Two blocks A and B of mass `m` and `2m` are connected together by a light spring of stiffness `k`. The system is lying on a smooth horizontal surface with block A in contact with a fixed vertical wall as shown in the figure. The block B is pressed towards the wall by a distance `x_0` and then released. There is not friction anywhere. If spring takes time `Deltat` to aquire its natural length then average force on the block A by the wall is

Options

A. (a) zero
B. (b) `(sqrt(2mk))/(Deltat)x_0`
C. (c) `(sqrt(mk))/(Deltat)x_0`
D. (d) `(sqrt(3mk))/(Deltat)x_0`

Answer

B

Explanation

Let v is the velocity of mass `2m` in natural length of spring, then from conservation of energy we have,
`1/2kx_0^2=1/2(2m)v^2`
`:. v=sqrt((k)/(2m))x_0`
Velocity of centre of mass at this instant,
`v_(CM)=("Total momentum")/("Total mass")`
`=(2mv)/(3m)=2/3v=1/3sqrt((2k)/(m))x_0`
Now, impulse on system=change in momentum of system.
`:. F_(av)Deltat=(3m)v_(CM)=(sqrt(2mk))x_0`
`:. `F_(av)=(sqrt(2mk)x_0)/(Deltat)`