Question

A 7.25 kg bowling ball moving at 5.75 m/s strikes a single pin (mass 1.50 kg) head on. The pin goes flying straight (same direction as the ball) at 9.75 m/s, what is the final velocity of the ball?

Answer 1

Let's write down the momentum conservation law (assuming that the collision was perfectly elastic):

$$m_1v_1 = m_1v_1' + m_2v_2'$$

where $m_1 = 7.25kg$ is the mass of the ball, $m_2 = 1.5kg$ is the mass of the pin, $v_1 = 5.75m/s$ is the velocity of the ball before the collision, $v_2' = 9.75m/s$ is the velocity of the pin after the collision, and $v_1'$ is the final velocity of the ball. Thus, obtain:

$$v_1' = \dfrac{m_1v_1-m_2v_2'}{m_1}\\ v_1' = \dfrac{7.25\cdot 5.75-1.5\cdot 9.75}{7.25} \approx 3.73m/s$$

Answer. 3.73 m/s.