
- Momentum conservation is especially important in collisions, where the total momentum just before the collision is the same as the total momentum after the collision.

#### Example 1

A truck with mass 500 kg and originally carrying 200 kg of dirt is rolling forward with the transmission in neutral and shooting out the dirt backwards at 2 m/s (so that the dirt is at relative speed of zero compared with the ground). If the truck is originally moving at 2 m/s, how fast will it be moving after it has shot out all the dirt. You may ignore the effects of friction.

##### Solution

To solve this problem we will apply conservation of momentum to the truck when it is full of dirt and when it has dumped all the dirt.

\begin{align*} m_iv_i&=m_fv_f && \text{start by setting the initial momentum equal to the final momentum}\\ (m_t+m_d)v_i&=m_tv_f && \text{substitute the mass of the truck plus the mass of the dirt in the truck at the initial and final states}\\ v_f&=\frac{(m_t+m_d)v_i}{m_t} && \text{solve for the final velocity}\\ v_f&=\frac{(500\;\text{kg} + 200\;\text{kg})*2\;\text{m/s}}{500\;\text{kg}} && \text{plug in the numerical values}\\ v_f&=2.8\;\text{m/s}\\ \end{align*}

#### Example 2

John and Bob are standing at rest in middle of a frozen lake so there is no friction between their feet and the ice. Both of them want to get to shore so they simultaneously push off each other in opposite directions. If John's mass is 50 kg and Bob's mass is 40 kg and John moving at 5 m/s after pushing off Bob, how fast is Bob moving?

##### Solution

For this problem, we will apply conservation of momentum to the whole system that includes both John and Bob. Since both of them are at rest to start, we know that the total momentum of the whole system must always be zero. Therefore, we know that the sum of John's and Bob's momentum after they push off each other is also zero. We can use this to solve for Bob's velocity.

\begin{align*} 0&=m_jv_j+m_bv_b\\ -m_bv_b&=m_jv_j\\ v_b&=-\frac{m_jv_j}{m_b}\\ v_b&=-\frac{50\;\text{kg}*5\;\text{m/s}}{40\;\text{kg}}\\ v_b&=-6.25\;\text{m/s}\\ \end{align*}

The answer is negative because Bob is traveling in the opposite direction to John.

### Watch this Explanation

### Explore More

- You find yourself in the middle of a frozen lake. There is no friction between your feet and the ice of the lake. You need to get home for dinner. Which strategy will work best?
- Press down harder with your shoes as you walk to shore.
- Take off your jacket. Then, throw it in the direction opposite to the shore.
- Wiggle your butt until you start to move in the direction of the shore.
- Call for help from the great Greek god Poseidon.

- You and your sister are riding skateboards side by side at the same speed. You are holding one end of a rope and she is holding the other. Assume there is no friction between the wheels and the ground. If your sister lets go of the rope, how does your speed change?
- It stays the same.
- It doubles.
- It reduces by half.

- You and your sister are riding skateboards (see Problem 3), but now she is riding behind you. You are holding one end of a meter stick and she is holding the other. At an agreed time, you push back on the stick hard enough to get her to stop. What happens to your speed? Choose one. (For the purposes of this problem pretend you and your sister weigh the same amount.)
- It stays the same.
- It doubles.
- It reduces by half.

- An astronaut is using a drill to fix the gyroscopes on the Hubble telescope. Suddenly, she loses her footing and floats away from the telescope. What should she do to save herself?
- A \begin{align*}5.00\;\mathrm{kg}\end{align*} firecracker explodes into two parts: one part has a mass of \begin{align*}3.00 \;\mathrm{kg}\end{align*} and moves at a velocity of \begin{align*}25.0 \;\mathrm{m/s}\end{align*} towards the west. The other part has a mass of \begin{align*}2.00 \;\mathrm{kg}\end{align*}. What is the velocity of the second piece as a result of the explosion?
- A firecracker lying on the ground explodes, breaking into two pieces. One piece has twice the mass of the other. What is the ratio of their speeds?
- While driving in your pickup truck down Highway \begin{align*}280\end{align*} between San Francisco and Palo Alto, an asteroid lands in your truck bed! Despite its \begin{align*}220 \;\mathrm{kg}\end{align*} mass, the asteroid does not destroy your \begin{align*}1200\;\mathrm{kg}\end{align*} truck. In fact, it landed perfectly vertically. Before the asteroid hit, you were going \begin{align*}25 \;\mathrm{m/s}\end{align*}. After it hit, how fast were you going?
- An astronaut is \begin{align*}100\;\mathrm{m}\end{align*} away from her spaceship doing repairs with a \begin{align*}10.0 \;\mathrm{kg}\end{align*} wrench. The astronaut’s total mass is \begin{align*}90.0 \;\mathrm{kg}\end{align*} and the ship has a mass of \begin{align*}1.00 \times10^4\;\mathrm{kg}\end{align*}. If she throws the wrench in the opposite direction of the spaceship at \begin{align*}10.0 \;\mathrm{m/s}\end{align*} how long would it take for her to reach the ship?

#### Answers to Selected Problems

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- \begin{align*}37.5 \;\mathrm{m/s}\end{align*}
- \begin{align*}v_1 = 2v_2\end{align*}
- \begin{align*}21 \;\mathrm{m/s}\end{align*}
- a. \begin{align*}90 \;\mathrm{sec}\end{align*}