Two Body Collision Calculator
Find the speeds after a head-on collision, and see how much kinetic energy is kept or lost, in metric or imperial units.
What it does
Two things billiard balls do in every game of pool: they keep their total momentum, and they lose their kinetic energy. The momentum rule holds for every collision there is, with no exceptions, and the energy rule holds only for the collisions that bounce. Understanding the difference is what tells you whether a crash was a bounce or a crumple.
The OpenStax College Physics text sets out the two conservation laws that settle a one dimensional collision. Momentum is conserved because the net force during the collision is zero, which gives m1 v1 + m2 v2 = m1 v1 prime + m2 v2 prime. Kinetic energy is conserved when the collision is elastic, which gives half m1 v1 squared plus half m2 v2 squared equal to the same sum after the collision.
Two equations with two unknowns, and the solution comes out the same way every time. The text works the case of a light body striking a heavy one at rest, finds the quadratic that follows, and discards the root that simply reproduces the starting condition. What is left is the physical answer, and in that example the small body bounces backwards at 3.00 m per s while the heavy one takes over at 1.00 m per s.
This page runs both kinds of collision. Give it the two masses and the two speeds, pick an elastic or a perfectly inelastic collision, and it returns the speed of each body afterwards along with the momentum and the kinetic energy on both sides, so you can see exactly which law held and how much energy was lost.
The limits are worth stating. The calculation is for a head-on collision along one line, with no rotation, no friction and no energy spent on sound or heat except in the inelastic case, where the loss is assumed to be exactly the difference between the energies. A real collision also depends on how hard the objects are and on the coefficient of restitution between them.
How to use
- Choose metric or imperial for the masses, the speeds and the answer.
- Pick the type of collision. Elastic keeps both momentum and kinetic energy, perfectly inelastic means the bodies stick together.
- Enter the two masses.
- Enter the two speeds. A negative number means the body moves in the opposite direction to the first one.
- Read the two speeds after the collision. A negative speed for the first body means it bounced back.
- Read the energy row to see how much kinetic energy was lost, and the momentum rows to confirm that momentum did not change.
Formula
- v1 prime
- Speed of the first body after the collision.
- m1
- Mass of the first body.
- v2
- Speed of the second body before the collision.
- m2
- Mass of the second body.
The two formulas come from solving the momentum and the energy equations together, and they give the same result as the quadratic route in the text once the root that reproduces the starting condition is thrown away. The energy lost in an inelastic collision is the difference between the two kinetic energies, and it goes into heat, sound and any permanent change in shape.
Worked example
Example: a 0.5 kg body at 4 m per s hits a 3.5 kg body at rest
Defaults: metric units, an elastic collision, masses of 0.5 kg and 3.5 kg, and speeds of 4 m per s and 0.
- The momentum before the collision is 0.5 x 4 + 3.5 x 0 = 2.0000 kg m per s, and that has to be the same afterwards.
- The speed of the first body afterwards is ((0.5 - 3.5) x 4 + 2 x 3.5 x 0) over 4 = -3.0000 m per s, so it bounces backwards.
- The speed of the second body afterwards is ((3.5 - 0.5) x 0 + 2 x 0.5 x 4) over 4 = 1.0000 m per s.
- The kinetic energy before is half of 0.5 x 16 plus half of 3.5 x 0, which is 4.0000 J.
- Afterwards it is half of 0.5 x 9 plus half of 3.5 x 1, which is 2.25 + 1.75 = 4.0000 J, so nothing was lost and the collision is elastic.
- The momentum afterwards is 0.5 x -3 + 3.5 x 1 = 2.0000 kg m per s, which matches the value before.
- The tool shows -3.0000 m per s, 1.0000 m per s, 2.0000 kg m per s either side, 4.0000 J of kinetic energy either side and no energy lost.
Reference table
| Mass of the second body | Speed of the first body after | Speed of the second body after | Energy kept |
|---|---|---|---|
| 0.5 kg | 4.0000 m per s | 0.0000 m per s | 100.00 % |
| 1.5 kg | -2.0000 m per s | 2.0000 m per s | 100.00 % |
| 3.5 kg | -3.0000 m per s | 1.0000 m per s | 100.00 % |
| 7.5 kg | -3.5000 m per s | 0.5000 m per s | 100.00 % |
Limits
- The calculation is for a head-on collision along one line, with no rotation and no sideways motion.
- It assumes no outside force acts during the collision, which is what lets the momentum be conserved.
- In the perfectly inelastic case the lost energy is assumed to disappear into heat and sound, with no way to get it back.
- It says nothing about what the objects are made of, and a real collision between soft objects is somewhere between the two cases.
Common mistakes
- Forgetting that a negative speed means the body bounced back, and reading it as a slow forward motion.
- Using the elastic formulas for a collision that is really inelastic, which is why the energy does not add up.
- Mixing up the masses, so the heavy body takes the light body's share of the speed.
- Forgetting that momentum has the units of kg m per s while energy has the units of joules, and mixing the two in one calculation.
- Expecting the speeds to be symmetric. They only are when the two masses are equal.
Try it yourself
Quick checks. Answers sit tucked under each question.
What happens if the second body has the same mass as the first?
The first body stops dead and the second takes over the full speed, so 0.0000 m per s and 4.0000 m per s. That is the case of two equal balls meeting head on, and no energy is lost.
How much energy is lost in a perfectly inelastic collision?
In the example the two bodies leave together at 0.5000 m per s, the kinetic energy falls from 4.0000 J to 0.5000 J, and 3.5000 J is lost, which is 87.50 percent of the original. The momentum is unchanged at 2.0000 kg m per s.
Why does a light body bounce back from a heavy one?
Because the heavy body hardly slows down, and the momentum has to be shared, so the light body has to reverse. With a second body of 7.5 kg the first one leaves at -3.5000 m per s, and with a lighter partner of 1.5 kg it leaves at -2.0000 m per s while the other body takes 2.0000 m per s.
Questions
What is an elastic collision?
One that conserves internal kinetic energy as well as momentum, so the total kinetic energy before and after is the same. The OpenStax text says an elastic collision is one that also conserves internal kinetic energy, the sum of the kinetic energies of the objects in the system.
Why is momentum always conserved but energy not always?
Momentum is conserved whenever the net external force during the collision is zero, which is almost always true. Kinetic energy is only conserved if nothing is permanently changed, and in a real crash a lot of it ends up as heat, sound and deformation.
What is a perfectly inelastic collision?
One where the objects stick together and move off with a common speed, which is the weighted average of the two starting speeds, 2.0000 over 4 = 0.5000 m per s in the example.
Does the mass of the second body matter?
Very much. A heavier second body takes only a little speed and sends the first one back, while a lighter one takes most of the speed and the first one nearly stops.
How do I know which answer to trust?
Check the momentum row. If the value before and the value after agree, the calculation is consistent, and if the kinetic energy is unchanged the collision is elastic.
Does this work for a car crash?
The momentum part is exactly right and is the basis of every crash investigation. The energy part needs the coefficient of restitution, which says how much of the kinetic energy the structure managed to keep.
Sources
- OpenStax College Physics 2e, 8.4: elastic collisions in one dimension, with the conservation equations
- Physics LibreTexts, 8.4: elastic collisions in one dimension, with the worked example