Students will learn how to solve problems involving both energy and momentum conservation and where one is allowed to use either momentum conservation or energy conservation or both.

### Key Equations

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

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

Guidance

When attacking these problems remember that in a collision, energy is transferred out of the system to the environment, via heat, sound, breaking of molecular bonds, etc. Thus this is not a closed system during a collision of two objects. Thus you cannot use energy conservation immediately before the collision to immediately after the collision. However, momentum is conserved during the collision. Thus break the problem into two parts. For the collision part use momentum conservation. You can use energy conservation for the other parts, where it is conserved.

#### Example 1

### Watch this Explanation

### Explore More

- You throw a \begin{align*}0.5\;\mathrm{kg}\end{align*} lump of clay with a speed of \begin{align*}5\;\mathrm{m/s}\end{align*} at a \begin{align*}15\;\mathrm{kg}\end{align*} bowling ball hanging from a vertical rope. The bowling ball swings up to a height of \begin{align*}0.01\;\mathrm{m}\end{align*} compared to its initial height. Was this an elastic collision? Justify your answer.
- The \begin{align*}20\;\mathrm{g}\end{align*} bullet shown above is traveling to the right with a speed of \begin{align*}20\;\mathrm{m/s}\end{align*}. A \begin{align*}1.0\;\mathrm{kg}\end{align*} block is hanging from the ceiling from a rope \begin{align*}2.0\;\mathrm{m}\end{align*}in length.
- What is the maximum height that the bullet-block system will reach, if the bullet embeds itself in the block?
- What is the maximum angle the rope makes with the vertical after the collision?

- A new fun foam target on wheels for archery students has been invented. The arrow of mass, \begin{align*}m\end{align*}, and speed, \begin{align*}v_0\end{align*}, goes through the target and emerges at the other end with reduced speed, \begin{align*}v_0/2\end{align*}. The mass of the target is \begin{align*}7\;\mathrm{m}\end{align*}. Ignore friction and air resistance.
- What is the final speed of the target?
- What is the kinetic energy of the arrow after it leaves the target?
- What is the final kinetic energy of the target?
- What percent of the initial energy of the arrow was lost in the shooting?

#### Answers to Selected Problems

- .
- a. \begin{align*}.008\;\mathrm{m}\end{align*} b. 5.\begin{align*}12^\circ\end{align*}
- a. \begin{align*}v_0 /14\end{align*} b. \begin{align*}mv_{0}{^{2}}/8\end{align*} c. \begin{align*}7mv_{0}{^{2}}/392\end{align*} d. \begin{align*}71\end{align*}%