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A block of mass 1 kg moving with a speed...

A block of mass `1 kg` moving with a speed of `4 ms^(-1)`, collides with another block of mass `2 kg` which is at rest. The lighter block comes to rest after collision. The loss in `KE` of the system is

A

`8 J`

B

`4xx10^(-7)` J

C

`4 J`

D

`0 J`

Text Solution

Verified by Experts

The correct Answer is:
`(3)`
`m_(1)u_(1) + m_(2)u_(2) = m_(1)v_(1) + m_(2)v_(2)`
`Delta KE = 1/2 m_(1)u_(1)^(2) - 1/2 m_(2)v_(2)^(2)`
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