Start — bumper cars. Picture two bumper cars rolling toward each other. When they hit, neither one simply stops dead while the other sails on untouched — they share the crash. Some of the motion gets passed across, and how springy the rubber bumpers are decides whether the cars bounce apart or clump together and trundle off as one. Every collision in the universe, from billiard balls to crashing galaxies, is some version of that single shared shove.
Build — count the motion. Physicists measure the "amount of motion" as momentum, $p = mv$ — mass times velocity. The headline rule is that in any collision with no outside push, the total momentum before equals the total after. If a $2\text{ kg}$ cart rolling at $3\text{ m/s}$ (so $p = 6\ \text{kg·m/s}$) strikes a stationary $1\text{ kg}$ cart, the two velocities afterward must still add up to $6\ \text{kg·m/s}$ of momentum — not a scrap more or less. A second number, the coefficient of restitution $e$, captures the bounciness: $e=1$ means the carts separate exactly as fast as they approached (a perfect bounce), while $e=0$ means they stick and move off together.
Deepen — the two equations. Momentum conservation alone can't pin down both final speeds, so we add the restitution rule, $v_2' - v_1' = -e\,(v_2 - v_1)$ — separation speed equals $e$ times approach speed. Solving the pair gives $v_1' = \frac{(m_1 - e\,m_2)v_1 + (1+e)\,m_2 v_2}{m_1+m_2}$ and the mirror-image expression for $v_2'$. The kinetic energy that disappears into heat, sound, and dents is $\Delta KE = \tfrac12\,\mu\,(v_1-v_2)^2(1-e^2)$, where $\mu = \tfrac{m_1 m_2}{m_1+m_2}$ is the reduced mass — note it vanishes when $e=1$ and is largest when $e=0$. The sliders feed straight into these formulas: $m_1, m_2$ set the cart masses, $v_1, v_2$ the starting speeds, and $e$ the bounciness.
Try this in the sim above. (1) Set $m_1 = m_2$, $e=1$, and start one cart at rest — watch the two velocities swap exactly (the trick behind Newton's cradle). (2) Drag $e$ down to $0$: the carts stick, the kinetic-energy bar drops, yet the momentum bar holds rock-steady. (3) Make $m_1$ huge and $m_2$ tiny and crash them — the light cart rockets off at nearly twice the heavy cart's speed.
$$\Delta KE = \tfrac12\,\frac{m_1 m_2}{m_1+m_2}\,(v_1-v_2)^2\,(1-e^2)$$
Symbol Definitions
Symbol
Meaning
SI Unit
$m_1, m_2$
Masses of the two bodies
kg
$v_1, v_2$
Velocities before collision
m s⁻¹
$v_1', v_2'$
Velocities after collision
m s⁻¹
$e$
Coefficient of restitution (1 = elastic, 0 = perfectly inelastic)
dimensionless
$p = mv$
Linear momentum
kg m s⁻¹
$KE = \tfrac12 m v^2$
Kinetic energy
J
$\mu = \tfrac{m_1 m_2}{m_1+m_2}$
Reduced mass
kg
Step 1Apply momentum conservation. No external impulse acts during the brief collision, so total momentum is conserved: $m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$.
Step 2Apply Newton's experimental law. The relative velocity of separation equals $e$ times the relative velocity of approach: $v_2' - v_1' = -e(v_2 - v_1)$.
Step 3Solve the linear system. Two equations, two unknowns ($v_1', v_2'$). Substituting and rearranging yields the boxed result above.
Step 4Energy check. When $e=1$, $\Delta KE = 0$ (elastic). When $e=0$, $v_1' = v_2' = (m_1 v_1 + m_2 v_2)/(m_1+m_2)$ (perfectly inelastic — bodies stick).
Step 52D extension. Decompose velocities along the line-of-centres and tangentially. The above 1D formula applies along the line-of-centres; tangential components remain unchanged in a smooth collision.
How simulation variables map to the equations
The sliders m₁, m₂ set the masses; v₁, v₂ set the initial velocities used in the conservation equations. The slider e directly enters the formulas for $v_1'$ and $v_2'$. In the 2D mode, the angle slider rotates the line-of-centres by $\theta$ relative to $\hat{x}$. The Newton's cradle mode chains five equal-mass elastic collisions to demonstrate sequential momentum transfer.
Reference: Halliday, Resnick & Walker — Fundamentals of Physics, 10th Ed., Ch. 9 §9-7 to §9-11: "Conservation of Linear Momentum" and "Elastic Collisions in One Dimension".
Section 04
Frequently Asked Questions
Car crash safety (crumple zones increase $\Delta t$ to reduce force), billiards and pool (nearly elastic), Newton's cradle on executive desks, particle accelerators (deep inelastic scattering of quarks), and sports — a tennis ball hitting a racquet has $e \approx 0.7-0.9$, a baseball off a bat $\approx 0.55$.
💡 Same equations govern a crashing car and a colliding proton.
Two coloured discs slide toward each other along a horizontal track. The vertical bars beneath show momentum (conserved — bar length stays constant) and kinetic energy (decreases when $e<1$). Set $e=1$ for elastic, $e=0$ for "stick together", or anywhere between. The 2D mode shows oblique impact along a line-of-centres.
