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Question

A bullet of mass 10 g 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 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

A
120 m/s
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B
160 m/s
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C
100 m/s
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D
80 m/s
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Solution

The correct option is A 120 m/s

Given,
Mass of the wooden block (m1) is 2 kg
Mass of the bullet (m2) is 10 g
Velocity =400 m/s
Length of the string =5 m

Let the velocity of the bullet and the block after the collision will be v2 and v1 respectively

The block rises 10 cm=0.1 m

According to conservation of energy,

P.E at the maximum height = K.E at the bottom of the block,

m1gh=12m1v21
v1=2gh=2×9.8×0.1=1.4 m/s

Let u1 be the initial velocity of the bullet. Then, according to conservation of momentum,

m2u2=m1v1+m2v2
v2=m2u2m1v1m2
=0.01×4002×1.40.01=120 m/s

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