1.

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

Answer»

1 J
2 J
3 J
4 J

Solution :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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