How Newton's Cradle Works, and Why Momentum Is Conserved • The Science 360

Lift the end ball of a Newton's cradle a few centimetres and let go. It clacks into the row, stops dead, and the ball at the far end flies out to almost the same height. Lift two, and two fly out. Nobody is surprised any more, but it is worth asking why it is never one ball out at double speed, or two balls out at half speed. Both of those would carry exactly the right momentum.

The answer needs two conservation laws working together, and it is the same physics that decides every collision, from snooker balls to car crashes. The collision and Newton's cradle simulator lets you change masses, speeds and the bounciness of every collision and watch the momentum and energy bars.

Quick answerA Newton's cradle works because its steel balls collide almost elastically, so both momentum and kinetic energy pass down the row. For equal masses an elastic collision swaps velocities: the moving ball stops and the next one takes its speed. The only way to leave with the same momentum and the same energy is for the same number of balls to swing out at the same speed.

Why is momentum conserved in a collision?

Newton's third law. While two balls are touching, ball A pushes on ball B with some force and B pushes back on A with an equal force in the opposite direction, for exactly the same time. Force times time is the change in momentum (the impulse), so whatever momentum B gains, A loses:

\begin{aligned}\Delta p_A &= -F\,\Delta t \\ \Delta p_B &= +F\,\Delta t \\ \Rightarrow\ m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2\end{aligned}

Nothing in that argument cares whether the balls bounce, stick, or crumple. Momentum is conserved in every collision as long as no outside force acts during the impact. Kinetic energy is a different story.

Worked example: two trolleys that stick
A 1 kg trolley at 2 m/s hits an identical trolley at rest, and Velcro holds them together.
Momentum before:

1 \times 2 + 1 \times 0 = 2\ \text{kg m/s}

. After, 2 kg share it:

v = 2/2 = 1\ \text{m/s}

.
Kinetic energy before:

\tfrac12 \times 1 \times 2^2 = 2\ \text{J}

. After:

\tfrac12 \times 2 \times 1^2 = 1\ \text{J}

.
Momentum is unchanged; half the kinetic energy has become heat and sound.

Perfectly inelastic collision of two 1 kg trolleys with Velcro bumpers: they stick and move off together at 1.00 m/s, momentum stays 2.00 kg m/s but kinetic energy falls from 2.00 J to 1.00 J, half lost — Figure 1. Two 1 kg trolleys that stick. The momentum bars match before and after; the energy bar loses 1.00 J (dashed). Open this setup in the simulator.

What decides how much a collision bounces?

One number, the coefficient of restitution $e$ : the speed the bodies separate at, divided by the speed they approached at.

v_2 - v_1 = e\,(u_1 - u_2)

With $e = 1$ they fly apart as fast as they came together and kinetic energy is conserved (an elastic collision). With $e = 0$ they stick. Everything real sits in between. Together with momentum conservation, this second equation fixes both final velocities, and the kinetic energy lost comes out as

\begin{aligned}\Delta KE &= (1 - e^2)\,\tfrac12\,\mu\,(u_1 - u_2)^2 \\ \mu &= \frac{m_1 m_2}{m_1 + m_2}\end{aligned}

For a ball bouncing on a hard floor, $e$ is easy to measure: the rebound speed over the impact speed is the square root of the rebound height over the drop height. Sports bodies specify it, in effect, in their ball rules.

Worked example: the restitution of a tennis ball
The ITF drops a tennis ball from 254 cm and requires a rebound of 135 to 147 cm.

e = \sqrt{135/254} = 0.73

\sqrt{147/254} = 0.76

.
The share of kinetic energy kept in the bounce is

e^2

: 53% to 58%. Almost half goes into flexing the rubber and the gas inside.

