Question

A gun of mass $20kg$ has bullet of mass $0.1kg$ in it. The gun is free to recoil $804J$ of recoil energy are released on firing the gun. The speed of bullet $\left(m{s}^{-1}\right)$ 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,
$m_1=20kg$
$m_2=0.1kg$
$v_1=$ 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
$=\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}\times 20{\left(\frac{v_2}{200}\right)}^2$
$804=\frac{10v_2^2}{4\times 10^4}=\frac{v_2^2}{4\times 10^3}$
$v_2=\sqrt{804\times 4\times 10^3}m{s}^{-1}$