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Question

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:


A
0.5
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B
0.25
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C
0.4
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D
0.8
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Solution

The correct option is D $$0.25$$
Coefficient of Restitution (e) -

e = $$\dfrac{v_2 - v_1}{u_2 - u_1}$$

Ratio of relative velocity after collision to relative velocity before collision.

 Let velocity of 4m mass is v after collision

then mv = 4mv

v = $$\dfrac{v}{4v}$$

$$e ={\dfrac{velocity\ of\ separation}{velocity\ of\ approach} = \dfrac{\cfrac{v}{4}}{v}}$$ 

$$= \cfrac{1}{4}$$

$$e = 0.25$$



Physics

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