Inelastic Collision: Conservation of Momentum

Two carts on an air track collide and stick together. The blue and red arrows are the momenta of the carts, the golden one is their sum Σp: in the impact the single momenta change, but the sum stays the same. See how much kinetic energy turns into heat and why the centre of mass moves just as it did before the impact.

→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: a 3 kg cart moving at 4 m/s hits a 2 kg cart coming the other way at 6 m/s. The momenta are equal and opposite, Σp = 0, so after the impact the carts stop dead and all 60 J of kinetic energy becomes heat. Then pick “Bullet into a block”: when a light, fast body hits, 91 % of the energy becomes heat. And switch the collision to elastic — momentum is conserved the same way, but there is no heat.

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.

Perfectly inelastic collision

$$v = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$

The bodies stick together and move as one; v₁, v₂ are signed velocity components

▶ Detailed Explanation

After the impact the bodies share one velocity v, so momentum conservation m₁v₁ + m₂v₂ = (m₁ + m₂)v gives the answer at once. The common velocity equals the velocity of the centre of mass — it was the same before the impact.

Textbook problem: m₁ = 3 kg, v₁ = 4 m/s, m₂ = 2 kg, v₂ = −6 m/s (head-on). Σp = 12 − 12 = 0, so v = 0 — the bodies stop. Pick this preset and see that the golden arrow has zero length from the start and the centre of mass marker stands still.

Energy turned into heat

$$Q = E_{k} - E_{k}^{\prime} = \frac{m_1 m_2\,(v_1 - v_2)^2}{2\,(m_1 + m_2)}$$

Kinetic energy decreases in an inelastic collision; the loss depends only on the relative velocity

▶ Detailed Explanation

Substitute v = (m₁v₁ + m₂v₂)/(m₁ + m₂) into the energy difference ½m₁v₁² + ½m₂v₂² − ½(m₁ + m₂)v² and you get this formula. Momentum is always conserved, mechanical energy in an inelastic collision is not: it goes into deformation, heating and sound.

For a target at rest the lost fraction Q/E_k equals m₂/(m₁ + m₂): a 0.5 kg bullet stuck in a 5 kg block loses 91 % of the energy, while a heavy body hitting a light one loses little. In the head-on cart problem (Σp = 0) 100 % is lost: all 60 J become heat. The energy bars under the scene show this clearly.

For comparison: elastic collision

$$v_1^{\prime} = \frac{(m_1 - m_2)v_1 + 2m_2 v_2}{m_1 + m_2}, \qquad v_2^{\prime} = \frac{(m_2 - m_1)v_2 + 2m_1 v_1}{m_1 + m_2}$$

Both momentum and kinetic energy are conserved; Q = 0

▶ Detailed Explanation

These velocities follow from two equations: conservation of momentum and conservation of energy. A handy check: the relative velocity after the impact has the same size as before and the opposite direction, v₂′ − v₁′ = v₁ − v₂.

Switch the collision type in the simulator: with the same masses and velocities the golden arrow Σp does not change, but the heat Q becomes zero. Momentum is conserved in any collision, energy only in an elastic one.

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.

Momentum and a closed system. The momentum of a body p = mv is a vector pointing along the velocity. If no external forces act on a system (or they balance, like gravity and the reaction of a smooth track), its total momentum does not change: m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′. In the impact the carts push each other with equal and opposite forces for the same time (Newton's third law), so whatever momentum one cart gains, the other loses.

Perfectly inelastic collision. The bodies stick together and move on with a common velocity v = (m₁v₁ + m₂v₂)/(m₁ + m₂). Velocities are signed: motion the other way gets a minus sign. Momentum is conserved, kinetic energy is not: the part Q = m₁m₂(v₁ − v₂)²/(2(m₁ + m₂)) goes into deforming the plasticine and heating it. If the target is at rest, the fraction m₂/(m₁ + m₂) of the energy is lost: a bullet stuck in a block gives almost all its energy to heat, while a heavy body hitting a light one loses little.

Compared with an elastic collision. Replace the plasticine with a spring bumper and the impact becomes elastic: both momentum and energy are conserved, and the bodies fly apart. Identical carts simply swap velocities. The main lesson: momentum is conserved in any collision, mechanical energy only in an elastic one.

Centre of mass. The velocity of the centre of mass is v_c = Σp/Σm. Since Σp does not change, the centre of mass (the golden marker) glides at the same speed before and after the impact, and the stuck carts move exactly with it. When Σp = 0, the centre of mass stands still — and after the impact the bodies stop.

Things to try. Choose speeds that make Σp = 0. Compare “Bullet into a block” and “Heavy into light” on the energy bars under the scene. The same law explains cannon recoil and rocket jet propulsion. The model is ideal: a frictionless track and an instantaneous impact.

Inelastic Collision: Conservation of Momentum

KEY FORMULAS

Conservation of momentum

Perfectly inelastic collision

Energy turned into heat

For comparison: elastic collision

Velocity of the centre of mass

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Other simulators

Problems on this topic