Recoil of a gun

Problem

A $1\;\text{kg}$ gun fires a $10\;\text{g}$ bullet at $400\;\text{m/s}$. Calculate the recoil velocity of the gun and the kinetic energy of the recoil; compare it with the kinetic energy of the bullet.

Solution

Firing the gun is like an explosion: before the shot the system is at rest, so total momentum is zero and it is conserved. The stated mass of the gun ($1\;\text{kg}$) includes the bullet: after the shot the recoiling part has mass $M - m_p = 1 - 0{,}01 = 0{,}99\;\text{kg}$. Data: $m_p = 0{,}01\;\text{kg}$, $v_p = 400\;\text{m/s}$. Conservation of momentum (positive = direction of the bullet): $0 = m_p\,v_p + (M - m_p)\,v_r$ $v_r = -\frac{m_p\,v_p}{M - m_p} = -\frac{0{,}01\cdot 400}{0{,}99} = -4{,}04\ \mathrm{m/s}$ The negative sign indicates recoil, opposite to the bullet. Kinetic energies: $E_r = \tfrac{1}{2}(M - m_p)\,v_r^2 = \tfrac{1}{2}\cdot 0{,}99\cdot (4{,}04)^2 \approx 8{,}1\ \mathrm{J}$ $E_p = \tfrac{1}{2}\,m_p\,v_p^2 = \tfrac{1}{2}\cdot 0{,}01\cdot 400^2 = 800\ \mathrm{J}$ Although momentum is equal and opposite, the bullet carries about $100$ times more energy than the gun: indeed $E = \dfrac{p^2}{2m}$, and at equal $p$, energy is inversely proportional to mass. That’s why the recoil is manageable with the shoulder while the bullet is lethal. $\ev{v_r \approx -4{,}04\ \mathrm{m/s},\quad E_r \approx 8{,}1\ \mathrm{J}\ \text{versus}\ E_p = 800\ \mathrm{J}}$

Links

Topics: Momentum and collisions Concepts: Explosion · Conservation of momentum · Kinetic energy Skills: Conservation of momentum Objects: Bullet