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Question

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 and B. C collides with A elastically. Then:


A
The kinetic energy of AB system at maximum compression of the spring is zero.
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B
The kinetic energy of the AB system at maximum compression of the spring is mv24.
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C
The maximum compression of the spring is vmk.
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D
The maximum compression of the spring is vm2k.
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Solution

The correct options are
B The kinetic energy of the AB system at maximum compression of the spring is mv24.
D The maximum compression of the spring is vm2k.

Since C and A are of equal mass and collision is elastic, C transfers its total momentum to A. Therefore, A has the velocity v immediately after collsion. At maximum compression, the velocity of both the bodies ae equal, say v.


Conservation of linear momentum yields
mv=(m+m)v
v=(v2).
Initial K.E =12mv2
The K.E. of the system at max. compression
=12(m+m)v2
=12(2m)v24=mv24
If the max. compression of spring is x,
12kx2=ΔKE=mv22mv24
x=(m2k)v
Hence B & D are correct options.

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