Momentum, Impulse & Collision Calculator
p = mv, impulse and impact force, and head-on collisions with any restitution e.
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About the Momentum, Impulse & Collision Calculator
Momentum is mass times velocity, p = mv: a heavy truck and a fast bullet can both have a lot of it. A force acting for a time changes it — the impulse J = FΔt equals the change of momentum Δp — which is why stretching out a stop (airbags, crumple zones, bending your knees on landing) lowers the force. In a collision between two bodies the total momentum stays the same, while some kinetic energy may turn into heat, sound and dents.
This calculator solves p = mv for any unknown; finds the impulse and the average force of an impact from the mass, the velocities before and after and the contact time (also as a multiple of the body’s weight); and solves a head-on collision that is elastic, perfectly inelastic (the bodies stick) or anything between, set by the coefficient of restitution e. It gives both final velocities, the momentum before and after, and the kinetic energy lost. A fourth mode finds e from measured speeds or from how high a dropped ball bounces, and predicts the following bounces.
Without a pass, the free preview shows the value you solve for in full, with the first rows of each table, the bounce chart as a watermarked picture and the first step of the working; the other figures, the whole working, Copy result and the CSV need a Pro pass, or unlocking this one result.
How to use it
- Choose Momentum, Impulse, Collision or Restitution.
- Enter velocities with a sign for their direction: positive one way, negative the other. In a collision, body 1 starts on the left and positive is to the right; use the ± buttons on a phone.
- For a collision pick Elastic, Sticks together or Partly elastic (then type e between 0 and 1).
- Read the results and the working. Copy them, or download the bounce table as CSV, with a Pro pass or after unlocking this result.
Examples
1,500 kg at 100 km/h
p = 41,667 kg·m/s; kinetic energy 578.7 kJ
87.8 kg driver, 27 m/s to rest in 2.5 s — or in 0.20 s against a wall
Average force 948 N (1.1 × his weight) — or 11,853 N (13.8 × his weight)
OpenStax University Physics Vol. 1, Example 9.2.
0.057 kg ball from rest to 58 m/s in 5.0 ms
J = 3.3 N·s; average force 661 N
OpenStax Example 9.5.
0.172 kg hitting the floor at 5.42 m/s, stopped in 0.026 s
Average force 36 N, over 20 times its weight
OpenStax Example 9.4.
12 g at 2.5 m/s hits a 15 g puck at rest, elastic
The 12 g puck bounces back at 0.28 m/s; the other leaves at 2.22 m/s
OpenStax Example 9.11.
A proton at 7.0 × 10⁶ m/s meets a neutron at −4.0 × 10⁶ m/s; they stick
The deuteron moves at 1.5 × 10⁶ m/s
OpenStax Example 9.10.
1,200 kg at 27.5 m/s into 3,000 kg at rest, e = 0.4
The car stops dead, the truck moves off at 11.0 m/s; 272 kJ (60 %) of the energy is lost
OpenStax Example 9.13 (the car stops, so e = 11.0 ÷ 27.5 = 0.4).
Dropped from 2 m, bounces back to 1.28 m
e = 0.8; then 0.82 m, 0.52 m…; it travels 9.11 m and comes to rest after 5.75 s in all
Common uses
- Physics homework on momentum, impulse, conservation of momentum and elastic and inelastic collisions, with every substitution shown.
- Estimating impact forces: falls, crashes, catches, kicks and hits, from the speeds and the contact time.
- Lab work: the coefficient of restitution of a ball from its bounce height, or of a collision from measured speeds.
- Seeing how much kinetic energy a collision destroys, and why sticking together loses the most.
The formulas
- Momentum: p = m v, a vector: its sign follows the velocity. The kinetic energy is K = p² ÷ (2m). - Impulse: J = F Δt = Δp = m (v₂ − v₁), so the average force over the contact time is F = m (v₂ − v₁) ÷ Δt. With a constant force the body also covers d = ½ (v₁ + v₂) Δt meanwhile. - Collision: momentum is conserved, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, and the bodies separate at e times the speed they approached, v₂ − v₁ = e (u₁ − u₂). Together: v₁ = (m₁u₁ + m₂u₂ + m₂e(u₂ − u₁)) ÷ (m₁ + m₂) and v₂ = (m₁u₁ + m₂u₂ + m₁e(u₁ − u₂)) ÷ (m₁ + m₂). - Energy lost: ΔK = ½ · m₁m₂ ÷ (m₁ + m₂) · (1 − e²) · (u₁ − u₂)² — nothing when e = 1, the most when e = 0. - A bounce: a ball dropped from H that rises to h has e = √(h ÷ H); bounce n reaches H e^(2n), the total path is H (1 + e²) ÷ (1 − e²) and the total time √(2H ÷ g) × (1 + e) ÷ (1 − e).
