A bullet of mass 0.012 kg and horizontal speed 70 km/hr strikes a block of mass 0.4 kg and instantly comes to rest with respect to the block.The block is suspended from the ceiling by means of thin wires.Calculate the height to which the block rises.Also,estimate the amount t of heat produced in the block.

Asked by sudipt.kumar | 9th Oct, 2010, 11:46: AM

Expert Answer:

Dear Student,
Let us consider the bullet and the block as one system. Applying law of conservation of momentum to the system:
MBullet uBullet + MBlock uBlock = MBullet vBullet + MBlock vBlock
(0.012)(70)   +  0                =   0                +  (0.4)vBlock
vBlock = 0.012*70 / 0.4 = 2.1 km/h = 0.58 m/s
this velocity makes the block rise to a hight where the velocity becomes zero.
Thus its kinetic energy gets converted to potential energy.
(MBlock vBlock2) / 2  = MBlock g h
h = vBlock2 / (2g) = (0.582) / 20 = 0.017 m
Amount of heat produced is equal to difference of energy of (bullet+block) before and after it hit the block. 
Heat Produced = (MBullet uBullet2)/2    -  (MBlock vBlock2)/2
                     = 2.19 J
Hope that solves the problem.

Answered by  | 9th Oct, 2010, 07:10: PM

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