Question

A 0.170 kg cue ball strikes a 0.160 kg 11-ball at 3.54 m/s. The cue ball stops passing all of its momentum to the 11-ball. What is the velocity of the 11-ball after impact?

Answer 1

The total momentum of the cue ball (before the impact) was:

$$p_c = m_cv_c = 0.170kg\times 3.54m/s = 0.6018\space kg\cdot m/s$$

The cue ball passed all of its momentum. Thus, the momentum of the 11-ball (after impact) is:

$$p_{11} = m_{11}v_{11} = p_c$$

where $m_{11} = 0.16kg$ is the mass of the 11-ball and $v_{11}$ is its required velocity after impact. Expressing $v_{11}$ from the last equality, obtain:

$$v_{11} = \dfrac{p_c}{m_{11}} = \dfrac{0.6018}{0.16} = 3.76125 \space m/s$$

Answer. 3.76125 m/s.