Two blocks A and B each of mass m are connected by a mass less spring of natural length I 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. (7). A third identical block C, also of mass m, moving on the floor with a speed v along the line joining A and 8, 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 A-B system, at maximum compression of the spring, is (m v2 / 4)
c
the maximum compression of the spring is V(m/2k)
d
2 and 3 both are correct
answer is D.
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Detailed Solution
Let the velocity acquired by,4 and B be v'then mv=mv′=mv′ or v′=v/2 And 12mv2=12mv′2+12mv′2+12kx2 where x = displacement of the block mv2=mv22+mv22+kx2 or mv22=kx2 or x=vm2k1/2 At maximum compression, the kinetic energy of AB system will be 12mv′2+12mv′2=mv′2=mv22=mv24