When a shell is fired horizontally, the barrel of a cannon…

When a shell is fired horizontally, the barrel of a cannon (mass $m_1 = 1000\ \mathrm{kg}$) recoils by $l = 0.25\ \mathrm{m}$. The shell has mass $m_2 = 40\ \mathrm{kg}$ and leaves at $v_0 = 800\ \mathrm{m/s}$. Find the average force of the recoil absorbers.

Explanation of Method

Average force of a cannon's recoil absorbers.

Step 1: Recoil speed

\[ m_1 v_1 = m_2 v_0 \;\Longrightarrow\; v_1 = \frac{40 \cdot 800}{1000} = 32\ \mathrm{m/s}. \]

Step 2: Braking the barrel

The absorbers remove the barrel's kinetic energy over the distance $l$: \[F\,l = \tfrac12 m_1 v_1^2 = \tfrac12 \cdot 1000 \cdot 32^2 = 512\ \mathrm{kJ}.\]

\[F = \frac{512\,000}{0.25} \approx 2.05 \cdot 10^6\ \mathrm{N} = 2.05\ \mathrm{MN}\]

$40\ \mathrm{kg},\ 800\ \mathrm{m/s}$ $v_1 = 32\ \mathrm{m/s}$ $l = 0.25\ \mathrm{m}$ $1000\ \mathrm{kg}$ $F\,l = \tfrac12 m_1 v_1^2 \Rightarrow F \approx 2.05\ \mathrm{MN}$

The barrel recoils at 32 m/s and the recoil absorbers must remove its 512 kJ of kinetic energy within 25 cm (dashed: the barrel's final position).

Result

\[\boxed{\,F = \frac{512\,000}{0.25} \approx 2.05 \cdot 10^6\ \mathrm{N} = 2.05\ \mathrm{MN}\,}\]

Remark

Check. The original result is correct (the momentum of the powder gases is neglected). The barrel stops in $t = 2l/v_1 \approx 16\ \mathrm{ms}$.