How much force does a 12 gauge exert on your shoulder when firing a deer slug?

Mass of gun = 3.5kg

Mass of slug = 0.22kg

$\vec{F}_{net}$ on shoulder

${\Delta t} \longrightarrow 1/24s$ - Based on when a gun is fired in a movie, it usually occurs at about one movie frame, therefore, the collision time is less than 1/24s.

$\vec{V}_{Slug} \longrightarrow 500m/s$ This is a conservative estimate based on an internet search.

System: Gun + Slug

Surroundings: Nothing

$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$

$\vec{p}_{sys,f} = \vec{p}_{sys,i}$

$\vec{p}_{1,f} + \vec{p}_{2,f} = \vec{p}_{1,i} + \vec{p}_{2,i}$

$m_1\vec{v}_{1,f} + m_2\vec{v}_{2,f} = m_1\vec{v}_{1,i} + m_2\vec{v}_{2,i}$

We know that the momentum of the system (gun + slug) does not change due to their being no external forces acting on the system, therefore, the change in momentum in the x-direction is 0.

${\Delta p_x} = 0$

The total momentum of the system in x direction is also 0.

$P_{tot,x} = 0$

This is because the initial momentum of the system is 0 and therefore the final momentum of the system is zero.

$P_{tot,i,x} = 0$

We can relate the momentum before to the momentum after then giving us the following equation.

$0 = M_G * V_G + m_S * V_S \longrightarrow M_G * V_G$ is negative and $m_S * V_S$ is positive (see diagram).

To find the force acting on the shoulder of the shooter me need to know $V_G$ in order to find change in momentum for the gun and relate this to the force using $\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$. Rearrange the previous equation.

$V_G = {\dfrac{-m_s}{M_G}} V_S$

Fill in the values for the corresponding variables.

$V_G = - {\dfrac{0.22kg}{3.5kg}}{500m/s} = -31.4m/s$

Use the value found for $V_G$ to find the change in momentum and hence find what kind of force that is on your shoulder.

$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$

Fill in values for known variables.

$\vec{F}_{net} =\dfrac{(3.5kg)(-31.4m/s + 0m/s)}{(1/24s)}$

$\vec{F}_{net} = 2637.6N$ (at least)