Elastic Collision of Two Bodies

Two balls roll along a track and collide. Set the masses and velocities (the sign is the direction) and watch what happens in the impact: how the balls exchange momentum, how the energy hides for a moment in their squeezing, and why the total momentum and the total energy stay the same.

The idea in 30 seconds

An impact is an exchange of momentum

While the balls touch, they push on each other equally hard but in opposite directions (Newton’s third law). The momentum one ball loses, the other one gains — so the total momentum does not change.

Energy does not vanish

In an elastic impact the balls squeeze for a moment like two springs: the energy of motion hides in the squeezing and then comes back in full. So the total kinetic energy is the same before and after. (Plasticine balls would stick together and part of the energy would turn into heat — that is an inelastic impact.)

Approach speed = separation speed

The balls separate at the same speed at which they approached. This rule together with the conservation of momentum gives the speeds after the impact without any heavy calculation.

Try it yourself

Press “Equal masses”: the balls simply swap speeds — the moving one stops and the one at rest leaves with its speed.
Press “Light → heavy”: the light ball bounces back while the heavy one barely stirs — like a ball off a wall.
Press “Heavy → light”: the light ball shoots ahead faster than the heavy one — almost twice as fast.
Switch on “View from the centre of mass”: in this frame of reference the balls just reverse their velocities — the picture becomes symmetric.

Where you meet it

Billiards and bowling: a ball hitting a ball is almost elastic — that is why both the “speed swap” and the bounces work.
Gas molecules collide almost elastically — the molecular-kinetic theory and the explanation of gas pressure are built on this.
Nuclear reactors: neutrons are slowed down by elastic impacts on light nuclei (water, graphite) — the closer the masses, the stronger the slowing.

m₁ mass 1,0 kg

v₁ velocity +2,0 m/s

m₂ mass 1,0 kg

v₂ velocity −1,0 m/s

The sign of the velocity is the direction: plus — right, minus — left. You can also change a ball’s velocity on the scene: press “To the start” and drag the dot on the arrow.

Try it in one click

Playback speed

Setup

Walls: the balls bounce off the ends

View from the centre of mass

Show

Velocity arrows

Strobe: prints at equal time intervals

Centre of mass

Impact forces

Slow down the impact (display only)

Test yourself: guess the speed after the impact

The balls are perfectly elastic and there is no friction. You can slow the show down or speed it up — the numbers do not change. A real impact lasts a fraction of a second: in the “movie” it is stretched so that the squeezing is visible.

Graphs: what changes and how

The speed of each ball changes at the impact, but the total momentum and energy — the gold lines — stay flat. The green area on the energy graph is the energy “hidden” for a moment in the squeezing of the balls.

Speeds of the balls: they change only during the impact

Momentum: it changes for each ball, the total does not

Energy: motion + squeezing, the total is constant

KEY FORMULAS

Law of conservation of momentum

$$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$$

The total momentum of the two balls before the impact equals the total momentum after it

▶ Detailed Explanation

Momentum is a vector quantity: p = mv (in a one-dimensional problem the sign gives the direction). During the impact the balls push on each other with equal and opposite forces for the same time (Newton’s third law), so the momentum lost by one ball is gained by the other and the sum does not change. This holds for any masses and velocities, and even for an inelastic impact. In the simulator it is the gold line: on the “p” gauge under the scene and on the momentum graph it does not move during the approach, during the impact or afterwards (until a ball hits a wall — the wall adds external momentum).

Conservation of kinetic energy

$$\frac{m_1v_1^2}{2} + \frac{m_2v_2^2}{2} = \frac{m_1v_1'^2}{2} + \frac{m_2v_2'^2}{2}$$

In an elastic impact the sum of the kinetic energies of the balls is the same before and after

▶ Detailed Explanation

This is exactly what distinguishes an elastic impact from an inelastic one: in an inelastic impact part of the energy goes into heat and deformation and E decreases. In an elastic impact energy is not lost but changes form for a moment: the balls squeeze like springs, kinetic energy turns into compression energy and then comes back. In the simulator the total energy E = E₁ + E₂ + E_comp (the “E” gauge under the scene) is constant at every moment, while the kinetic energy E₁ + E₂ only dips during the impact and rises back to its former value.

Velocities after an elastic impact

$$\begin{aligned} v_1' &= \frac{(m_1 - m_2)v_1 + 2m_2v_2}{m_1 + m_2} \\ v_2' &= \frac{(m_2 - m_1)v_2 + 2m_1v_1}{m_1 + m_2} \end{aligned}$$

Velocities of both balls right after a central elastic impact

▶ Detailed Explanation

The formulas follow from two conditions: conservation of momentum and conservation of kinetic energy. The sign of a velocity gives its direction: plus means right, minus means left. In the simulator these values are visible in advance — in the line under the scene and in the before/after table (the “by the formula” column) — and after the impact the table shows what came out of the simulation: the numbers match. Put m₁ = m₂ and you get v₁′ = v₂, v₂′ = v₁ (the speeds are swapped).

