Question

A gun of weight of $10\,{\text{kg}}$ fires a shot of $0.5\,{\text{g}}$ with a velocity $230\,{\text{m/s}}$. Velocity of recoil gun is

Answer 1

Use the law of conservation of linear momentum. Also, use the formula for linear momentum of an object. First determine the initial linear momentum of the gun-bullet system and then determine the final linear momentum of the gun-bullet system. Use these values in the formula for law of conservation of linear momentum and determine the recoil velocity of the gun.

Formulae used:
The expression for law of conservation of linear momentum is
${P_i} = {P_f}$ …… (1)

Here, ${P_i}$ is the initial linear momentum of the object and ${P_f}$ is the final linear
momentum of the object.
The linear momentum $P$ of an object is given by
$P = mv$ …… (2)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.

Complete step by step answer:
We have given that mass of the gun which is $10\,{\text{kg}}$ and the mass of bullet which is
$0.5\,{\text{g}}$.
${m_g} = 10\,{\text{kg}}$
${m_b} = 0.5\,{\text{g}}$
The velocity of the bullet after firing is $230\,{\text{m/s}}$.
${v_b} = 230\,{\text{m/s}}$

We have asked to determine the recoil velocity of the gun after firing the bullet.
We can determine the recoil velocity of the gun using law of conservation of linear momentum.

The initial velocity of the gun and the bullet before firing is zero as they both are at rest.
Hence, the initial momentum of the gun and bullet according to equation (1) is zero.
${P_i} = 0\,{\text{kg}} \cdot {\text{m/s}}$
The final momentum of the gun-bullet system is the sum of final momentum ${P_g}$ of gun and
final momentum ${P_b}$ of the bullet.
${P_f} = {P_g} + {P_b}$
The final momentum of the gun and bullet according to equation (1) are
${P_g} = {m_g}{v_g}$
${P_b} = {m_b}{v_b}$

Therefore, the final momentum becomes
${P_f} = {m_g}{v_g} + {m_b}{v_b}$

Let us now apply the law of conservation of linear momentum to the gun-bullet system.
${P_i} = {P_f}$
Substitute $0$ for ${P_i}$ and ${m_g}{v_g} + {m_b}{v_b}$ for ${P_f}$ in the above equation.
$0 = {m_g}{v_g} + {m_b}{v_b}$
Rearrange the above equation for ${v_g}$.
${v_g} = - \dfrac{{{m_b}{v_b}}}{{{m_g}}}$
Substitute $0.5\,{\text{g}}$ for ${m_b}$, $10\,{\text{kg}}$ for ${m_g}$ and
$230\,{\text{m/s}}$ for ${v_b}$ in the above equation.

\right)}}{{\left( {10\,{\text{kg}}} \right)}}$$ $$ \Rightarrow {v_g} = - \dfrac{{\left[ {\left( {0.5\,{\text{g}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{kg}}}}{{1\,{\text{g}}}}} \right)} \right]\left( {230\,{\text{m/s}}} \right)}}{{\left( {10\,{\text{kg}}} \right)}}$$ $$ \Rightarrow {v_g} = - 11.5 \times {10^{ - 3}}\,{\text{m/s}}$$ $$ \therefore {v_g} = - 1.15\,{\text{cm/s}}$$ **Hence, the recoil velocity of the gun is $$1.15\,{\text{cm/s}}$$.** The negative sign indicates that the gun moves in the backward direction after the bullet is fired. Note: The students should always keep in mind that the law of conservation of linear momentum is applicable to systems only when there is no exchange of mass and there must be no external force acting on the system. The system on which the law of conservation of system is to be applied must be a closed system.