Question

A gun of mass M fires a bullet of mass m with a velocity v relative to the gun. The average force required to bring the gun to rest in 0.5 sec. is

• A. $\frac{2MmV}{m+M}$
• B. $\frac{MmV}{2(M+m)}$
• C. $\frac{3MmV}{2(M+m)}$
• D. $\frac{MmV}{(M+m)}$

Answer 1

The correct option is A $\frac{2MmV}{m+M}$

Velocity of gun $=V_1$

Velocity of bullet $=V-V_1$

$\begin{array}{c}m\left(V-V_1\right)=M{V}_1\\ V_1=\frac{mV}{m+M}\end{array}$

distance travelled by gun in 0.5 sec

$\begin{array}{c}0=V_1-a\times \left(0.5\right),V=u+at\\ -a=retardation\\ a=2V_1\\ S=V_1\times 0.5-\frac{1}{2}a{\left(0.5\right)}^2\\ S=\frac{V_1}{4}\end{array}$

from work energy theorem

$\begin{array}{c}F.S=\frac{1}{2}M{V}_1^2-0\\ F=2M{V}_1\\ F=\frac{2MmV}{m+M}\end{array}$