Q.

Two blocks A and B each of mass m are connected by a massless 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 figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A to B and collides with A. Then

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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 mv24

c

the maximum compression of the spring is vmk

d

the maximum compression of the spring is vm2k

answer is B.

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Detailed Solution

Here, at maximum compression xmax, we introduce the concept of reduced mass μ of the system (details discussed in Rotational Dynamics).        1μ=1m+1m⇒ μ=m2Further, by Law of Conservation of Energy       12μv2=12kxmax2⇒ xmax=vm2k and          KE=12μv2=12m2v2⇒ KE=mv24
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