Cannon Recoil: Conservation of Momentum

A cannon fires a shell — and rolls back. In slow motion you can see that the horizontal momenta of the shell and the cannon are equal and opposite, Σpₓ = 0, while their energies differ by tens of times. Raise the barrel, change the masses and friction, and measure the recoil distance with the ruler.

→p₁ — momentum of cart 1 →p₂ — momentum of cart 2 →Σp — total momentum ◉centre of mass of the system

Presets

m₁ mass of cart 1

v₁ velocity of cart 1

m₂ mass of cart 2

v₂ velocity of cart 2

“+” — to the right, “−” — to the left. Cart 1 is on the left, cart 2 on the right

Collision type

Slow motion

Presets

M cannon mass

m shell mass

v shell speed

Relative to the ground, right after leaving the barrel

α barrel angle °

μ friction coefficient (wheels braked)

Slow-motion shot

Momentum and velocity arrows

Centre of mass

Test yourself

The calculation is exact: the motion is given by formulas of time, and the moments of impact, shot, lift-off and burnout are found from equations, so the numbers match the formulas and do not depend on the playback speed. The impact and the shot are instantaneous, there is no air drag, g = 9.8 m/s².

Before the impact

p₁ = m₁v₁

p₂ = m₂v₂

Σp = p₁ + p₂

v_c = Σp / (m₁ + m₂)

velocity of the centre of mass — the same before and after the impact

After the impact

v = (m₁v₁ + m₂v₂) / (m₁ + m₂)

E_k before

E_k′ after

Q = m₁m₂(v₁ − v₂)² / 2(m₁ + m₂)

turned into heat

Q / E_k

V = m v cosα / M

recoil speed right after the shot

m v cosα

horizontal momentum of the shell

−M V

momentum of the cannon

m v sinα

vertical momentum — given by the support

E_sh / E_c = M / (m cos²α)

E_sh = mv²/2

E_c = MV²/2

s = V² / (2μg)

recoil distance

t = V / (μg)

recoil time

μ = (m₀ − m_f) / t_burn

fuel flow rate

F = μu

thrust

Now

m = m₀ − μt

v − u

speed of the just-ejected gas relative to the launch point

Start with the textbook problem: an 800 kg cannon fires a 10 kg shell horizontally at 400 m/s — the recoil speed is V = mv/M = 5 m/s. Raise the barrel to 45° and 60°: the recoil gets weaker, because only the horizontal part of the momentum is conserved. Pick “Shell as heavy as the cannon” — the recoil becomes as fast as the shot. And on ice (μ = 0) the cannon never stops.

KEY FORMULAS

Conservation of momentum

$$\vec p = m\vec v, \qquad m_1\vec v_1 + m_2\vec v_2 = m_1\vec v_1^{\,\prime} + m_2\vec v_2^{\,\prime}$$

In a closed system the total momentum before an interaction equals the total after it

▶ Detailed Explanation

Momentum is mass times velocity, a vector pointing along the velocity. When two bodies interact, the forces they exert on each other are equal and opposite (Newton's third law) and act for the same time. So one body gets the momentum FΔt, the other −FΔt, and the sum does not change.

Signs matter: for motion along one line choose a positive direction and give velocities pointing the other way a minus sign. In the simulator the arrows p₁, p₂ are the momenta of the bodies and the golden arrow Σp is their sum: during the impact, the shot and the rocket burn in space it changes neither its length nor its direction.

Cannon recoil

$$MV = mv\cos\alpha \;\Rightarrow\; V = \frac{mv\cos\alpha}{M}, \qquad s = \frac{V^2}{2\mu g}$$

v is the shell's speed relative to the ground, α the barrel angle; s is the recoil distance with friction μ

▶ Detailed Explanation

Before the shot Σp = 0. The powder gases push the shell forward and the cannon back, and horizontally there are practically no external forces during the shot, so mv·cos α − MV = 0. Vertically momentum is not conserved: with a raised barrel the shell carries the momentum mv·sin α upwards, the cannon is pressed into the ground, and the Earth answers with the impulse of the normal force.

