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Generated on: July 20, 2026 at 09:33 PM

Elastic Collision Calculator

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Mass 12 kg
Initial velocity 13 m/s
Mass 24 kg
Initial velocity 2-1 m/s

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Physics

Elastic Collision Calculator

Solve a one-dimensional elastic collision between two bodies. Enter the masses and initial velocities to find both final velocities, with momentum and kinetic energy checks.

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\begin{aligned} v_1 &= \dfrac{(m_1 - m_2)\,u_1 + 2 m_2 u_2}{m_1 + m_2} \\ &= \dfrac{\left(2\,\text{kg} - 4\,\text{kg}\right)\left(3\,\text{m/s}\right) + 2\left(4\,\text{kg}\right)\left(-1\,\text{m/s}\right)}{2\,\text{kg} + 4\,\text{kg}} \\ &= ?\,\text{m/s} \end{aligned}
\begin{aligned} v_2 &= \dfrac{(m_2 - m_1)\,u_2 + 2 m_1 u_1}{m_1 + m_2} \\ &= \dfrac{\left(4\,\text{kg} - 2\,\text{kg}\right)\left(-1\,\text{m/s}\right) + 2\left(2\,\text{kg}\right)\left(3\,\text{m/s}\right)}{2\,\text{kg} + 4\,\text{kg}} \\ &= ?\,\text{m/s} \end{aligned}

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Elastic collisions in one dimension

An elastic collision is one in which both momentum and kinetic energy are conserved. Nothing is lost to heat, sound, or deformation, so the two bodies separate carrying exactly the kinetic energy they brought in. Hard spheres like billiard balls, gas molecules, and colliding subatomic particles are all close to perfectly elastic. This calculator takes the two masses and their initial velocities along a single line and returns both final velocities, along with the momentum and kinetic energy before and after so the conservation laws can be checked directly.

Because everything happens along one axis, velocity is captured by a signed number: choose one direction as positive, and a body moving the other way gets a negative value.

The two conservation laws

Two equations govern a one-dimensional elastic collision. Conservation of momentum:

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

and conservation of kinetic energy:

12m1u12+12m2u22=12m1v12+12m2v22\tfrac{1}{2} m_1 u_1^2 + \tfrac{1}{2} m_2 u_2^2 = \tfrac{1}{2} m_1 v_1^2 + \tfrac{1}{2} m_2 v_2^2

Solving the pair for the unknown final velocities v1v_1 and v2v_2 gives the closed-form result the calculator uses:

v1=(m1−m2) u1+2m2u2m1+m2,v2=(m2−m1) u2+2m1u1m1+m2v_1 = \frac{(m_1 - m_2)\,u_1 + 2 m_2 u_2}{m_1 + m_2}, \qquad v_2 = \frac{(m_2 - m_1)\,u_2 + 2 m_1 u_1}{m_1 + m_2}

A useful intermediate fact drops out of the same algebra: in an elastic collision the relative velocity of approach equals the relative velocity of separation, u1−u2=−(v1−v2)u_1 - u_2 = -(v_1 - v_2). The bodies part as fast as they came together.

Special cases worth knowing

SituationResult
Equal masses (m1=m2m_1 = m_2)The bodies swap velocities: v1=u2v_1 = u_2, v2=u1v_2 = u_1.
Light body hits a heavy wall (m2≫m1m_2 \gg m_1, u2=0u_2 = 0)The light body rebounds at nearly the same speed: v1≈−u1v_1 \approx -u_1.
Heavy body hits a light one (m1≫m2m_1 \gg m_2, u2=0u_2 = 0)The heavy body barely slows; the light one shoots off near 2u12u_1.

The equal-mass case is the principle behind a Newton's cradle and a clean head-on pool shot: the cue ball stops and the struck ball leaves at the cue ball's speed.

Worked example

A 2 kg cart moving right at 3 m/s strikes a 4 kg cart approaching at 1 m/s (so u2=−1u_2 = -1 m/s). Taking rightward as positive:

v1=(m1−m2) u1+2m2u2m1+m2=(2−4)(3)+2(4)(−1)2+4=−6−86≈−2.33 m/s\begin{aligned} v_1 &= \frac{(m_1 - m_2)\,u_1 + 2 m_2 u_2}{m_1 + m_2} \\ &= \frac{(2 - 4)(3) + 2(4)(-1)}{2 + 4} \\ &= \frac{-6 - 8}{6} \approx -2.33\ \text{m/s} \end{aligned} v2=(m2−m1) u2+2m1u1m1+m2=(4−2)(−1)+2(2)(3)2+4=−2+126≈1.67 m/s\begin{aligned} v_2 &= \frac{(m_2 - m_1)\,u_2 + 2 m_1 u_1}{m_1 + m_2} \\ &= \frac{(4 - 2)(-1) + 2(2)(3)}{2 + 4} \\ &= \frac{-2 + 12}{6} \approx 1.67\ \text{m/s} \end{aligned}

