Question

The gun of mass M fires a bullet of mass m with velocity V relative to the gun.The average force required to bring gun to rest in 0.5second is ​

Answer 1

Let $\sf{v_b}$ and $\sf{- v_g }$ be velocities of bullet and gun wrt the earth. Negative sign in $\sf{-v_g}$ shows that the recoil of gun is in opposite direction to the motion of the bullet.

Given that $\sf{V}$ is the velocity of the bullet wrt the gun, i.e.,

$\sf{\longrightarrow V=v_b-(-v_g)}$

$\sf{\longrightarrow V=v_b+v_g}$

$\sf{\longrightarrow v_b=V-v_g}$

By conservation of linear momentum,

$\sf{\longrightarrow mv_b-Mv_g=0}$

$\sf{\longrightarrow mv_b=Mv_g}$

$\sf{\longrightarrow m\left(V-v_g\right)=Mv_g}$

$\sf{\longrightarrow \dfrac{V-v_g}{v_g}=\dfrac{M}{m}}$

By rule of componendo and invertendo,

$\sf{\longrightarrow \dfrac{v_g}{V}=\dfrac{m}{M+m}}$

$\sf{\longrightarrow v_g=\left(\dfrac{m}{M+m}\right)V }$

Initially the gun was in rest, so its initial linear momentum,

• $\sf{p_i=0}$

After firing the gun recoils backward with a velocity $\sf{-v_g,}$ so its final momentum,

• $\sf{p_f=-Mv_g}$

Hence the change in linear momentum of the gun,

$\sf{\longrightarrow \Delta p=-Mv_g-0}$

$\sf{\longrightarrow \Delta p=-Mv_g}$

$\sf{\longrightarrow \Delta p=-\left(\dfrac{Mm}{M+m}\right)v_g}$

The average force exerted on the gun due to the recoil in $\sf{\Delta t=0.5\ s}$ is,

$\sf{\longrightarrow F=\dfrac{\Delta p}{\Delta t}}$

$\sf{\longrightarrow F=-\dfrac{\left(\dfrac{Mm}{M+m}\right)v_g}{0.5}}$

$\sf{\longrightarrow F=-\left(\dfrac{2Mm}{M+m}\right)v_g}$

Hence the average force required to bring the gun to rest will be,

$\sf{\longrightarrow\underline{\underline{F'=\left(\dfrac{2Mm}{M+m}\right)v_g}}}$

Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

Let $\sf{v_b}$ and $\sf{- v_g }$ be velocities of bullet and gun wrt the earth. Negative sign in $\sf{-v_g}$ shows that the recoil of gun is in opposite direction to the motion of the bullet.

Given that $\sf{V}$ is the velocity of the bullet wrt the gun, i.e.,

$\sf{\longrightarrow V=v_b-(-v_g)}$

$\sf{\longrightarrow V=v_b+v_g}$

$\sf{\longrightarrow v_b=V-v_g}$

By conservation of linear momentum,

$\sf{\longrightarrow mv_b-Mv_g=0}$

$\sf{\longrightarrow mv_b=Mv_g}$

$\sf{\longrightarrow m\left(V-v_g\right)=Mv_g}$

$\sf{\longrightarrow \dfrac{V-v_g}{v_g}=\dfrac{M}{m}}$

By rule of componendo and invertendo,

$\sf{\longrightarrow \dfrac{v_g}{V}=\dfrac{m}{M+m}}$

$\sf{\longrightarrow v_g=\left(\dfrac{m}{M+m}\right)V }$

Initially the gun was in rest, so its initial linear momentum,

$\sf{p_i=0}$

After firing the gun recoils backward with a velocity $\sf{-v_g,}$ so its final momentum,

$\sf{p_f=-Mv_g}$

Hence the change in linear momentum of the gun,

$\sf{\longrightarrow \Delta p=-Mv_g-0}$

$\sf{\longrightarrow \Delta p=-Mv_g}$

$\sf{\longrightarrow \Delta p=-\left(\dfrac{Mm}{M+m}\right)v_g}$

The average force exerted on the gun due to the recoil in $\sf{\Delta t=0.5\ s}$ is,

$\sf{\longrightarrow F=\dfrac{\Delta p}{\Delta t}}$

$\sf{\longrightarrow F=-\dfrac{\left(\dfrac{Mm}{M+m}\right)v_g}{0.5}}$

$\sf{\longrightarrow F=-\left(\dfrac{2Mm}{M+m}\right)v_g}$

Hence the average force required to bring the gun to rest will be,

$\sf{\longrightarrow\underline{\underline{F'=\left(\dfrac{2Mm}{M+m}\right)v_g}}}$

${\huge{\underline{\small{\mathbb{\pink{HOPE \ THIS \ HELPED \ UH♡}}}}}}$