Question

A bullet of mass m = 50 gm strikes a bag of mass M = 5 kg hanging from a fixed point, with a horizontal velocity ${\overline{V}}_p$. If bullet sticks to the sand bag then just after collision the ratio of final & initial kinetic energy of the bullet is approximately :

• A. ${10}^{-2}$
• B. ${10}^{-3}$
• C. ${10}^{-6}$
• D. ${10}^{-4}$

Answer 1

The correct option is D ${10}^{-4}$

By momentum conservation we have

$m{V}_p=(m+M)V$ ,m=0.05 kg

$0.05\times V_p=(5+0.05)V\Rightarrow V_p=101V$

initial KE of bullet=$\frac{m{V}_p^2}{2}$

Final KE=$\frac{m{V}^2}{2}$

So ratio =$\frac{m{V}^2}{m{V}_p^2}={10}^{-4}$