Two blocks of mass `m_(1)` and `m_(2)` `(m_(1) lt m_(2))` are connected with an ideal spring on a smooth horizontal surface as shown in figure. At t = 0 `m_(1)` is at rest and `m_(2)` is given a velocity v towards right. At this moment, spring is in its natural length. Then choose the correct alternative :

Answer

Kinetic energyof `m_(1)>` initial mechanical energy of system

Explanation

If velocity of `m_(2)` is zero then by momentum conservation ltbr. `m_(1)v'=m_(2)v`
`v'=(m_(2)v)/(m_(1)`
Now kinetic energy of `m_(1)`
`=(1)/(2)m'v^('2)=(1)/(2)m_(1)((m_(2))/(m_(1)))^(2)v^(2)`
`=(1)/(2)((m_(2))/(m_(1)))m_(2)v^(2)=((m_(2))/(m_(1)))(1)/(2)m_(2)v^(2)=(m_(2))/(m_(1))`
`x` initial Kinetic energy
Kinetic energy of `m_(1) gt` initial mechanical energy of system
Hence proved