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Q.

A steel ball of mass 0.5 kg is fastened to a cord 20 cm long and fixed at the far end and is released when the cord is horizontal. At the bottom of its path the ball strikes a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. The speed of the block just after the collision will be.Take g=10ms-2

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a

103ms−1

b

23ms−1

c

5ms−1

d

53ms−1

answer is B.

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Detailed Solution

Let the ball strike the block with a speed u1. Since the initial speed (speed before collision) of the block = u2 = 0 for the perfectly elastic collision.m1u1=m1v1+m2v2 0.5u1=0.5v1+2.5v2----(1)e=1=v2−v1u1−u2⇒v2−v1=u1∵u2=0----(2)From (1) & (2) ⇒v2=u13=2gl3=23ms−1
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