A block `m_(1)` strikes a stationary block `m_(3)` inelastically. Another block `m_(2)` is kept on `m_(3)`. Neglecting the friction between all contacting surfaces, calculate the fractional decrease in `KE` of the system in collision.

Explanation

Impact takes place between `m_(1)` and `m_(3)` horizontal. Since `m_(2)` is kept on `m_(3)` and all the surface in contact are smooth, friction does not act between `m_(2)` and `m_(3)` during the displacement of `m_(3)` in the impact. Even though there is a friction between `m_(2)` and `m_(3)`, which is very less than the impace force, the frictional force is assumed as non-impulsive force here.
Since the impact between `m_(1)` and `m_(2)` is inelastic. `m_(1)` and `m_(3)` will move together toward right and `m_(2)` will not due to the absence of friction.
The velocity of the combined mass `=v^(')=(m_(1)v)/(m_(1)+m_(3))`
`(|/_\KE|)/(KE)=(1/2m_(1)v^(2)-1/2(m_(1)+m_(3))v^('2))/(1/2m_(1)v^(2))`
`=1-((m_(1)+m_(3))/(m_(1)))((v^('))/v)^(2)`
`=1-((m_(1)+m_(3))/(m_(1)))((m_(1))/(m_(1)+m_(3)))^(2)`
`=1-(m_(1))/(m_(1)+m_(3))=(m_(3))/(m_(1)+m_(3))`
If `m_(2)` and ` m_(3)` are rigidly attached, both together behave as a single mass `(m_(2)+m_(3))` and the answer would have been `(m_(2)+m_(3))/(m_(1)+m_(3)+m_(3))`