A cannon of mass m_1 = 1.8 t fires a shell of mass m_2 = 1…

A cannon of mass $m_1 = 1.8\ \mathrm{t}$ fires a shell of mass $m_2 = 1\ \mathrm{kg}$ at $\alpha = 30^\circ$ to the horizontal with speed $v_2 = 360\ \mathrm{m/s}$. Find (a) the recoil speed of the cannon and (b) its average deceleration if it stops after sliding $s = 2\ \mathrm{m}$.

Explanation of Method

Recoil of a cannon firing at 30°.

Step 1: (a) Recoil speed

During the shot the ground can push vertically but not horizontally, so only the horizontal momentum is conserved: $m_1 v_1 = m_2 v_2\cos\alpha \;\Longrightarrow\; v_1 = \frac{1 \cdot 360 \cdot 0.866}{1800}$.

\[v_1 \approx 0.173\ \mathrm{m/s} \ \text{(backward)}\]

Step 2: (b) Average deceleration

\[ 0 = v_1^2 - 2as \;\Longrightarrow\; a = \frac{v_1^2}{2s} = \frac{0.173^2}{4}. \]

\[a \approx 7.5 \cdot 10^{-3}\ \mathrm{m/s^2}\]

$v_2 = 360\ \mathrm{m/s}$ $30^\circ$ $v_1 = 0.173\ \mathrm{m/s}$ $m_1 = 1800\ \mathrm{kg}$ $\text{stops in } 2\ \mathrm{m}$

Only the horizontal momentum is conserved: the vertical push of the shell is taken up by the ground. The cannon recoils backward and then brakes to a stop within 2 m.

Result

\[\boxed{\,v_1 \approx 0.173\ \mathrm{m/s} \ \text{(backward)}\,}\]

\[\boxed{\,a \approx 7.5 \cdot 10^{-3}\ \mathrm{m/s^2}\,}\]

Remark

Check. The original results are correct. The braking takes $t = v_1/a \approx 23\ \mathrm{s}$, i.e. the cannon only creeps back. (A real gun barrel recoils far faster, because its recoil mechanism lets only the barrel move at first.)