The coefficient of restitution
e is the speed at which two bodies separate after a collision divided by the speed at which they approached. e = 1 is a perfectly elastic collision: no kinetic energy is lost and the bodies bounce apart as fast as they came together. e = 0 is perfectly inelastic: they stick together and move on as one. Real collisions lie between, and e depends on the materials, the shapes and the impact speed, so measure it for your case: from the speeds before and after, or by dropping a ball on the surface and timing or filming the bounce height.
Signs and directions
Velocity, momentum, impulse and force all have a direction; in one dimension it is the sign. Choose a positive direction and keep to it: in the collision mode body 1 starts on the left and positive means to the right, so a body moving left has a negative velocity. A negative force or impulse acts in the negative direction — against the motion if the body was moving the positive way. For collisions the calculator checks that the bodies really approach each other (u₁ greater than u₂).
Sources
- OpenStax, University Physics Volume 1, chapter 9 Linear Momentum and Collisions: §9.1 Linear Momentum, §9.2 Impulse and Collisions (Eq. 9.5 and 9.7, Examples 9.2, 9.4 and 9.5), §9.3 Conservation of Linear Momentum and §9.4 Types of Collisions (Examples 9.10–9.13).
- Wolfram ScienceWorld, Coefficient of Restitution: the ratio of the speed of separation to the speed of approach.
Limitations
- Classical mechanics: above about 1 % of the speed of light the calculator says how much relativistic momentum (γmv) differs.
- Collisions are head-on (one-dimensional). Glancing collisions need the momentum split into two directions.
- The force from an impulse is the average over the contact time; the peak force during the impact is higher.
- Bodies are treated as points, and outside forces during the brief collision (friction, gravity) are neglected.
- The bounce series assumes the same e at every bounce and no air resistance; real balls usually stop sooner.
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Frequently asked questions
What do I get without a pass?
Without a pass, Momentum, Impulse & Collision Calculator shows the value you solve for in full, with the first rows of each table, a watermarked chart and the first step of the working; every other figure is hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
What is the formula for momentum?
p = m v, mass times velocity, in kg·m/s (the same as N·s). A 1,500 kg car at 100 km/h (27.78 m/s) has 41,667 kg·m/s; a 0.057 kg tennis ball served at 58 m/s has 3.3 kg·m/s.
What is impulse?
Impulse is a force multiplied by the time it acts, J = F Δt, and it equals the change in momentum it causes (the impulse–momentum theorem). It is measured in N·s, the same unit as momentum.
How do I calculate the force of an impact?
Divide the change in momentum by the contact time: F = m (v₂ − v₁) ÷ Δt. An 87.8 kg driver stopping from 27 m/s in 0.20 s feels an average 11,853 N, about 14 times his weight; stretched over 2.5 s by a seatbelt and airbag it is only 948 N.
What is the difference between elastic and inelastic collisions?
Both conserve momentum. In an elastic collision kinetic energy is conserved too and the bodies bounce apart; in an inelastic one some kinetic energy becomes heat, sound and deformation. When the bodies stick together (perfectly inelastic) the loss is the largest possible.
What is the coefficient of restitution?
The speed of separation after a collision divided by the speed of approach before it: 1 for a perfectly elastic collision, 0 when the bodies stick together. For a ball dropped on a hard floor, e = √(bounce height ÷ drop height): a ball dropped from 2 m that bounces to 1.28 m has e = 0.8.
Is momentum conserved in an inelastic collision?
Yes. With no outside force the total momentum is the same before and after any collision. What an inelastic collision loses is kinetic energy, not momentum: in OpenStax Example 9.13 the 33,000 kg·m/s of the car becomes the truck’s, but 60 % of the kinetic energy is gone.