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 0.1 m. What is the speed of the bullet after it emerges from the block?
Text Solution
Verified by ExpertsThe correct answer is:
B
Conservation of linear momentum
mv = Mv' + mv' or 0.01 × 500 = 2v' + 0.01 v''
or 5 = 2v' + 0.01 v'' .... (1)
Block rises upto a height of 0.1 m
v' = 
= 
= 1.4 m/s
from eq. (1) v'' = 220 m/s
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