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 (ms^-1) is

Here, ${\mathrm{m}}_1=20\mathrm{kg}$

         ${\mathrm{m}}_2=0.1\mathrm{kg}$

where, v₁ = velocity of recoil of gun

and      v₂ = velocity of bullet.

As,        ${\mathrm{m}}_1{\mathrm{v}}_1={\mathrm{m}}_2{\mathrm{v}}_2$           ($\because$momentum is conserved)

                         ${\mathrm{v}}_1=\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}{\mathrm{v}}_2=\frac{0.1}{20}{\mathrm{v}}_2=\frac{{\mathrm{v}}_2}{200}$

Recoil energy of gun $=\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2=\frac{1}{2}\times 20{\left(\frac{{\mathrm{v}}_2}{200}\right)}^2$

                                    $804=\frac{10{\mathrm{v}}_2^2}{4\times {10}^4}=\frac{{\mathrm{v}}_2^2}{4\times {10}^3}$

                                    ${\mathrm{v}}_2=\sqrt{804\times 4\times {10}^3}{\mathrm{ms}}^{-1}$