a block of mass m moving at speed v collides with another block of mass 2 m at rest. the lighter block come to rest after the collision. find the coefficient of restitution ?
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Suppose the second block moves at speed v' towards right after the collision. from the principal of conservation of momentum,
or
hence, the velocity of separation = v/2 and the velocity of approach = v.
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Answered by
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Answer: The value of e is 0.5
Explanation:
Given that,
Mass of first block= m
Mass of second block = 2m
Initial velocity of first block=u1 =v
Initial velocity of second block=u2 =0
The lighter block comes to rest after collision.
So,
Final velocity of first block=v1 =0
We know that,
From conservation of momentum
m1u1+m2u2= m1v1+m2v2
mv+0= 0+2mv2
v2= v/2
The coefficient of restitution is
e=v2-v1/u1-u2
e= v/2-0/v-0
e= 1/2
e= 0.5
Hence, the value of e is 0.5
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