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?
"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}."
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