One-Dimensional Collisions

Momentum is conserved across a collision for a reason you can check with numbers, and the same numbers say when to stop trusting it. A 1.2 kg cart arriving at 2.0 m s⁻¹ on a track with 0.60 N of rolling friction strikes a second cart; the contact lasts 8.0 ms and brings the first cart to rest, an impulse of 2.4 N s. Friction's impulse during those 8.0 ms is 0.60 N × 0.0080 s = 4.8 × 10⁻³ N s, a fifth of a percent of that, so the momentum of the two-cart system just after the contact equals its momentum just before. Over the 3.0 s the cart rolled beforehand, friction delivered 1.8 N s, three-quarters of the momentum it arrived with. So “before” and “after” mean the instants bracketing the contact, never the start and end of the run. Then fix a positive direction and give every velocity its sign: a cart moving left at 1.5 m s⁻¹ enters the equation as −1.5 m s⁻¹, and a negative answer is a rebound, not a mistake.

Send a 2.0 kg cart at +3.0 m s⁻¹ into a 1.0 kg cart at rest. Conservation says 2.0v₁′ + 1.0v₂′ = 6.0 kg m s⁻¹, a straight line of candidate outcomes rather than an answer. Some candidates are physical: (2.0, 2.0) m s⁻¹, the carts moving together; (1.0, 4.0), the rear cart slowing and the front one leaving fast; (1.5, 3.0), in between. Others satisfy momentum just as well and are impossible. In (3.0, 0) the rear cart passes through the front one, so the first constraint is v₂′ ≥ v₁′. In (0, 6.0) the carts leave with 18 J of kinetic energy from a collision that brought in 9.0 J, so the second constraint, for bodies with no stored energy to release, is K′ ≤ K. Together the two constraints bracket v₁′ between 1.0 and 2.0 m s⁻¹. Where inside that bracket the carts actually land is not a question momentum can answer: it depends on what the bumpers are made of.

The missing information is how springy the contact is, and the coefficient of restitution measures it directly: e = (v₂′ − v₁′)/(v₁ − v₂), the speed at which the bodies separate divided by the speed at which they approached. In the line of outcomes above, (1.0, 4.0) has e = 3.0/3.0 = 1, (1.5, 3.0) has e = 1.5/3.0 = 0.50, and (2.0, 2.0) has e = 0. Solving momentum and restitution together gives each final velocity as the centre-of-mass velocity V plus or minus a share of e(v₁ − v₂), weighted by the other body's mass. Because e involves only a relative velocity, it has the same value in any frame moving at constant velocity, so a value found on a bench holds on a moving train. Measure it by dropping a ball onto a heavy floor: e = √(h′/h), and a ball dropped from 1.00 m that rebounds to 0.64 m has e = 0.80. Treat that as a value at one impact speed. Most real contacts dissipate a larger share when struck harder, so e falls as the impact speed rises.

Split the kinetic energy into two parts that behave differently: K = ½(m₁ + m₂)V² + ½μ(v₁ − v₂)², where μ = m₁m₂/(m₁ + m₂) is the reduced mass. The first part is the motion of the centre of mass; momentum conservation fixes V, so it cannot change. The second is the relative motion, and restitution multiplies the relative velocity by −e, so a fraction e² of it survives. That is the classification: e = 1 keeps all of K, 0 < e < 1 loses part of the relative share while the bodies still separate, and e = 0 loses all of it, the most momentum allows. For the 2.0 kg and 1.0 kg carts, μ = 0.667 kg and the relative share is ½ × 0.667 kg × (3.0 m s⁻¹)² = 3.0 J of the 9.0 J: sticking loses 3.0 J and e = 0.50 loses 2.25 J. So judge a collision by e, not by K′/K. A 1.0 kg cart at 5.0 m s⁻¹ that catches and sticks to a 1.0 kg cart at 4.0 m s⁻¹ keeps 20.25 J of 20.5 J, 99%, and is still completely inelastic: almost all of that energy was centre-of-mass energy that no collision could remove.

With the target at rest the general result becomes v₁′ = (m₁ − e m₂)v₁/(m₁ + m₂) and v₂′ = (1 + e)m₁v₁/(m₁ + m₂). Read them before substituting. Equal masses with e = 1 exchange velocities: the striker stops dead and the target leaves at v₁, which is why a cue ball struck without spin stops almost on the spot after a full-face hit. A light striker bounces back whenever e m₂ exceeds m₁: a 0.10 kg ball at 5.0 m s⁻¹ hitting a 2.0 kg block with e = 0.80 returns at (0.10 − 1.6) × 5.0/2.1 = −3.6 m s⁻¹, while the block creeps away at 0.43 m s⁻¹. A heavy striker barely slows and flings a light target off at up to (1 + e)v₁, twice its own speed when the collision is elastic. Let m₂ grow without limit and the target becomes a wall: v₁′ → −e v₁, the bounce rule. If your algebra breaks one of these limits, the algebra is wrong.

Each ingredient of the model has a visible way to fail. Momentum conservation needs a brief contact: a 1500 kg car skidding on locked wheels, with kinetic friction coefficient 0.70, through a 0.12 s crash takes a road-friction impulse of 0.70 × 1500 kg × 9.81 m s⁻² × 0.12 s = 1.2 × 10³ N s, about 5% of the 2.3 × 10⁴ kg m s⁻¹ it carries at 15 m s⁻¹, and crash reconstruction has to correct for it. One dimension needs the push to act along the line joining the centres; an off-centre blow sends the bodies sideways or into spin, and rotational kinetic energy then looks like an unexplained loss. Two bodies are essential: with three in contact at once, as in a Newton's cradle, momentum and energy give two equations for three final velocities, and the outcome depends on how the contact forces travel along the chain. And 0 ≤ e ≤ 1 holds only for passive bodies. A spring-loaded plunger released on impact adds energy and gives e above 1, while a negative e would mean one body had passed through the other.