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 kintic energy of A-B system, at maximum compressioonof the sprin is `(mv^(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

At maximum compression `vecv_(A) = vecv_(B)` & kinetic energy of A-B system will be minimum
so `v_(A) = v_(B) = v/2 rArr K_(AB) = 1/4 mv^(2)`
From energy conservation
`1/2 mv^(2) = 1/2 m (v/2)^(2) + 1/2m(v/2)^(2) + 1/2kx_(m)^(2) rArr x_(m) = vsqrt((m)/(2k))`