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 `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

(b,d) In situation 1, mass C is moving towards right with
velocity v. A and B are at rest.
In situation 2, which is just after the collision of C with
A, C stop and A acquires a velocity v. [head-on elastic
collision between identical masses]
When A starts moving towards right, the spring suffer
a compression due to which B also starts moving
towards right. The compression of the spring continues
till there is relative velocity between A and B. When
this relative velocity becomes zero. both A and B move
with the same velocity v' and the spring is in a state of
maximum compression.
Applying momentum coservation in situation 1
and 3,

`mv=mv'+mv' rArr v'=v/2`
`:.` K.E. of the system in situation 3 is
`1/2mv'^2+1/2mv'^2=mv'^2=(mv^2)/4 (:' v'=v/2)`
This is the kinetic energy possed by A-B system (since,
C is at rest).
Let x be the maximum compression of the spring.
Applying energy conservation
`1/2mv^2=1/2mv'^2+1/2mv'^2++1/2Kx^2`
`rArr 1/2mv^2=1/4mv^2+1/2Kx^2`
`rArr 1/2Kx^2=1/4mv^2 :. x=vsqrt(m/(2K)`