What this calculator does
Find out how fast two objects move after they crash head-on, whether they bounce apart, stick together or land somewhere in between. When two objects collide and nothing else pushes on them, their total momentum stays the same. Momentum is mass times velocity, where velocity is speed in a given direction. This rule, called conservation of momentum, is the starting point for every collision problem. On its own, though, it can't find two unknown final velocities. You need a second fact: how much of the kinetic energy (energy of motion) survives the crash.
In an elastic collision, such as two hard steel balls, no kinetic energy is lost. In a perfectly inelastic collision, the objects stick together and move off as one. That loses as much energy as momentum conservation allows. Most real collisions fall in between. They're described by the coefficient of restitution e, a bounce factor from 0 to 1. This calculator handles all three cases along a straight line. It gives the final velocities, the total momentum, and the kinetic energy before, after and lost.
Where it's used
Why this math matters outside the classroom:
- Billiards. In pool, a cue ball hitting a still ball of equal mass head-on nearly stops and the struck ball moves off at the cue ball's speed.
- Crash reconstruction. Crash investigators use conservation of momentum to work back from where two vehicles came to rest to their speeds before impact.
- Coupling carts. A 2 kg cart at 3 m/s that couples with a still 1 kg cart moves on at 6 ÷ 3 = 2 m/s.
- Newton's cradle. A Newton's cradle shows nearly elastic collisions, with momentum passed through the line of balls to the last one.
- Sports equipment. Equipment testers measure the coefficient of restitution of balls and bats to check how lively they are.
Grade level and standards: Grades 11–12; Common Core State Standards for Mathematics HSA.REI.C.6. © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.
How to use it
- Choose the type of collision: elastic, perfectly inelastic, or partly elastic with a coefficient of restitution.
- Enter each object's mass and its velocity before the collision. Count moving right as positive; object 1 starts on the left.
- For a partly elastic collision, enter e between 0 and 1.
- Read both final velocities and the energy figures. A negative final velocity means the object is moving left.
What the inputs mean
- Initial velocity of object 1 (u₁)
- Velocities include direction. If object 1 moves right at 3 m/s, enter 3. If object 2 comes toward it at 1 m/s, enter −1.
- Initial velocity of object 2 (u₂)
- For a target sitting still, enter 0. Object 1 must be moving right faster than object 2, or the two will never meet.
- Coefficient of restitution (e)
- How fast the objects move apart afterwards, compared with how fast they came together. A golf ball on a hard floor is near 0.8; a lump of clay is near 0.
Formula and method
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Elastic: v₁ = ((m₁ − m₂)u₁ + 2m₂u₂) ÷ (m₁ + m₂), v₂ = ((m₂ − m₁)u₂ + 2m₁u₁) ÷ (m₁ + m₂) Perfectly inelastic: v = (m₁u₁ + m₂u₂) ÷ (m₁ + m₂) With restitution: v₂ − v₁ = e × (u₁ − u₂) KE = ½m₁u₁² + ½m₂u₂²
Here's what the letters mean. m₁ and m₂ are the two masses. u₁ and u₂ are their velocities before the crash, and v₁ and v₂ are their velocities after. v is the shared velocity when they stick together. e is the coefficient of restitution, and KE is the total kinetic energy.
Momentum conservation gives one equation linking the two unknown final velocities. The type of collision gives the second. In an elastic collision, kinetic energy also stays the same, which leads to the pair of formulas above. In a perfectly inelastic one, both objects share one final velocity. In between, e sets how fast the objects move apart afterwards, as a fraction of how fast they came together.
Solving the momentum equation and the restitution equation together gives two results. They are v₁ = (p + m₂e(u₂ − u₁)) ÷ (m₁ + m₂) and v₂ = (p + m₁e(u₁ − u₂)) ÷ (m₁ + m₂), where p is the total momentum. Setting e = 1 gives back the elastic formulas, and e = 0 gives the perfectly inelastic one.
Worked example
A 2 kg cart rolling right at 3 m/s meets a 1 kg cart rolling left at 1 m/s in an elastic collision. How fast does each move afterwards?
- Total momentum: 2 × 3 + 1 × (−1) = 5 kg·m/s.
- Cart 1: ((2 − 1) × 3 + 2 × 1 × (−1)) ÷ (2 + 1) = 0.333333 m/s.
- Cart 2: ((1 − 2) × (−1) + 2 × 2 × 3) ÷ (2 + 1) = 4.333333 m/s.
- Kinetic energy before: 0.5 × 2 × 3² + 0.5 × 1 × (−1)² = 9.5 J, and after: 9.5 J.
Result: The heavier cart slows to about 0.333 m/s, still moving right. The lighter cart bounces off to the right at about 4.333 m/s. Momentum (5 kg·m/s) and kinetic energy (9.5 J) both stay the same.
Understanding the result
Momentum is always conserved in these calculations; only the kinetic energy changes. In an inelastic collision, the lost energy turns into heat, sound and dents (permanent changes in shape).
Two patterns are worth knowing. Equal masses in an elastic collision just swap velocities, which is what you see in a row of hanging balls. A light object hitting a much heavier one at rest bounces back at almost its original speed, while the heavy one barely moves.
Assumptions and limitations
- Motion is along one straight line (a head-on collision).
- No outside forces, such as friction, act during the brief collision.
- The objects are solid enough to treat as points, and their masses don't change.
- Glancing collisions in two or three dimensions need momentum to be conserved separately in each direction.
- Spinning objects, and collisions that pass on spin, aren't modeled.
- The coefficient of restitution can change with impact speed, so a single value is an approximation.
Frequently asked questions
What is the difference between an elastic and an inelastic collision?
Both keep momentum the same. An elastic collision also keeps kinetic energy the same, so the objects bounce apart with the same total energy. An inelastic collision loses some kinetic energy. When the objects stick together, it's called perfectly inelastic, and the loss is as large as it can be.
Is kinetic energy ever gained in a collision?
Not in ordinary collisions. A gain needs stored energy to be released, as in an explosion or a spring-loaded toy. That would mean a coefficient of restitution above 1, which is outside this calculator's range.
Why must u₁ be greater than u₂?
Object 1 is on the left, and moving right counts as positive. So the two only meet if object 1 is catching up to object 2. If u₁ is less than or equal to u₂, they move apart or keep the same gap, and never collide.
How do I find the coefficient of restitution for a bouncing ball?
Drop it onto a hard floor from height h and measure how high it bounces, h′. Because the floor doesn't move, e = √(h′ ÷ h). A ball that bounces back to 64% of its drop height has e = 0.8.
How this calculator is checked
Method: Conservation of linear momentum with Newton's experimental law of restitution (v₂ − v₁ = e(u₁ − u₂)) for one-dimensional collisions; e = 1 elastic, e = 0 perfectly inelastic. The calculation is covered by automated tests with known reference values and edge cases. Last updated 2026-10-18. Found a problem? Tell us.
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