A ball of mass `4 kg` moving with a velocity of `12 m//s` impinges directly on another ball of mass `8 kg` moving with velocity of `4 m//s` in the same direction. Find their velocities after impact and calculate the loss of `KE` due to impact if `e =0.5`.

Explanation

`u_(1)=12m//s m_(1)=4kg`
`u_(2)=4m//s m_(2)=8kg`
Let `v_(1)` and `v_(2)` be the velocities after impact.
conservation of momentum:
`m_(1)v_(1)+m_(2)v_(2)=m_(1)u_(1)+m_(2)u_(2)` ………….i
`implies 4v_(1)+8v_(2)=80`.................i
Newoton's experimental law: `v_(1)-v_(2)=-e(u_(1)-u_(2))`
`v_(1)-v_(2)=-0.5(12-4)=-4`.........ii
Solving eqn i and ii we get
`v_(1)=4m//s, v_(2)=9m//s`
Loss in `KE = 1/2m_(1)u_(1)^(2)+1/2m_(2)u_(2)^(2)-(1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2))`
`=1/2[4(144)+8(16)]-1/2[4(16)+8(64)]=64J`