A 1 kg ball moving at 12 m s^-1 collides with a 2 kg ball moving in opposite direction at 24 m s^-1. If the coefficient of restitution is 2/3, then their velocities after the collision are

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Accepted Answer

Correct answer: B. (b) : Given, m_1=1 kg ,v_1=12 m/s m_2=2 kg ,v_2=-24 m/s ,e=2/3 Let the velocities after collision is V_1 and V_2. By using conservation of momentum, m_1v_1+m_2v_2=m_1V_1+m_2V_2 1 12-2 24=V_1+2V_2 ∴ V_1+2V_2=-36 (i) Coefficient of restitution, e= V_1-V_2v_2-v_1 ∴ 23= V_1-V_2-24-12 or V_1-V_2=-24 (ii) By solving (i) and (ii), V_2=-4 m/s and V_1=-28 m/s

$A 1 kg ball moving at 12 m s^{- 1} collides with a 2 kg ball moving in opposite direction at 24 m s^{- 1} . If the coefficient of restitution is 2/3, then their velocities after the collision are$

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