Question

What is the recoil velocity of the gun? An artillery gun of mass \[{\text{1500kg}}\] fires a shell of mass ${\text{10kg}}$ at a velocity of $400{\text{m/s}}$. What is the recoil velocity of the gun?

Answer 1

Hint: When a gunman fires a projectile, the recoil velocity is the backward velocity encountered by the shooter. A backward jerk occurs as a result of the recoil velocity. The system's linear momentum is conserved, which results in the recoil velocity.

Complete step by step answer:
We are given that:
${{\text{m}}_g}\,:\,$mass of the gun, ${{\text{m}}_g}\, = \,1500kg$
${v_g}:\,$velocity of the gun before the shell is fired, ${v_g}\, = \,0$
${{\text{m}}_s}\,:\,$mass of the shell, \[{{\text{m}}_s}\, = \,10kg\]
${v_s}\,:\,$velocity of the shell, ${v_s} = \,400\,m/s$
${v_r}\,:\,$the recoil velocity of the gun, ${v_r}\, = \,?$
We can use the theory of conservation of momentum to solve this problem. Prior to firing the shell, we can measure the total momentum. Let ${\overrightarrow {\text{P}} _b}$be the total momentum before the shell is fired. Since the gun's velocity is zero before the shell is shot, its momentum is also zero. i.e; ${{\text{P}}_g}\, = \,0$

The momentum of the shell is zero since its velocity is zero before it is shot. i.e; ${{\text{P}}_{\text{s}}}\, = \,0$
$\Rightarrow \,{{\text{P}}_{\text{b}}}\,{\text{ = }}\,{{\text{P}}_{\text{g}}}\,{\text{ + }}\,{{\text{P}}_{\text{s}}}\,{\text{ = }}\,{\text{0}}\,{\text{ + }}\,{\text{0}}\,{\text{ = }}\,{\text{0}}$
Let $\overrightarrow {{{\text{P}}_a}} $ be the total momentum after the shell is fired.
${{\text{P}}_{\text{g}}}\,{\text{ = }}\,{{\text{m}}_{\text{g}}}\,{{ \times }}\,\,{{\text{v}}_{\text{r}}}\,$ the gun’s momentum after shell is fixed
\[\Rightarrow {{\text{P}}_{\text{g}}}\,{\text{ = }}\,{\text{1500}}\,\,{{ \times }}\,\,{{\text{v}}_{\text{r}}}\]
$\Rightarrow {{\text{P}}_{\text{s}}}\,{\text{ = }}\,{{\text{m}}_{\text{s}}}\,{{ \times }}\,{{\text{v}}_{\text{s}}}$ shell’s momentum after
$\Rightarrow {{\text{P}}_{\text{s}}}\,{\text{ = }}\,{\text{10}}\,\,{{ \times }}\,\,{\text{400}}\,{\text{ = }}\,{{4000\, {\rm K}gm/s}}$
$\Rightarrow \overrightarrow {{{\text{P}}_{\text{a}}}} \,{\text{ = }}\,\overrightarrow {{{\text{P}}_{\text{g}}}} \,{\text{ + }}\,\overrightarrow {{{\text{P}}_{\text{s}}}} \\
\Rightarrow \overrightarrow {{{\text{P}}_{\text{a}}}} \,{\text{ = }}\,{\text{1500}}\,\,{{ \times }}\,{v_r}\, + \,4000 \\ $
As the momentum is conserved,
\[{{\text{P}}_{\text{b}}}\,{\text{ = }}\,{{\text{P}}_{\text{a}}} \\
\Rightarrow {\text{0}}\,{\text{ = }}\,{\text{1500}}\,\,{{ \times }}\,{{\text{v}}_{\text{r}}}\,{\text{ + }}\,{\text{4000}} \\ \]
On further solving, we get
${\text{1500}}\,\,{{ \times }}\,{{\text{v}}_{\text{r}}}{\text{ = }}\,\,{\text{ - }}\,{\text{4000}} \\
{{\text{v}}_{\text{r}}}{\text{ = }}\,{\text{ - }}\,\dfrac{{{\text{4000}}}}{{{\text{1500}}}}\,\,{\text{ = }}\,\,{\text{ - 2}}{\text{.67m/s}} \\ $
Hence, the recoil velocity of the gun is ${\text{ - 2}}{\text{.67}}\,{\text{m/s}}$.

Note:The negative sign means that the gun's velocity is opposite that of the bullet's velocity. The recoil velocity is determined by the system's linear momentum conservation. In reality, whether it's a gun, crossbow, bow and arrow, or rocket launcher, every projection device has a recoil velocity.