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Q.

A pendulum bob of mass m connected to the end of an ideal string of length l is released from rest from horizontal position as shown in figure.At the lowest point, the bob makes an elastic collision with a stationary block of mass 5m, which is kept on a frictionless surface. Mark out the correct statement (s) for the instant just after the impact.

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a

Tension in the string is (17/9) mg

b

Tension in the string is 3 mg.

c

The velocity of the block is 2gl/3

d

The maximum height attained by the pendulum bob after impact is (measured from the lowest position) 4l/9.

answer is A.

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Detailed Solution

The velocity of bob just before the impact is v=2gl along the horizontal direction. From momentum conservation,mv=−mv1+5mv2From coefficient of restitution equation,1=v1+v2v⇒v1+v2=vSolving above equations, we get v1=2v3v2=v3For tension in string, T−mg=mv12l⇒ T=179mgLet the maximum height attained by the bob be h, then mv122=mgh⇒ h=4l9
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