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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 in 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 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 mv2.

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C

The maximum compression of the spring is.

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D

The maximum compression of the spring is.

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Solution

The correct option is D The maximum compression of the spring is. Due to equal mass, C transfers its total momentum to A. Therefore A has the velocity v. At max. compression the velocity of both the bodies are equal say, v'. Conservation of linear momentum yields mv=(m+m)v′ ⇒v′=(v2) ∴ The K.E. of the system =12(m+m)v′2 =12(2m)v24=mv24 To calculate the maximum compression x, 12kx2=△KE=mv22−mv24 ⇒x=(√m2k) v.

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