Two identical blocks A and B, each of mass m resting on smooth floor are connected by a light spring of natural length L and spring constant k with the spring at its natural length. A third identical block C (mass m) movimng with a speed v along the line joining A and B collides with A, the maximum compression in the spring is

Options

A. (a) The KE of the AB system at maximum compression of the spring is zero.
B. (b) The KE of the AB system at maximum compression of the spring is `(1//4)mv^2`.
C. (c) The maximum compression of the spring is `vsqrt(m/k)`.
D. (d) The maximum compression of the spring is `vsqrt((m)/(2k))`

Answer

B, D

Explanation

Initially, there will be collision between C and A which is elastic.
So by conservation of momentum, we have
`mv=mv_A+mv_C`
`v=v_A+v_C` (i)
And as in elastic collision, KE after collision is same as before collision, hence
`1/2mv^2=1/2mv_A^2+1/2mv_C^2`
i.e., `v^2=v_A^2+v_C^2`
Subtracting Eq. (ii) from the square of Eq. (i), we have
`2v_Av_C=0`
So, either `v_A=0` or `v_C=0`
`v_A=0` corresponds to no interaction between A and C, so the only physically possible solution is `v_C=0`, which is the light of Eq. (i) gives `v_A=v`, i.e., after collision C stops and A starts moving with velocity v. Now A will move and compress the spring which in turn accelerates B and retards A and finally both A and B will move with same velocity (say V). In this situation, compression of the spring will be maximum.
As external force is zero, momentum of the system `(A+B+spri ng)` is conserved, i.e.,
`mv=(m+m)VimpliesV=v/2`
By conservation of mechanical energy,
`1/2mv^2=1/2(m+m)V^2+1/2kx_0^(2_2)`
or `mv^2-2m(v_0/2)^2=kx^2`
i.e., `k=(mv^2)/(2x_0^2)` or `x_0=vsqrt((m)/(2k))`