A bullet of mass 0.01 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 center of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet after it emerges from the block ?

580 ms^-1

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220 ms^-1

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1.4 ms^-1

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7.8 ms^-1

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Solution

The correct option is B 220 ms^-1
The speed acquired by block, on account of collision of bullet with it, be .Since the block rises by 0.1 m, hence

Now as per law of conservation of momentum for collision between bullet and block,
$V_{0} m s^{- 1}$

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Q. A bullet of mass

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and travelling at a speed of

500 m/s

strikes a block of

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which is suspended by a string of length 5 m. The centre of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet in

m/s

after it emerges from the block?

A bullet of mass 0.01 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 center of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet after it emerges from the block ?

Q. A bullet of mass 10g moving horizontally with a velocity of 400 m/s strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity the block is found to rise a vertical distance of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be:

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A bullet of mass 0.01 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 center of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet after it emerges from the block ?

The correct option is B 220 ms-1The speed acquired by block, on account of collision of bullet with it, be .Since the block rises by 0.1 m, hence Now as per law of conservation of momentum for collision between bullet and block, V0ms−1

A bullet of mass 0.01 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 center of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet after it emerges from the block ?

The correct option is B 220 ms^-1
The speed acquired by block, on account of collision of bullet with it, be .Since the block rises by 0.1 m, hence

Now as per law of conservation of momentum for collision between bullet and block,
$V_{0} m s^{- 1}$