Question

Two bowling balls collide. A red 5 kg bowling ball is traveling with a velocity of 1 m/s to the left. A blue 4 kg bowling ball is traveling with a velocity of 2 m/s to the right. The 5 kg ball moves 0.8 m/s to the right after the collision. What is the final velocity of the 4kg ball?

Answer 1

We can find the final velocity of the blue ball from the law of conservation of momentum. Let's choose the right as a positive direction. Then, we can write:

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

here, $m_1 = 4 kg$ is the mass of the blue ball, $m_2 = 5 kg$ is the mass of the red ball, $v_{1i} = 2 m/s$ is the initial velocity of the blue ball, $v_{2i} = 1 m/s$ is the initial velocity of the red ball, $v_{1f}$ is the final velocity of the blue ball, $v_{2f} = 0.8 \dfrac{m}{s}$ is the final velocity of the red ball.

Then, from this formula we can find the final velocity of the blue ball:

$$v_{1f} = \dfrac{m_1v_{1i} - m_2v_{2i} - m_2v_{2f}}{m_1},$$ $$v_{1f} = \dfrac{4kg \cdot 2 \dfrac{m}{s} - 5kg \cdot 1 \dfrac{m}{s} - 5kg \cdot 0.8 \dfrac{m}{s}}{4kg} = -0.25 \dfrac{m}{s}.$$

The sign minus indicates that the blue ball moves to the left after the collision.

Answer:

$v_{1f} = 0.25 \dfrac{m}{s}$, to the left.