# A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of 2 kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1m. What is the speed of the bullet after it emerges from the block? a) 200 m/s b) 220 m/s c) 204 m/s d) 284 m/s

## Question ID - 100190 :- A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of 2 kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1m. What is the speed of the bullet after it emerges from the block? a) 200 m/s b) 220 m/s c) 204 m/s d) 284 m/s

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 (b) Suppose the velocity of the bullet of mass is and it strikes the block of mass . After collision, the linear velocity of the block is and that of the bullet is Applying  law of conservation of linear momentum, we get Or  500 0.01 = 2 +0.01 Or  5 = 2 + 0.01 (i) Now, the kinetic energy gained by the block raises it to a height where it gains gravitational potential energy By conservation of energy, we get Or m/s Putting the value of in Eq. (i), we get or m/s

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