Question

A bullet of mass 0.01 kg travelling at a speed of 500 m/s strikes a block of mass 2 kg which is suspended by a string of length 5 m. The centre of gravity of the block is found to rise a vertical distance of 0.2 m. What is the speed of the bullet after it emerges from the block?

• A. 50m/s
• B. 75m/s
• C. 100m/s
• D. 125 m/s

Answer 1

The correct option is C

100m/s

Conservation of momentum just before and after the collsion yields,

Mu = m $V_1$ + m $V_2$ ... (1)

Conservation of energy of the block between the points 1 and 2 after the bullet pierces,

It yields,

$\Delta$ K.E + $\Delta$ P.E = 0

$\Rightarrow$ - $\frac{1}{2}$ m ${V_2}^2$ + Mgh = 0

$\Rightarrow$ = $V_2$ = $\sqrt{2gh}$ ... (2)

By using (1) and (2) we obtain

u = $V_1$ + $\frac{M}{m}$ $\sqrt{2gh}$ $\Rightarrow$ $u_1$ - $\frac{M}{m}$$\sqrt{2gh}$ = 100 m/s