Conservation of Momentum: Elastic and Inelastic Collision Simulator

Click the first ball of a Newton's cradle into the row and, instead of all five swinging away together, one ball pops out the far side at almost exactly the speed the first one arrived with. The cradle is performing a calculation in steel and string: it is conserving momentum. The same bookkeeping governs car crashes, rocket launches, billiard breaks and the recoil of a rifle.

This page is an interactive collision simulator built around the law of conservation of momentum. Set the masses and velocities of two blocks, slide the elasticity anywhere between a perfect bounce and a sticky crash, and watch the before/after bars prove the law that examiners in AP Physics, A-Level, JEE and NEET never tire of testing: total momentum is always conserved. Kinetic energy is a different story.

What Is the Law of Conservation of Momentum?

Momentum is mass times velocity, $p = mv$ — a vector pointing along the motion, measured in kg·m/s. The conservation law states:

In any collision or interaction with no external force, the total momentum of the system is the same before and after: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where $u$ denotes velocities before the collision and $v$ after.

Why must this be true? It follows directly from Newton's third law. During the collision, block 1 pushes on block 2 as hard as block 2 pushes back on block 1, for the same duration. Equal and opposite forces for equal times mean equal and opposite impulses ($J = F\Delta t = \Delta p$), so whatever momentum one block gains, the other loses. The total cannot change; it can only be redistributed.

Two ice skaters at rest who push apart demonstrate the vector nature perfectly: the light skater glides away fast, the heavy one slowly, and the total momentum stays exactly what it was: zero.

What Is the Difference Between Elastic and Inelastic Collisions?

Momentum is conserved in every collision. The type of collision is decided by what happens to kinetic energy:

The coefficient of restitution $e$ measures the "bounciness" — the ratio of separation speed to approach speed:

$e = \frac{v_2 - v_1}{u_1 - u_2}$

A dropped superball ($e \approx 0.9$) rebounds to about 81% of its height (height scales with $e^2$); a beanbag ($e \approx 0$) just thuds.

Interactive Collision Simulator

Set each block's mass and initial velocity, choose the elasticity, and press play. The blocks are labelled with their live velocities; the bar chart below is the whole point of the lesson: the momentum bars stay equal before and after, while the kinetic-energy bars drop by the fraction lost to heat, sound and deformation. The velocity–time graph at the bottom shows the collision the way examiners draw it: two flat lines, one instantaneous step, with the dotted marker tracking the animation in real time. Three experiments worth running: equal masses with e = 1 (they swap velocities), any masses with e = 0 (they stick and move together), and a heavy block hitting a light one (the light one rockets away at up to nearly twice the impact speed).

How Do You Solve Collision Problems Step by Step?

Every 1D collision problem uses the same recipe:

1. Choose a positive direction and write every velocity with its sign.
2. Write momentum conservation: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
3. Add the second equation your problem type provides:
   • Perfectly inelastic: the objects share one final velocity, $v_1 = v_2 = v$, giving $v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$
   • Perfectly elastic: kinetic energy is also conserved, which (after algebra) yields $v_1 = \frac{(m_1 - m_2)\,u_1 + 2 m_2 u_2}{m_1 + m_2}, \qquad v_2 = \frac{(m_2 - m_1)\,u_2 + 2 m_1 u_1}{m_1 + m_2}$
4. Sanity-check the special cases. Equal masses in an elastic collision swap velocities. A heavy object hitting a light stationary one barely slows down, while the light one flies off at up to $2u_1$. A light object bouncing off a heavy one reverses direction with almost unchanged speed.

The simulator implements the general case using the restitution form: $v_2 - v_1 = e(u_1 - u_2)$ combined with momentum conservation, which smoothly interpolates between the elastic ($e = 1$) and perfectly inelastic ($e = 0$) formulas above.

Why Does a Newton's Cradle Release Exactly as Many Balls as You Lift?

The equal-mass velocity swap is the entire operating principle of the desk toy below. Lift one, two, three or four balls and release: each collision in the chain hands the incoming velocity to the next ball, the impulse walks invisibly through the stationary row, and the same number of balls exits the far side. The alternative — two balls popping out at half speed after lifting one — would conserve momentum just fine, and it never happens, because it would not conserve kinetic energy ($\frac{1}{2}m v^2$ punishes splitting speed across more mass). The cradle enforces both laws at once, every swing.

The timing doesn't depend on how many balls move together, since a pendulum's period doesn't care about mass. It does depend on how far you lift them, though: large-angle pendulums run measurably slower than the small-angle formula predicts.

Watch the swings shrink over several passes, too. This demo isn't a perfect, frictionless idealisation — every real cradle bleeds off a little kinetic energy each swing to air resistance and to ball-on-ball collisions that are very nearly, but not quite, perfectly elastic. Same story as the KE bars above, just playing out more slowly.

Why Is Kinetic Energy Lost in Inelastic Collisions?

The "lost" kinetic energy hasn't vanished. Metal bends, materials heat up, sound waves carry energy away — it's gone somewhere, just not into motion anymore. Total energy is always conserved; only the kinetic portion shrinks (our kinetic and potential energy visualiser explores these conversions).

Car crumple zones are this physics used deliberately. The collision is made as inelastic as possible so the car body absorbs kinetic energy by deforming — and, just as importantly, the crumpling stretches the collision over a longer time. The occupants' momentum change is fixed by the crash, but impulse $J = F\Delta t$ means a longer $\Delta t$ requires a smaller force. Rigid old-fashioned cars bounced — transferring the energy and the short, violent force to the passengers instead.

Worked Examples for Physics Exams

Example 1: Railway trucks coupling (perfectly inelastic)

A 4,000 kg truck moving at 3 m/s couples with a stationary 2,000 kg truck. Find their common speed and the kinetic energy lost.

$v = \dfrac{4000 \times 3 + 2000 \times 0}{6000} = 2$ m/s. KE before $= \frac{1}{2}(4000)(3^2) = 18{,}000$ J; KE after $= \frac{1}{2}(6000)(2^2) = 12{,}000$ J. One third of the kinetic energy became sound, heat and deformation of the coupling. Try it in the simulator: m₁ = 4, m₂ = 2, u₁ = 3, u₂ = 0, e = 0 (same numbers, scaled ×1000).

Example 2: Equal masses, elastic collision

A 2 kg ball moving at 5 m/s strikes an identical stationary ball perfectly elastically. Find both final velocities.

Using the elastic formulas with $m_1 = m_2$: $v_1 = 0$, $v_2 = 5$ m/s — the balls exchange velocities. This is the entire secret of Newton's cradle: each ball hands its momentum and energy to the next, and only the last one is free to move.

Example 3: Ballistic block (bullet embeds)

A 10 g bullet travelling at 400 m/s embeds itself in a stationary 1.99 kg wooden block. Find the block's speed and the fraction of kinetic energy lost.

$v = \dfrac{0.01 \times 400}{0.01 + 1.99} = \dfrac{4}{2} = 2$ m/s. KE before $= \frac{1}{2}(0.01)(400^2) = 800$ J; KE after $= \frac{1}{2}(2.0)(2^2) = 4$ J. 99.5% of the energy is lost to splintering wood and heat — yet momentum is perfectly conserved. This is the principle of the ballistic pendulum, historically used to measure bullet speeds.