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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 |
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Answer» 1 J `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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