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
B
0.25
C
0.4
D
0.8

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 collisionthen mv = 4mvv = $$\dfrac{v}{4v}$$$$e ={\dfrac{velocity\ of\ separation}{velocity\ of\ approach} = \dfrac{\cfrac{v}{4}}{v}}$$ $$= \cfrac{1}{4}$$$$e = 0.25$$Physics

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