The energies split inversely to the masses: E_shell/E_cannon = M/m (for α = 0). After the shot friction slows the cannon down with acceleration μg, and it travels s = V²/(2μg) in a time V/(μg) — now the cannon's momentum is no longer conserved; the Earth takes it away through friction.

Velocity of the centre of mass

$$\vec v_c = \frac{m_1\vec v_1 + m_2\vec v_2 + \dots}{m_1 + m_2 + \dots} = \frac{\sum \vec p}{\sum m}$$

With no external forces the centre of mass of a system moves uniformly in a straight line

▶ Detailed Explanation

The numerator is the total momentum of the system, the denominator its mass. Since Σp does not change, neither does the velocity of the centre of mass. Internal forces cannot move the centre of mass: when a person walks to the stern of a boat, the boat moves forward; when a cannon fires, the centre of mass of 'cannon + shell' stays where it was.

In the simulator the centre of mass is the golden marker. In a collision it glides at the same speed before and after the impact, and the stuck carts move along with it. For a rocket in space the centre of mass of 'rocket + exhaust gas' stays at the launch point, even though the rocket itself flies hundreds of kilometres away.

The energy of a shot is split unequally

$$E = \frac{p^2}{2m}, \qquad \frac{mv^2/2}{MV^2/2} = \frac{M}{m\cos^2\alpha}$$

The momenta of the shell and the cannon are equal, while energy is inversely proportional to mass

▶ Detailed Explanation

Kinetic energy is easy to express through momentum: E = mv²/2 = p²/(2m). In a horizontal shot the momenta of the shell and the cannon have the same size, so the energies are in inverse ratio to the masses: E_shell/E_cannon = M/m. An 800 kg cannon and a 10 kg shell: the shell gets 80 times more energy, although the momenta are equal.

If the barrel is raised by an angle α, the cannon gets only the horizontal part of the momentum, and the ratio grows to M/(m·cos²α). The energy for both bodies comes from the powder gases — conservation of energy does not help to find the speeds here, but conservation of momentum does. That is why recoil problems always start with momentum.

Why the cannon recoils. Before the shot the cannon and the shell are at rest, their total momentum is zero. The powder gases push the shell forward and the cannon back, and these forces are equal (Newton's third law). The shot lasts thousandths of a second; during that time the friction of the wheels is negligible compared with the gas pressure, so the horizontal momentum of the system is conserved: mv·cos α − MV = 0, which gives V = mv·cos α / M. Here v is the shell's speed relative to the ground.

A raised barrel. The vertical component of the shell's momentum mv·sin α is not conserved: at the moment of the shot the cannon is pressed into the ground, and the Earth answers with the impulse of the normal force (the green arrow at the wheels). So the steeper the barrel, the weaker the recoil: at 60° it is half as strong as in a horizontal shot.

Equal momenta, unequal energies. Energy can be written through momentum: E = p²/(2m). With equal momenta the light shell gets M/m times more energy than the heavy cannon (M/(m·cos²α) times at an angle α). That is why the shell is dangerous, not the recoil, and why a rifle is pressed to the shoulder — to “add” the shooter's mass to it.

After the shot. Friction slows the cannon down with acceleration μg, and it stops after a distance s = V²/(2μg) in a time V/(μg). Now an external force acts on the system, so Σpₓ starts to grow — the cannon's momentum goes into the Earth. The centre of mass of 'cannon + shell' stays put at the moment of the shot and moves only because of external forces: friction and the shell hitting the ground.

Where else. The kick of a rifle, the jerk of a fire hose, a boat that a person jumps off — it is all the same law. Fire gas continuously and you get rocket jet propulsion; if the bodies stick together instead of flying apart, it is an inelastic collision. The model is ideal: an instantaneous shot, no air drag, g = 9.8 m/s².

Cannon Recoil: Conservation of Momentum

KEY FORMULAS

Conservation of momentum

Cannon recoil

Velocity of the centre of mass

The energy of a shot is split unequally

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