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