Two blocks A and B, each of mass m, are connected by a masslesss spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in fig. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then

Options

A. the kinetic energy of the A – B system at maximum compression of the spring, is zero.
B. the kinetic energy of the A – B system, at maximum compression of the spring, is `(m v^2)/(4)`
C. the maximum compression of the spring is `vsqrt((m)/(K))`
D. the maximum compression of the spring is `vsqrt((m)/(2K))`

Answer

B, D

Explanation

In situation (i), mass C is moving towards right with velocity v. A and B are at rest. In situation (ii), which is just after the collision of C with A, C stops and A acquires a velocity v. When A starts moving towards right, the spring suffers a compression due to which B also starts moving towards right. The compression of the spring continues till there is a relative velocity between A and B. When this relative velocity becomes zero, spring is in a state of maximum compression.
Applying moment conservation in situations (ii) and (iii)
`mv = m v + mv' implies v' = v/2`
Therefore, KE of the system in situation (iii) is
`1/2 mv'^2 + 1/2 m v'^2 = mv'^2 = (mv^2)/4`
Applying energy conservation, we get
`1/2 mv^2 = 1/2 mv'^2 + 1/2 mv'^2 + 1/2 mv'^2 + 1/2 Kx^2`
Solve to get `x = v sqrt((m)/(2K))`