Approach speed equals separation speed

$$v_2' - v_1' = -(v_2 - v_1), \qquad e = \frac{v_2' - v_1'}{v_1 - v_2} = 1$$

In an elastic impact the relative velocity of the balls changes sign but not size

▶ Detailed Explanation

The coefficient of restitution e is the ratio of the separation speed to the approach speed. In an elastic impact e = 1: the balls separate at the same speed at which they approached. In an inelastic one 0 ≤ e < 1, and for e = 0 the balls stick together. This rule together with conservation of momentum gives the same formulas for v₁′ and v₂′ without any squaring — sometimes it is quicker this way. In the before/after table it is the row “approach → separation speed”.

Centre of mass: the velocity is “reflected”

$$v_c = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}, \qquad v_1' = 2v_c - v_1, \quad v_2' = 2v_c - v_2$$

The velocity of the centre of mass does not change in the impact, and the velocity of each ball is “reflected” about it

▶ Detailed Explanation

The centre of mass of the two balls moves uniformly and does not “notice” the impact — it is the teal marker on the scene (when there are no walls). Seen from a frame moving along with it (“View from the centre of mass”), the total momentum is zero, the balls approach with momenta of equal size, and after the impact each one simply reverses its velocity: u′ = −u. Returning to the usual frame we get v′ = 2v_c − v. This is the shortest way to find the velocities after an elastic impact.

Special case: one ball at rest

$$v_2 = 0: \qquad v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1, \qquad v_2' = \frac{2m_1}{m_1 + m_2}\,v_1$$

If ball 2 stood still, ball 1 moves after the impact with (m₁ − m₂)/(m₁ + m₂)·v₁ and ball 2 with 2m₁/(m₁ + m₂)·v₁

▶ Detailed Explanation

Three important cases. Equal masses (m₁ = m₂): v₁′ = 0, v₂′ = v₁ — the incoming ball stops and the one at rest leaves with its speed. Light onto heavy (m₁ ≪ m₂): v₁′ ≈ −v₁, v₂′ ≈ 0 — the light ball bounces back at almost the same speed, like a ball off a wall. Heavy onto light (m₁ ≫ m₂): v₁′ ≈ v₁, v₂′ ≈ 2v₁ — the heavy ball hardly slows down and the light one shoots ahead at almost twice the speed. Try all three with the “Try it in one click” buttons.

What is going on. Two balls roll along a smooth track and collide. “Elastic” means the balls do not stick together and do not lose energy as heat: after the impact they fly apart so that both the total momentum p = m₁v₁ + m₂v₂ and the total kinetic energy E = ½m₁v₁² + ½m₂v₂² are exactly what they were before. The velocity carries a sign: plus means moving right, minus means left, so no separate angle is needed.

What happens during the impact. While the balls touch, they push on each other with forces equal in size and opposite in direction (Newton’s third law). So the momentum one ball loses, the other one gains. For a short time the balls squeeze like two springs: the energy of motion “hides” in the squeezing (the green part of the energy bar) and then comes back in full. A real impact lasts a fraction of a second; in the simulator the show slows down around the impact (“movie mode”) so that the squeezing can be seen — this does not affect the numbers or the formulas.

How to find the velocities after the impact. The two conservation laws give v₁′ = ((m₁ − m₂)v₁ + 2m₂v₂)/(m₁ + m₂) and v₂′ = ((m₂ − m₁)v₂ + 2m₁v₁)/(m₁ + m₂). There is also a shortcut: find the velocity of the centre of mass v_c = (m₁v₁ + m₂v₂)/(m₁ + m₂) — after the impact each velocity is “reflected” about it: v′ = 2v_c − v. One more handy rule: the approach speed equals the separation speed, v₂′ − v₁′ = −(v₂ − v₁).

Special cases. Equal masses: the balls simply swap velocities (like billiard balls). A light ball hits a heavy one: it bounces back at almost the same speed while the heavy one barely moves (like a ball off a wall). A heavy ball hits a light one: the heavy one hardly slows down and the light one shoots ahead at almost twice the speed.

How to use it. Set the masses and velocities or press a ready-made experiment — and follow the before/after table, the momentum and energy gauges under the scene, and the graphs. Switch on “View from the centre of mass” to see that in this frame the balls just reverse their velocities. The “Test yourself” mode gives a task: guess the speed after the impact, then check it with the experiment.