Two blocks A and B of masses m and 2m respectively are placed on a smooth floor. They are connected by a spring. A third block C of mass m moves with a velocity `v_0` along the line joing A and B and collides elastically with A, as shown in figure. At a certain instant of time `t_0` after collision, it is found that the instantaneous velocities of A and B are the same. Further, at this instant the compression of the spring is found to be `x_0`. Determine (i) the common velocity of A and B at time `t_0`, and (ii) the spring constant.

Answer

B, C

Explanation

Initially, the blocks A and B are at rest and C is moving with velocity `v_0` to the right. As masses of C and A are same and the collision is elastic the body C treansfers its whole momentum `mv_0` to body A and as a result the body C stops and A starts moving with velocity `v_0` to the right. As this instant the spring is uncompressed and the body B is still at rest.
The momentum of the system at this instant `=mv_0`
Now, the spring is compressed and the body B comes in motion. After time `t_0`, the compression of the spring is `x_0` and common velocity of A and B is v(say).
As external force on the system is zero, the law of conservation of linear momentum gives
`mv_0=mv+(2m)v` or `v=(v_0)/(3)`
The law of conservation of energy gives
`1/2mv_0^2=1/2mv^2+1/2(2m)v^2+1/2kx_0^2`
or `1/2mv_0^2=3/2mv^2+1/2kx_0^2`
`1/2mv_0^2=3/2m(v_0/3)^2+1/2kx_0^2`
`:. 1/2kx_0^2=1/2mv_0^2-1/6mv_0^2`
or `1/2kx_0^2=1/3mv_0^2`
`k=2/3(mv_0^2)/(x_0^2)`