💡 Watch the momentum bar — it never changes during a collision.
Conservation laws come from symmetries (Noether's theorem). Momentum conservation arises from spatial translation symmetry — physics is the same here as one metre away — and the collision's internal forces are equal-and-opposite by Newton's third law, so they cancel internally. Energy can convert to heat, sound, deformation — none of which violate spatial symmetry, but they do remove KE from the macroscopic motion.
💡 Internal forces redistribute energy but cannot change total momentum.
It means the bodies separate at half the speed they approached. If two objects approach at relative speed $10$ m/s and $e=0.5$, they separate at $5$ m/s. This loses $1-e^2 = 75\%$ of the kinetic energy of relative motion. $e$ depends on materials (steel-on-steel ≈ 0.6, glass-on-glass ≈ 0.95) and impact speed.
💡 $e$ measures how "bouncy" two materials are together.
Set $m_1 = m_2 = m$, $v_2 = 0$, $e = 1$ in the formulas: $v_1' = 0$ and $v_2' = v_1$. The two are completely interchanged. This is the principle behind Newton's cradle — the impulse propagates through the chain almost losslessly because each ball is identical.
No — by definition, in a perfectly inelastic collision the bodies stick together, which forces them to share a single final velocity. Substituting $v_1' = v_2' = v_f$ into momentum conservation and computing $\Delta KE$ gives $\Delta KE = -\tfrac12\mu(v_1 - v_2)^2$, which is zero only when $v_1 = v_2$ initially (i.e., no actual collision).
💡 Sticking together always loses kinetic energy — except for a trivial "non-collision".
Ping-pong balls have $e \approx 0.9$ at low speeds because of their thin elastic plastic shell that stores and releases energy efficiently. Tennis balls have a felt cover, pressurized rubber core, and lose energy to internal viscoelastic deformation, giving $e \approx 0.75$. The ITF actually regulates the rebound height — and hence $e$ — for official tournament balls.
💡 Bounciness is a property of the material pair, not of either alone.
Resource: Khan Academy — "Elastic and Inelastic Collisions"; HyperPhysics — Collisions (hyperphysics.phy-astr.gsu.edu/hbase/colsta.html); MIT OCW 8.01 Lecture 15.
Section 05
Common Misconceptions
❌ In every collision, kinetic energy is conserved.
✅ Only elastic collisions conserve KE. In real life, almost no macroscopic collision is perfectly elastic — energy is always lost to sound, heat, and deformation. Momentum is conserved in every collision (in the absence of external forces) regardless of $e$.
📖 HRW 10th Ed., §9-9: "Elastic Collisions in One Dimension".
❌ A heavy object always pushes a light object out of the way unchanged.
✅ When a heavy object collides head-on elastically with a stationary light object, the heavy one slows only slightly while the light one shoots forward at almost twice the heavy object's speed. Setting $m_1 \gg m_2$ in the elastic formulas gives $v_2' \to 2v_1$. This is why a moving truck launches a stationary tennis ball at huge speed.
📖 HRW 10th Ed., §9-9, Eq. 9-67 to 9-68.
❌ In a perfectly inelastic collision, all kinetic energy is lost.
✅ Only the kinetic energy of relative motion is lost. The kinetic energy of the centre-of-mass motion is preserved — it cannot disappear, because momentum is conserved. The fraction of KE lost is $\tfrac{m_2}{m_1+m_2}$ (when $m_2$ is initially at rest).
📖 Kleppner & Kolenkow, 2nd Ed., §4.3.
❌ "Action-reaction pairs" cancel each other out, so they have no effect.
✅ Internal action-reaction forces between two colliding bodies are equal and opposite, so they cancel in the system's total momentum. But each force acts on a different body, so neither body cancels itself out — each one is accelerated by the force on it. This is why both bodies change velocity even though the system's momentum is conserved.
📖 Arons — A Guide to Introductory Physics Teaching, Ch. 3 §3.18.
❌ Newton's cradle works because momentum "travels through" the middle balls.
✅ In a real Newton's cradle the middle balls do move slightly (microscopic compression waves propagate at the speed of sound in steel). The idealisation that they stay still works only for perfectly identical, perfectly elastic balls. With even tiny mass mismatch, multiple balls swing out at the far end — momentum and energy both must be conserved, and that uniquely determines the outcome only for identical balls.
📖 Hutzler et al. (2004), Am. J. Phys. 72, 1508 — "Rocking Newton's Cradle".
❌ In a 2D collision, treating each axis independently is always valid.
✅ Treating $x$ and $y$ as independent works for momentum (it is a vector). But the coefficient-of-restitution condition applies only along the line of impact (line-of-centres). The tangential component is governed by friction or — for smooth bodies — is unchanged. Mixing them up gives wrong post-collision angles.
📖 Marion & Thornton — Classical Dynamics, 5th Ed., §9.7.
Misconception research: Halloun & Hestenes (1985), Am. J. Phys. 53, 1056 — "The initial knowledge state of college physics students"; McDermott — "Physics by Inquiry" (1996), Vol. II.