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Center of Mass and Collision question

2018 · Q140
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Center of Mass and Collision question

2018 · Q140

NEETPhysicsCenter of Mass and CollisionMCQ+4 / −1
A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
  1. A
    0.5
  2. B
    0.25
  3. C
    0.8
  4. D
    0.4
View written solutionFree

Correct answer: B

From the law of conservation of linear momentum,

momentum before collision = momentum after collision, so

mv + 4m × 0 = m × 0 + 4mv′

mv = 4mv′ or v = 4v′

Coefficient of restitution,

e = v′v=v4v{{v'} \over v} = {{{v \over 4}} \over v}vv′​=v4v​​ = 0.25

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