Work Energy and Power - Result Question 46

49. 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

[2018]

(a) 0.5

(b) 0.25

(c) 0.4

(d) 0.8

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Answer:

Correct Answer: 49. (b)

Solution:

(b)

After Collision

According to law of conservation of linear momentum,

mv+4m×0=4mv+0v=v4

Coefficient of restitution,

e= Relative velocity of separation  Relative velocity of approach 

=v4V

or, e=14=0.25



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