A plank of mass `5kg` is placed on a frictionless horizontal plane. Further a block of mass `1kg` is placed over the plank. A massless spring of natural length `2m` is fixed to the plank by its one end. The other end of spring is compressed by the block by half of spring's natural length. They system is now released from the rest. What is the velocity of the plank when block leaves the plank? (The stiffness constant of spring is `100N//m`)

Answer

A, C

Explanation

Let the velocity of the block and the plank, when the block leaves the spring be `u` and `v` respectivley.
By conservation of energy `1/2kx^2=1/2m u^2+1/2Mv^2`
[M=mass of the plank, m=mass of the block]
`:. 100=u^2+5v^2` …(i)
By conservation of momentum
`m u+Mv=0`
`impliesu=-5y` ...(ii)
Solving Eqs. (i) and (ii)
`30v^2=100implies:.v=sqrt(10/3)m//s`
From this moment until block falls, both plank and block keep their velocity constant.
Thus, when block falls velocity of plank
`=sqrt(10/3)m//s`