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Question

A bullet of mass 0.012 kg and horizontal speed 70 ms1 strikes a 0.4 kg block of wood kept at rest and suspended from the ceiling by means of thin wire. If on stricking the block, the bullet instantly comes to rest with respect to the block, then calculate the height to which the block rises. Also estimate the amount of heat produced in the block.

A
0.212 m,28.54 J
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B
1.212 m,28.54 J
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C
0.212 m,48.54 J
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D
None
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Solution

The correct option is A 0.212 m,28.54 J
Here,
Mass of bullet (m)=0.012 kg
Mass of block (M)=0.4 kg
Speed of the bullet (u)=70 ms1
Let v be the velocity of the [block+bullet] system after collision
Now, using law of conservation of linear momentum, we can write that, mu=(M+m)v
or v=(muM+m)
=(0.012×700.4+0.012)=0.840.412
=2.04 ms1
Thus, the speed of bullet + block system after collision is 2.04 ms1.
Let h be the height up to which the block rises after striking.
Applying law of conservation of energy,
Increase in P.E. of the Bullet+Block system = Decrease in K.E. of the Bullet + Block system
Hence, we can write that,
(M+m)gh=12(M+m)v2
or h=v22g=(2.04)22×9.8
=0.212 m

Heat produced = Loss in K.E of Bullet + Block system.
=12mu212(M+m)v2
=12(0.012)×(70)212×0.412×(2.04)2
=28.54 J
Thus, option (a) is the correct answer.

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