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Question

Two mass-less strings of length 5m hang from the ceiling very near to each other as shown in the figure. Two balls A and B of masses 0.25 kg and 0.5 kg are attached to the string. The ball A is released from rest at a height 0.45 m, as shown in the figure. The collision between two balls is completely elastic. Immediately after the collision, the kinetic energy of ball B is 1J. The velocity of the ball A, just after the collision is

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A
5 ms1 to the right
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B
5 ms1 to the left
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C
1 ms1 to the right
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D
1 ms1 to the left
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Solution

The correct option is D 1 ms1 to the left
Given
mB=0.5Kg
mA=0.25Kg
h=0.45m
Velocity of ball A just before it collides with B is vAi

mAgh=12mA(vAi)2

Hence vAi=3
EB=1J
Now, after collision velocity of ball B can be obtained from

EB=12mB(vB)2

(vB)2=2mB

(vB)2=20.5

(vB)2=4
vB=2
Since, in collision elastic momentum is conserved hence,
mAvAi=mAvA+mBvB
.25×3=.25×vA+.5×2
.75=.25vA+1
vA=1ms1
Hence, the velocity of ball A just after the collision is 1ms1, to the left.

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