The lighter cart bounces back (negative v1v_1), while the heavier cart reverses to move right. Checking the books: momentum before is 2(3)+4(−1)=22(3) + 4(-1) = 2 kg·m/s and after is 2(−2.33)+4(1.67)=22(-2.33) + 4(1.67) = 2 kg·m/s; kinetic energy before is 12(2)(32)+12(4)(12)=11\tfrac{1}{2}(2)(3^2) + \tfrac{1}{2}(4)(1^2) = 11 J and after is 12(2)(2.332)+12(4)(1.672)=11\tfrac{1}{2}(2)(2.33^2) + \tfrac{1}{2}(4)(1.67^2) = 11 J. Both are preserved, as an elastic collision demands.

A 4 lb glider moving at 10 ft/s strikes an 8 lb glider drifting toward it at 3 ft/s (u2=−3u_2 = -3 ft/s). Taking the first glider's direction as positive:

v1=(m1−m2) u1+2m2u2m1+m2=(4−8)(10)+2(8)(−3)4+8=−40−4812≈−7.33 ft/s\begin{aligned} v_1 &= \frac{(m_1 - m_2)\,u_1 + 2 m_2 u_2}{m_1 + m_2} \\ &= \frac{(4 - 8)(10) + 2(8)(-3)}{4 + 8} \\ &= \frac{-40 - 48}{12} \approx -7.33\ \text{ft/s} \end{aligned} v2=(m2−m1) u2+2m1u1m1+m2=(8−4)(−3)+2(4)(10)4+8=−12+8012≈5.67 ft/s\begin{aligned} v_2 &= \frac{(m_2 - m_1)\,u_2 + 2 m_1 u_1}{m_1 + m_2} \\ &= \frac{(8 - 4)(-3) + 2(4)(10)}{4 + 8} \\ &= \frac{-12 + 80}{12} \approx 5.67\ \text{ft/s} \end{aligned}

The lighter glider rebounds, the heavier one moves off in the first glider's original direction, and the relative speed of separation (5.67−(−7.33)=135.67 - (-7.33) = 13 ft/s) equals the relative speed of approach (10−(−3)=1310 - (-3) = 13 ft/s), as it must in an elastic collision.

When the model applies

Real collisions are rarely perfectly elastic — some kinetic energy almost always escapes as heat or sound. The elastic equations are the right tool when that loss is negligible: steel ball bearings, near-frictionless air-track gliders with magnetic or spring bumpers, and atomic-scale scattering. If the bodies stick together or visibly deform, the collision is inelastic and kinetic energy is not conserved, so these formulas no longer hold. The momentum-before and momentum-after figures reported here stay equal in every isolated collision; the kinetic-energy pair staying equal is the signature that distinguishes the elastic case.

Frequently Asked Questions (FAQ)

What is an elastic collision?

An elastic collision is one in which both total momentum and total kinetic energy are conserved. No energy is lost to heat, sound, or permanent deformation, so the objects bounce apart carrying the same kinetic energy they had before.

Collisions between hard spheres such as billiard balls, and between atoms or subatomic particles, are close to perfectly elastic. Most everyday collisions are partly inelastic, but the elastic case is the clean limiting model.

What are the formulas for final velocities in a 1D elastic collision?

For bodies of mass m₁ and m₂ with initial velocities u₁ and u₂, the final velocities are v₁ = ((m₁ − m₂)u₁ + 2m₂u₂)/(m₁ + m₂) and v₂ = ((m₂ − m₁)u₂ + 2m₁u₁)/(m₁ + m₂). These come from solving the momentum equation m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ together with the kinetic-energy equation ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂². Velocities are signed, so keep a consistent positive direction.

What happens when two equal masses collide elastically?

When m₁ = m₂, the formulas simplify to v₁ = u₂ and v₂ = u₁: the two bodies simply exchange velocities. A moving ball striking an identical stationary ball stops dead while the second moves off at the first ball’s original speed — a result familiar from a Newton’s cradle and from a cue ball in a head-on pool shot.

How is an elastic collision different from an inelastic one?

Both conserve momentum, but only an elastic collision also conserves kinetic energy. In an inelastic collision some kinetic energy converts to heat or deformation, and in a perfectly inelastic collision the bodies stick together and move as one. This calculator assumes a perfectly elastic collision, so the kinetic-energy-before and kinetic-energy-after values it reports should always agree.

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