Object	Test	Coefficient of restitution
Table tennis ball	30.5 cm drop onto steel, 24 to 26 cm rebound	0.89 to 0.92
Golf driver	USGA limit for a ball fired at the club face	at most 0.83
Basketball	FIBA: 1,800 mm drop, 1,035 to 1,085 mm rebound	0.76 to 0.78
Tennis ball	ITF: 254 cm drop, 135 to 147 cm rebound	0.73 to 0.76
Cars in a crash	crumple, often lock together	close to 0

Why does one ball in make exactly one ball out?

For two equal masses with $e = 1$ , solving the two equations gives a neat result: the balls swap velocities. A ball arriving at speed $v$ stops, and the ball it hits leaves at $v$ . In a cradle with small gaps between the balls, that handover runs down the row, ball by ball, until the last ball has nobody to pass it to and swings out.

Now test the alternatives the toy never shows you. Suppose one ball of mass $m$ arrives at $v$ .

Worked example: why not two balls at half speed?
Coming in: momentum

mv

, kinetic energy

\tfrac12 mv^2

.
Two balls out at

v/2

: momentum

2m \times v/2 = mv

, which matches. Kinetic energy

2 \times \tfrac12 m (v/2)^2 = \tfrac14 mv^2

, half of what came in.
One ball out at

v

: momentum

mv

, kinetic energy

\tfrac12 mv^2

. Both match.
With the simulator's 30 cm strings and a 9 cm pull-back,

v = \sqrt{g/L} \times 0.09 = \sqrt{9.81/0.3} \times 0.09 = 0.51\ \text{m/s}

at the bottom of the swing.

Two balls out at half speed would quietly throw away half the energy, which nearly elastic steel cannot do. Only one ball out at full speed satisfies both laws. Lift two and the same logic demands two out.

Newton's cradle simulation with five steel balls on a desk: one ball was lifted and released, and the fifth ball swings out on the far side while the middle three stay still, with a velocity bar for each ball and 100 percent of the energy left — Figure 2. One ball in, one ball out. Only ball 5 is moving; the energy meter reads 100% with e = 1. Open this setup in the simulator.

Newton's cradle with two balls lifted: after the collisions the two right hand balls swing out together while the three left balls hang still, momentum and kinetic energy both conserved — Figure 3. Two balls in, two balls out. Open this setup in the simulator.

Is the ball by ball picture really what happens?

It is the standard explanation and it is right when the balls hang with tiny gaps, so each collision involves only two of them. In a real cradle the balls usually touch, and the push travels through the row as a compression wave in the steel. The full treatment uses Hertz's contact mechanics, which describes how two steel spheres squash slightly where they meet. That more careful picture predicts, and experiments confirm, that the middle balls are not perfectly still: the fourth ball moves a little and the first ball recoils slightly. Those small departures are one reason a real cradle's clean one in, one out pattern slowly smears.

The name, incidentally, is not Newton's. The principle was demonstrated by Edme Mariotte in the 17th century, and the name "Newton's cradle" was coined in 1967 by Simon Prebble for a wooden version his company sold.

Why does a real cradle end up swinging as one?

Because every clack is a little inelastic. Each impact turns some kinetic energy into the sound you hear and a little heat in the steel, so $e$ is slightly below 1. Once a collision is not perfectly elastic, the moving ball no longer stops dead; it keeps a little velocity, and the middle balls begin to move. The pattern drifts until all five swing together.

Worked example: where the cradle ends up
When all five balls move together, the original momentum

mv

is shared by

5m

, so

V = v/5

.
Kinetic energy:

\tfrac12 (5m) (v/5)^2 = \tfrac15 \times \tfrac12 mv^2

.
Exactly one fifth of the starting energy is left, with momentum fully conserved. The simulator, with

e = 0.9

, reaches this state within about 6 seconds and shows 20% on its energy meter.

Newton's cradle with a coefficient of restitution of 0.9 after 6.1 seconds: all five balls swing together and only 20 percent of the starting energy is left, because one ball's momentum shared by five carries one fifth of the energy — Figure 4. With e = 0.9 the cradle soon swings as one block with 20% of its energy, as the momentum argument predicts. A well made steel cradle gets there far more slowly. Open this setup in the simulator.

