Question

A 7.25 kg bowling ball moving at 6.75 m/s strikes a single pin (mass 1.50 kg) head on. The pin goes flying straight (same direction as the ball) at 10.5 m/s, what is the final velocity of the ball?

Answer 1

The initial momentum of the ball is:

$$p_{ball1} = m_{ball}v_{ball1} = 7.25kg\times 6.75m/s = 48.9375\space kg\cdot m/s$$

After impact pin carries the following momentum:

$$p_{pin} = m_{pin}v_{pin} = 1.5kg\times 10.5m/s = 15.75 \space kg\cdot m/s$$

Thus, the final momentum of the ball is:

$$p_{ball2} = p_{ball1} - p_{pin} = 33.1875\space kg\cdot m/s$$

Then, the velocity of the ball is:

$$v_{ball2} = \dfrac{p_{ball2}}{m_{ball}} = \dfrac{33.1875\space kg\cdot m/s}{7.25kg} \approx 4.58m/s$$

Answer. 4.58 m/s.