Car Crash Calculator

Learn how momentum and energy work in a two-vehicle collision: final velocities, delta-v, energy dissipated and the equivalent fall height.

🚗 Car Crash Calculator

Educational physics only. This simplified one-dimensional model is not for accident reconstruction, insurance or legal use.

Speed unit
Gravity (g)
m/s²
Vehicle 1 mass (m1)
kg
Vehicle 1 speed (v1)
Vehicle 2 mass (m2)
kg
Vehicle 2 speed (v2)
Impact type
Restitution (e)

Vehicle 1 starts on the left. Positive speed means moving right, negative means moving left, and vehicle 1 must be faster than vehicle 2 for a collision to occur.

Vehicle mass (m)
kg
Impact speed (v)
Crush distance (d)
m
Vehicle 1 velocity after
—
Vehicle 2 velocity after
—
Delta-v, vehicle 1
—
Delta-v, vehicle 2
—
Closing speed
—
Total momentum
—
Kinetic energy before
—
Kinetic energy after
—
Energy dissipated
—
Share of energy lost
—
Barrier-equivalent speed, vehicle 1
—
Barrier-equivalent speed, vehicle 2
—
Equivalent drop height, vehicle 1
—
Equivalent drop height, vehicle 2
—

Equivalent drop height
—
Speed
—
Drop height in storeys
—
Kinetic energy
—
Average deceleration
—
Average deceleration (g)
—
Stopping time
—
Average force
—
Impulse
—

🚗 What is a Car Crash Calculator?

A car crash calculator, in the sense used on this page, is a physics teaching tool that applies conservation of momentum and the coefficient of restitution to a collision between two vehicles moving along a straight line. Given the two masses, the two velocities and how bouncy the impact is, it finds the velocities right after the impact, the change of velocity (delta-v) of each vehicle, the kinetic energy that was turned into deformation, heat and sound, and an equivalent speed against a rigid barrier or an equivalent drop height. It also has a barrier mode that converts one vehicle's speed into a drop height and an average deceleration.

Important: this is an educational tool only. It is not for accident reconstruction, insurance claims, legal work or safety certification. Real crashes involve angled impacts, rotation, crush stiffness, tyre friction, braking before impact and structural details that a one-dimensional model ignores, and professional reconstruction uses measured evidence and validated methods. Use this page to understand the physics, for example in a school project, a lecture demonstration or simple curiosity about why speed matters so much.

Several ideas become clear when you play with it. Momentum is always conserved in the collision, whatever the impact type. Kinetic energy is not conserved unless the impact is perfectly elastic, and the energy lost depends on the closing speed squared, so doubling the closing speed quadruples the energy dissipated. The vehicle with the smaller mass always suffers the larger velocity change, since the impulses on the two vehicles are equal and opposite. Finally, a barrier impact at a given speed has the same energy as dropping the vehicle from a height of v squared divided by 2g, which gives an intuitive feel for what a speed means: about 9.84 m, roughly three storeys, for 50 km/h.

The two modes cover the two main situations. The collision mode handles two vehicles and uses a chart of final velocities against restitution so you can see how the outcome shifts from sticking together to bouncing apart. The barrier mode uses a crush distance to estimate the average deceleration in g and the stopping time, with a chart of deceleration against crush distance showing why longer crumple zones reduce the load on occupants. For related ideas see the conservation of momentum calculator.

📐 Formula

m1v1 + m2v2  =  m1v1′ + m2v2′   |   e  =  (v2′ − v1′) / (v1 − v2)
v1′ = ( m1v1 + m2v2 + m2 e (v2 − v1) ) / (m1 + m2)
v2′ = ( m1v1 + m2v2 + m1 e (v1 − v2) ) / (m1 + m2)
e = coefficient of restitution: 0 for vehicles that stick together (perfectly inelastic), 1 for a perfectly elastic bounce
Delta-v of each vehicle = |v′ − v|; for e = 0 it equals (other mass / total mass) × closing speed
Energy dissipated: ΔE = ½ μ (1 − e²) (v1 − v2)² with reduced mass μ = m1m2 / (m1 + m2)
Barrier-equivalent speed: vb,i = μ (v1 − v2) √(1 − e²) / mi, from sharing ΔE between the vehicles in the centre-of-mass frame
Equivalent drop height: h = v² / (2g); average deceleration a = v² / (2d) for crush distance d
Example: equal 1500 kg cars at 50 km/h head-on stop dead; each has delta-v = 50 km/h

