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Question

A bullet of mass 0.01kg is travelling at a speed of 500m/s strikes a block of mass 2kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.2m. What is the speed of the bullet after it emerges from the block?

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Solution

By law of conservation of energy,
12m2v22=m2ghv2=2gh=2×9.8×0.01=1.4ms1

If u1 is the initial velocity of bullet,
Applying law of conservation of momentum,

m1u1=m1v1+m2v2v1=0.01×5002×1.40.01=220ms1


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