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A 1 kg block moving with a velocity of 4...

A 1 kg block moving with a velocity of 4 mscollides with a stationary 2 kg block. The lighter block comes to rest after the collision. The loss of kinete energy of the system is

A

1 J

B

2 J

C

3 J

D

4 J

Text Solution

Verified by Experts

The correct Answer is:
d
Applying the law of conservation of momentum
`1xx4=2xxv` or `v=2 ms^(-1)`
`:.` Loss of K.E. is given by.
`DeltaE_k=1/2xx1xx4^2-1/2xx2xx(2)^2`
=(8-4)=4J
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