📖 How to Use This Calculator

Steps

1
Choose a mode. Pick Two-vehicle collision to study momentum and energy, or Barrier impact to convert one vehicle's speed to a drop height and an average deceleration.
2
Enter masses, speeds and impact type. Type the masses in kg and the speeds in your chosen unit (negative for a vehicle moving left), then choose perfectly inelastic, partly elastic, elastic or a custom restitution.
3
Read the results and the chart. See the final velocities, delta-v for each vehicle, energy dissipated and barrier-equivalent speeds, and check the chart for your mode.

💡 Example Calculations

Example 1 — Rear-End Collision, Vehicles Stick Together

1500 kg at 50 km/h hits a 1200 kg vehicle at 20 km/h in the same direction, e = 0

1
Momentum = 1500 × 13.89 + 1200 × 5.556 = 27 500 kg·m/s. Common velocity = 27 500 / 2700 = 10.19 m/s = 36.67 km/h
2
Delta-v: vehicle 1 slows by 13.33 km/h, vehicle 2 speeds up by 16.67 km/h. The lighter vehicle has the larger change.
3
Closing speed 30 km/h (8.33 m/s): ΔE = ½ × 666.7 × 8.33² = 23.15 kJ (14.2 % of the kinetic energy). Barrier-equivalent speeds are 13.33 and 16.67 km/h.
Both vehicles at 36.67 km/h after impact, energy dissipated 23.15 kJ
Try this example →

Example 2 — Head-On, Equal Cars

Two 1500 kg cars, one at +50 km/h and one at −50 km/h, e = 0

1
Total momentum = 0, so the wreck comes to rest: both final velocities are 0 km/h
2
Each car has delta-v = 50 km/h, and the closing speed is 100 km/h (27.78 m/s)
3
ΔE = ½ × 750 × 27.78² = 289.35 kJ, all of the kinetic energy. Each car's barrier-equivalent speed is 50 km/h, a drop of 9.84 m.
Delta-v = 50 km/h each, energy dissipated = 289.35 kJ
Try this example →

Example 3 — Head-On, Light Car Against Heavy Vehicle

1000 kg at +60 km/h meets 2000 kg at −40 km/h, e = 0

1
Momentum = 1000 × 16.67 − 2000 × 11.11 = −5556 kg·m/s, so the common velocity is −5556 / 3000 = −1.852 m/s = −6.67 km/h (toward the heavier vehicle's direction)
2
Delta-v: the light car changes by 66.67 km/h, the heavy one by 33.33 km/h, a ratio of 2 to 1, the inverse of the mass ratio
3
ΔE = ½ × 666.7 × 27.78² = 257.20 kJ (98.0 %). Equivalent drop heights are 17.48 m and 4.37 m.
Delta-v = 66.67 km/h (light) and 33.33 km/h (heavy)
Try this example →

Example 4 — Elastic Bumper Cars

200 kg bumper car at 4 m/s hits an identical car at rest, e = 1

1
v1′ = (800 + 200 × 1 × (0 − 4)) / 400 = 0 m/s and v2′ = (800 + 200 × 1 × 4) / 400 = 4 m/s
2
The cars swap velocities. Kinetic energy is 1.60 kJ before and after, so the energy dissipated is 0.00 kJ.
3
Each car has delta-v = 4 m/s and the barrier-equivalent speed is 0 because nothing is absorbed.
Final velocities 0 m/s and 4 m/s, energy dissipated 0.00 kJ
Try this example →

