QiLing Tools

🎱Collision Simulator - Elastic & Inelastic, Restitution, 2D Glancing

In 1D mode two carts collide on a track. Set their masses, their velocities and the coefficient of restitution e (1 is perfectly elastic, 0 means they stick together) and watch the collision play out. A table lists each cart's velocity, momentum and kinetic energy before and after, along with the energy lost, the centre-of-mass velocity and the impulse. The 2D mode is a glancing billiard collision. The impact parameter b sets how thin the hit is, and you see both balls' outgoing speeds and angles and the vector triangle in which the final momenta add up to the initial one. Both modes can switch to the centre-of-mass frame, where the physics is clearest.

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How to use Collision Simulator (Momentum & Energy)

  1. 1Choose the 1D cart collision or the 2D billiard collision, optionally starting from a preset
  2. 2Set the masses, velocities (or impact parameter b in 2D) and the coefficient of restitution e
  3. 3Press Play to watch, scrub the timeline, and tick “Centre-of-mass frame” to change viewpoint
  4. 4Read the momentum and kinetic-energy table, the energy lost and the outgoing angles

FAQ

Why is momentum always conserved but kinetic energy not?
During a collision the two objects push on each other with equal and opposite forces. As long as external forces (or their impulse) are negligible, total momentum is conserved. But objects deform, heat up and make noise, turning some kinetic energy into other forms, so KE usually decreases. Only a perfectly elastic collision (e = 1) conserves KE. A perfectly inelastic one (e = 0), where the objects move off together, loses the most.
How is the coefficient of restitution defined, and how much energy is lost?
The coefficient of restitution is the ratio of separation speed after to approach speed before: e = −(v₂′ − v₁′)/(v₂ − v₁). The energy lost is ΔKE = ½μ(v₁ − v₂)²(1 − e²), where μ = m₁m₂/(m₁+m₂) is the reduced mass. Total KE splits into the energy of the centre-of-mass motion, ½Mv_cm², and the energy of the relative motion, ½μv_rel². Momentum conservation locks in the first part, so no collision can remove it. A collision can only take away the fraction 1 − e² of the second.
Why do equal masses swap velocities in an elastic head-on collision?
Put m₁ = m₂ and e = 1 into the formulas and you get v₁′ = v₂ and v₂′ = v₁: the velocities are exchanged. This is how Newton's cradle works. The moving ball stops dead and the ball it hits leaves with its speed. When a heavy object hits a light one, both move forward and the light one goes faster. When a light object hits a heavier one at rest, the light one bounces back.
Why do billiard balls separate at 90°?
For equal masses, an elastic collision and a target at rest, momentum conservation gives p = p₁′ + p₂′ and energy conservation gives p² = p₁′² + p₂′². Both can hold only if p₁′ is perpendicular to p₂′, so the balls leave at 90° (in a head-on hit the cue ball simply stops). In the momentum diagram, the tip of p₁′ always lies on the circle with p₁ as its diameter. With unequal masses or energy loss, the angle is no longer 90°.
What is the impact parameter b?
It is the perpendicular distance from the incoming ball's line of motion to the centre of the target ball. b = 0 is a head-on hit, and the larger b, the thinner the hit, until the balls miss when b ≥ r₁ + r₂. For smooth balls the impulse acts along the line of centres at contact. That line makes an angle φ with the incoming direction where sin φ = b/(r₁+r₂), and the struck ball leaves along it. That's why the aiming point in pool decides where the object ball goes.