Question

A gun of mass $20\, kg$ has bullet of mass $0.1 \,kg$ in it. The gun is free to recoil $804 \,J$ of recoil energy are released on firing the gun. The speed of bullet $\left( ms ^{-1}\right)$ is

• A. $\sqrt{804 \times 2010}$
• B. $\sqrt{\frac{2010}{804}}$
• C. $\sqrt{\frac{804}{2010}}$
• D. $\sqrt{804 \times 4 \times 10^{3}}$

Answer 1

Answer: (D)

Here,
$m_{1}=20 \,kg$
$m_{2}=0.1 kg$
$v_{1}=$ velocity of recoil of gun.
$v_{2}=$ velocity of bullet
As $m_{1} \,v_{1}=m_{2}\, v_{2} $
$(\because$ momentum is conserved $)$
$v_{1}=\frac{m_{2}}{m_{1}} \,v_{2}=\frac{0.1}{20} v_{2}=\frac{v_{2}}{200}$
Recoil energy of gun
$=\frac{1}{2} m_{1} v_{1}^{2}=\frac{1}{2} \times 20\left(\frac{v_{2}}{200}\right)^{2}$
$804 =\frac{10 v_{2}^{2}}{4 \times 10^{4}}=\frac{v_{2}^{2}}{4 \times 10^{3}} $
$v_{2} =\sqrt{804 \times 4 \times 10^{3}} \,ms ^{-1}$