Example 5 — Partly Elastic Impact

1000 kg at 30 km/h hits an identical stationary 1000 kg vehicle, e = 0.3

1
v1′ = (30 + 1000 × 0.3 × (0 − 30) / 1000) / 2 = 10.50 km/h and v2′ = (30 + 0.3 × 30) / 2 = 19.50 km/h
2
They separate at 9.00 km/h, which is e × 30 km/h, and each has delta-v = 19.50 km/h
3
ΔE = ½ × 500 × (1 − 0.09) × 8.33² = 15.80 kJ (45.5 %). The barrier-equivalent speed is 14.31 km/h for each.
Final velocities 10.50 and 19.50 km/h, energy dissipated 15.80 kJ
Try this example →

Example 6 — Rigid Barrier at 50 km/h

1500 kg vehicle hits a fixed barrier at 50 km/h and crushes 0.6 m

1
v = 50 / 3.6 = 13.89 m/s. Drop height h = 13.89² / (2 × 9.80665) = 9.84 m (3.3 storeys of 3 m)
2
Kinetic energy = ½ × 1500 × 13.89² = 144.68 kJ
3
Average deceleration a = v² / (2d) = 192.9 / 1.2 = 160.8 m/s² = 16.4 g, over a stopping time of 2d / v = 86 ms. The average force is 241.1 kN.
Equivalent drop height = 9.84 m, average deceleration = 16.4 g
Try this example →

Example 7 — Low-Speed Barrier Contact

1200 kg vehicle at 30 km/h into a barrier with 0.5 m of crush

1
v = 8.33 m/s, so h = 69.44 / 19.6133 = 3.54 m (1.2 storeys)
2
Kinetic energy = ½ × 1200 × 8.33² = 41.67 kJ
3
a = 69.44 / (2 × 0.5) = 69.4 m/s² = 7.1 g over 120 ms. Compared with Example 6, the lower speed and similar crush distance cut the load a lot.
Equivalent drop height = 3.54 m, average deceleration = 7.1 g
Try this example →

❓ Frequently Asked Questions

How do you calculate the final velocity after a car collision?+
Use conservation of momentum with the restitution coefficient e: v1' = (m1 v1 + m2 v2 + m2 e (v2 - v1)) / (m1 + m2), and v2' = (m1 v1 + m2 v2 + m1 e (v1 - v2)) / (m1 + m2). For a perfectly inelastic crash e = 0 and both vehicles move together.
What is delta-v in a collision?+
Delta-v is the change in a vehicle's velocity caused by the collision. For e = 0 it equals (other mass / (m1 + m2)) times the closing speed, so the lighter vehicle always has the larger delta-v. For two equal cars head-on at 50 km/h each, delta-v is 50 km/h for both.
How much energy is lost in a car crash?+
The kinetic energy dissipated is 0.5 x mu x (1 - e^2) x closing speed squared, where mu = m1 m2 / (m1 + m2) is the reduced mass. It goes into deformation, heat and sound. A perfectly elastic collision (e = 1) loses none.
What is a perfectly inelastic collision?+
It is a collision in which the objects stick together and move with a common velocity afterwards, so e = 0. Momentum is still conserved but the maximum possible kinetic energy for the given momentum is dissipated.
What is the coefficient of restitution?+
It is the ratio of the separation speed to the approach speed along the line of impact, e = (v2' - v1') / (v1 - v2). It is 0 for objects that stick together and 1 for a perfectly elastic bounce. Real vehicle crashes are near the low end.
What does barrier-equivalent speed mean?+
It is the speed at which a vehicle would have to hit a rigid barrier to absorb the same energy as it did in the collision. In this calculator it is the square root of 2 E / m for the energy share assigned to each vehicle, using the centre-of-mass frame.
How is a crash speed related to a fall height?+
The speed gained in a fall is v = sqrt(2 g h), so the equivalent height is h = v^2 / (2 g). A barrier impact at 50 km/h (13.89 m/s) matches a fall from 9.84 m, about three storeys.
Why does the lighter car feel a bigger change in speed?+
Both vehicles experience equal and opposite impulses, so m1 times delta-v1 equals m2 times delta-v2 when they stick together. The lighter vehicle therefore has the larger velocity change. For 1000 kg against 2000 kg the delta-v ratio is 2 to 1.
What is the average deceleration in a barrier crash?+
If the vehicle stops uniformly over a crush distance d, the average deceleration is a = v^2 / (2 d). At 50 km/h with 0.6 m of crush it is 160.8 m/s2, about 16.4 g. The actual peak is higher than this average.
Can this calculator be used for accident reconstruction or legal cases?+
No. It is an educational physics tool for a simplified one-dimensional collision. It does not model crush stiffness, angles, rotation, tyre friction or post-impact motion, so it must not be used for accident reconstruction, insurance or legal purposes.
What happens in an elastic collision between two equal masses?+
With e = 1 and equal masses the vehicles swap velocities. A moving bumper car at 4 m/s that hits an identical stationary one stops, and the second car moves away at 4 m/s. Kinetic energy is unchanged.