After that, air drag and the flexing of the strings bring it to rest. A Newton's cradle is a beautiful demonstration of conservation laws, but not a perpetual motion machine.

The misconception: “momentum is lost in an inelastic collision”

It is very common to see momentum and kinetic energy treated as if they rise and fall together. They do not. Momentum is a vector and its conservation follows from Newton's third law, so it survives any collision, however crumpled the result. Kinetic energy is only conserved in the special, elastic case. The sticky trolleys above keep all 2 kg m/s of their momentum while losing half their kinetic energy, and the cradle that ends up swinging as one has kept all its momentum and lost four fifths of its energy.

What does this have to do with car crashes?

Everything. A car crash is a collision with $e$ close to zero: the metal crumples and the cars often lock together.

Worked example: 50 km/h into a parked car
A 1,500 kg car at 13.9 m/s (50 km/h) hits a parked 1,000 kg car and they move off together.
Momentum:

1500 \times 13.9 = 20{,}850\ \text{kg m/s}

, shared by 2,500 kg, so

v = 8.34\ \text{m/s}

(30 km/h).
Kinetic energy: 144,908 J before, 86,945 J after. The 57,963 J lost, 40%, goes into bending metal: about the energy needed to lift the bigger car 3.9 m.

A 1,500 kg blue car at 50 km/h crashes into a parked 1,000 kg orange car and they move off together at 8.34 m/s (30 km/h); momentum stays 20,850 kg m/s while 57,963 J of kinetic energy, 40 percent, is absorbed — Figure 5. A perfectly inelastic crash. Momentum bars match; the dashed block is the 57,963 J absorbed by the crumple zones. Open this setup in the simulator.

Momentum fixes how much each car's velocity changes, but not how fast. A crumple zone stretches the change over a longer time, so the force on the people inside is smaller for the same change of momentum. The same impulse idea shows up on a planetary scale in a gravity assist, which is an elastic collision between a spacecraft and a planet that never touch.

Elastic collision of two 1 kg trolleys with spring bumpers on a lab track: the first stops and the second leaves at 2.00 m/s, with momentum 2.00 kg m/s and kinetic energy 2.00 J the same before and after — Figure 6. The two ball version of the cradle: equal trolleys with spring bumpers swap velocities, and both bar charts balance. Open this setup in the simulator.

Try it yourself

Open the simulator on the elastic trolleys and slide $e$ slowly from 1 down to 0, watching the dashed "lost" block grow while the momentum bars stay level. Then load the light hits heavy preset to see a bounce back, and finish with the real cradle to watch the row settle into one block. For a kitchen-table version of these ideas, our physics experiments for kids post has a few.

Sources

Frequently asked questions

How does Newton’s cradle work?

The lifted ball hits the row and hands its velocity on from ball to ball. For equal steel balls in nearly elastic collisions, each ball stops and passes its full velocity to the next, so the last ball swings out with the speed the first one arrived with. Momentum and kinetic energy are both carried through.

Why does one ball in make exactly one ball out?

It is the only outcome that conserves both momentum and kinetic energy. Two balls leaving at half speed would carry the right momentum but only half the kinetic energy. One ball at full speed matches both.

Is momentum conserved in an inelastic collision?

Yes. Momentum is conserved in every collision where no outside force acts, because the two bodies push on each other with equal and opposite forces for the same time. Only kinetic energy is lost in an inelastic collision, into heat, sound and deformation.

Why does a Newton’s cradle stop?

Each impact turns a little kinetic energy into sound and heat, and air drag and the strings take a little more. As the collisions become less than perfectly elastic the middle balls start moving, the row ends up swinging together, and friction then brings it to rest. It is not a perpetual motion machine.

What is the coefficient of restitution?

The speed at which two bodies separate after a collision divided by the speed at which they approached. It is 1 for an elastic collision and 0 when they stick. For a ball dropped on a hard floor, it is the square root of the bounce height over the drop height.