    Question

# A ball of mass 4 kg moving with a velocity of 12 m/s impinges directly on another ball of mass 8 kg moving with velocity of 4 m/s in the same direction. Find their velocities after impact (if e = 0.5)

A

2 m/s, 4 m/s

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B

4 m/s, 8 m/s

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C

4 m/s, 6 m/s

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D

4 m/s, 12 m/s

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Solution

## The correct option is B 4 m/s, 8 m/s u1 = 12 m/s m1 = 4kg; u2 = 4 m/s m2 = 8 kg Let v1 & v2 be the velocities after impact Conservation of momentum: m1v1+m2v2 = m1u1+m2u2 ⇒ 4v1 + 8v2 = 80 ............(i) v1−v2 = −e(u1−u2) [∵ coefficient of restitution = velocity of separationvelocity of approach] v1−v2 = −0.5(12 − 4) = -4 ..............(ii) Solving (i) & (ii), we get: v1 = 4 m/s v2 = 8 m/s  Suggest Corrections  4      Similar questions  Related Videos   Conservation of Momentum
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