Question

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 ?

• A. 200 m/s
• B. 220 m/s
• C. 204 m/s
• D. 284 m/s

Answer 1

Answer: (B)

220 m/s

Answer 2

Conservation of linear momentum

mv = Mv' + mv'' or 0.01 × 500 = 2v' + 0.01 v''

or 5 = 2v' + 0.01 v'' ....(1)

Q Block rises upto a height of 0.1 m

\ v' = $\sqrt{2gh}$ = $\sqrt{2 \times 9.8 \times 0.1}$ = 1.4 m/s

from eq.(1) v'' = 220 m/s