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 :

Options

A. Block of mass m2 will be finally at rest after some time
B. Block of mass m2 will never come to rest
C. Both the blocks will be finally at rest
D. None of these

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