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Question

Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant k. A third block C of mass m moving with a speed v along the line joining A and B collides elastically with A. The maximum compression in the spring is:


A

mv2k

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B

m2k

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C

mvk

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D

vm2k

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Solution

The correct option is D

vm2k


Step 1: Given data

  1. Two blocks A and B each of mass m resting on the smooth horizontal floor connected by a light spring of natural length L and spring constant k.
  2. A third block C of mass m moving with a speed v along the line joining A and B collides elastically with A.

Step 2: To find

The maximum compression of the spring

Step 3: Formula and calculation

In this case the spring will compress until vA>vB. Therefore, maximum compression is when vA=vB

Since, here we have to find the maximum compression, let vA=vB=v1 be the velocity of blocks A and B at the maximum compression of the spring

From the law of conservation of momentum,

total momentum before collision = total momentum after collision.

C moves with a velocity v and hits the block A and then A starts to move with a velocity vA and since block A is connected to block B, B starts to move with a velocity vB. After collision C comes to rest.

mv=mvA+mvB⇒mv=mv1+mv1⇒v=2v1

From the law of conservation of energy,

Kinetic energy of C = sum of kinetic energies of the blocks A and B + the energy of the spring connecting A and B.

12mCvC2=12mAvA2+12mBvB2+12kx2(vA=vB=v1,mA=mB=mC=m)⇒12mv2=12m+mv12×2+12kx2⇒12mv2=122mv22+12kx2⇒12mv2=14mv2+12kx2⇒14mv2=12kx2⇒mv2=2kx2⇒x=mv22k⇒x=vm2k

The maximum compression of the spring, x=vm2k

Hence, option D is correct.


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