Answer to Question #146023 in Physics for Yousef

Question

A 0.18 kg billiard ball moving to the right at 1.2 m/s has a head-on elastic collision with another ball of an equal mass moving to the left at 0.85 m/s.

The first ball moves to the left at 0.85 m/s after the collision.

Find the velocity of the second ball after the collision. Show your work in detail.

Answer 1

Let's choose the right as the positive direction. Then, we can find the velocity of the second ball after the collision from the Law of Conservation of Momentum:

$$m_1v_{1i}-m_2v_{2i}=-m_1v_{1f}+m_2v_{2f},$$

here, $m_1=m_2=0.18\ kg$ are the masses of the first and second balls, respectively, $v_{1i}=1.2\ \dfrac{m}{s}$ is the velocity of the first ball before the collision, $v_{2i}=0.85\ \dfrac{m}{s}$ is the velocity of the second ball before the collision, $v_{1f}=0.85\ \dfrac{m}{s}$ is the velocity of the first ball after the collision and $v_{2f}$ is the velocity of the second ball after the collision.

Then, from this equation we can calculate $v_{2f}$:

$$v_{2f}=\dfrac{m_1v_{1i}-m_2v_{2i}+m_1v_{1f}}{m_2},$$ $$v_{2f}=\dfrac{0.18\ kg\cdot 1.2\ \dfrac{m}{s}-0.18\ kg\cdot 0.85\ \dfrac{m}{s}+0.18\ kg\cdot 0.85\ \dfrac{m}{s}}{0.18\ kg}=1.2\ \dfrac{m}{s}.$$

The sign plus means that the second ball moves to the right after the collision.

Answer:

$v_{2f}=1.2\ \dfrac{m}{s},$ to the right.