Two bodies A and B of masses m and 2 m respectively are placed on a smooth floor. They are connected by a spring. A third body C of mass m moves with velocity `v_0` along the line joining A and B and collides elastically with A as shown in Fig.

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

Since the collision between C and A is elastic and their
masses are equal and A was initially at rest, therefore the
result of collision will be that C will come to rest and A will
initially start moving with a velocity `v_0.` But since A is
connected to B with a spring, the spring will get compressed.


.At `t=t_0,` the velocity of A and B become same.
Applying energy conservation,
`1/2mV_0^2=1/2mv'^2+1/2 2mv'^2+1/2kx_0^2`
where `x_0` is the compression in the spring at `t=t_0`
`:. v_0^2=3v'^2+k/mx_0^2....(i)`
Applying momentum conservation, we get
`mv_0=mv'+2mv' :. v'=(v_0)/3....(ii)`
From (i) and (ii)
`v_0^2-3xx(v_0^2)/9=k/m x_0^2 rArr k=(2mv_0^2)/(3x_0^2)`