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Question

A bullet of mass 0.1 kg and travelling at a speed of 500 ms−1 strikes a block of mass 2 kg which is suspended by a string of length 5 m. The centre of gravity of the block is found to raise a vertical distance of 2.0 m. What is the speed of the bullet after is emerges from the block?

A
15ms1
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B
373.6ms1
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C
100ms1
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D
50ms1
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Solution

The correct option is C 373.6ms1
Given : m1=0.1kg m2=2kg u1=500 ms1 u2=0

Let the velocities of the bullet and the block just after the collision be v1 and v2 respectively.

The height upto which the block is raised h=2 m

From work-energy theorem for the block, 12m2v22=m2gh

v2=2gh

v=2×10×2=210 ms1

Applying conservation of momentum just before and just after the collision :

m1u1+0=m1v1+m2v2

0.1(500)=0.1v1+(2)×210

v1=373.51 ms1

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