How do you calculate the final velocity after a car collision?

Use conservation of momentum with the restitution coefficient e: v1' = (m1 v1 + m2 v2 + m2 e (v2 - v1)) / (m1 + m2), and v2' = (m1 v1 + m2 v2 + m1 e (v1 - v2)) / (m1 + m2). For a perfectly inelastic crash e = 0 and both vehicles move together.

What is delta-v in a collision?

Delta-v is the change in a vehicle's velocity caused by the collision. For e = 0 it equals (m other / (m1 + m2)) times the closing speed, so the lighter vehicle always has the larger delta-v. For two equal cars head-on at 50 km/h each, delta-v is 50 km/h for both.

How much energy is lost in a car crash?

The kinetic energy dissipated is 0.5 x mu x (1 - e^2) x closing speed squared, where mu = m1 m2 / (m1 + m2) is the reduced mass. It goes into deformation, heat and sound. A perfectly elastic collision (e = 1) loses none.

What is a perfectly inelastic collision?

It is a collision in which the objects stick together and move with a common velocity afterwards, so e = 0. Momentum is still conserved but the maximum possible kinetic energy for the given momentum is dissipated.

What is the coefficient of restitution?

It is the ratio of the separation speed to the approach speed along the line of impact, e = (v2' - v1') / (v1 - v2). It is 0 for objects that stick together and 1 for a perfectly elastic bounce. Real vehicle crashes are near the low end.

What does barrier-equivalent speed mean?

It is the speed at which a vehicle would have to hit a rigid barrier to absorb the same energy as it did in the collision. In this calculator it is the square root of 2 E / m for the energy share assigned to each vehicle.

How is a crash speed related to a fall height?

The speed gained in a fall is v = sqrt(2 g h), so the equivalent height is h = v^2 / (2 g). A barrier impact at 50 km/h (13.89 m/s) matches a fall from 9.84 m, about three storeys.

Why does the lighter car feel a bigger change in speed?

Both vehicles experience equal and opposite impulses, so m1 times delta-v1 equals m2 times delta-v2. The lighter vehicle therefore has the larger velocity change. For 1000 kg against 2000 kg the delta-v ratio is 2 to 1.

What is the average deceleration in a barrier crash?

If the vehicle stops uniformly over a crush distance d, the average deceleration is a = v^2 / (2 d). At 50 km/h with 0.6 m of crush it is 160.8 m/s2, about 16.4 g. The actual peak is higher than this average.

Can this calculator be used for accident reconstruction or legal cases?

No. It is an educational physics tool for a simplified one-dimensional collision. It does not model crush stiffness, angles, rotation, tyre friction or post-impact motion, so it must not be used for accident reconstruction, insurance or legal purposes.

What happens in an elastic collision between two equal masses?

With e = 1 and equal masses the vehicles swap velocities. A moving bumper car at 4 m/s that hits an identical stationary one stops, and the second car moves away at 4 m/s. Kinetic energy is unchanged.