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Question

A bullet of mass 0.01kg and traveling at a speed of 500m/sec strikes a block of which suspended by a string of length 5m . The centre of gravity of the block is found to vertical distance of 0.1m . What is the speed of the bullet after it emerges from the block?

A
359m/s
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B
220m/s
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C
204m/s
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D
284m/s
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Solution

The correct option is A 359m/s
By energy conservation of the suspended block we can say
initial KE= final potential energy
12mv2=mgH
12mv2=gH
v=2gH
v=2×10×0.1=1.41m/s
now, by momentum conservation
Pbullet=Pblock+Pbullet
0.01×500=1×1.41+0.01×vf
51.41=0.01×vf
0.01vf=3.59
vf=359m/s
Hence,
option (A) is correct answer.

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