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

Options

A. `m_(1)/(m_(1)+m_(2)+m_(3))`
B. `m_(1)/(m_(2)+m_(3))`
C. `m_3/(m_1+m_3)`
D. `(m_(2)+m_(3))/(m_(1)+m_(2)+m_(3))`

Answer

C

Explanation

Impact takes place between `m_(1)` and `m_(2)` herizontally. Since `m_(2)` is kept on `m_(3)` an the ball the surfaces in contact are smooth, friction does not act between `m_(2)` and `m_(3)` during the displacement of `m_(1)` in the impact. Even though if there is a friction between `m_(2)` and `m_(3)`, which is very less than the impact force the frictional force is assumed as non impulsive force here.
Since the impact between `m_(1)` and `m_(3)` is inelatic `m_(1)` and `m_(3)` will move combining toward right `m_(3)` does not move due to the absence of friction.
The velocity of the combined mass `=v'=(m_(1)v)/(m_(1)+m_(3))`
`implies(|/_\KE|)/(KE)=(1/2m_(1)v^(2)-1/2(m_(1)+m_(3))v^('2))/(1/2m_(1)v^(2))`
`=1-((m_(1)+m_(2))/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))`