Question

a stationary 0.16kg pool ball is struck by a 0.17kg cue ball moving at 2.5m/s, If the cue ball rolls backwards from the point of contact at 0.15m/s what is the final velocity of the pool ball?

Answer 1

$m_1=0.16\ \text{kg},\quad m_2=0.17\ \text{kg},\quad v_{01}=0\ \text{m/s},\quad v_{02}=2.5\ \text{m/s}, \quad v_2=-0.15\ \text{m/s},\quad v_1 =?$

Conservation of momentum: $m_1v_{01}+m_2v_{02}=m_1v_1+m_2v_2$

$0+0.17\cdot 2.5=0.16\cdot v_1-0.17\cdot 0.15$

$0.16v_1=0.4505$

$v_1=2.815625$

Answer: final velocity of the pool ball is $2.8\ \text{m/s}.$