Pull back k balls and let go, and k balls fly out of the other side at the same speed. See the momentum and kinetic-energy bars conserved through the collisions, and get a feel for why the motion decays when you lower the coefficient of restitution e.
Parameters
5
1
1.00
e = 1 is perfectly elastic; e < 1 decays (real steel balls are about 0.95–0.99).
45 °
Presets
You can also drag the leftmost ball up; releasing it starts the chain of collisions.
Live values
0
Moving balls
0.00
Total momentum p
0.00
Total kinetic energy
0.0%
Energy lost (vs initial)
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Period T [s]
Real-time Newton's cradle
Velocity vectorsTrailsCollision
Momentum conservation + kinetic-energy conservation → k balls in, k balls out.
For elastic collisions of equal masses, the only solution that satisfies both conservation laws at once is "the same number of balls leave the other side at the same speed".
Momentum & kinetic energy (conserved through collisions)
Total momentum p
0.00
Total kinetic energy
0.00
Dashed line = reference initial value. With e = 1, both p and KE return to this line every time the balls collide at the bottom (conserved across the collision). With e < 1 the height the KE bar reaches drops with every collision and itgradually decays.
Results
0.00
Kinetic energy J/kg
0.00
Potential energy J/kg
0.00
Total energy J/kg
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Period T (s)
Relation to CAE
It is the basic model for impact simulation (LS-DYNA, Abaqus/Explicit). Modelling the coefficient of restitution is central to impact-absorption design and crash-safety analysis, and ball mills and impact forming are governed by the same contact mechanics.
Conservation laws: $\sum m_i v_i = \text{const}$ (momentum), and for $e=1$, $\sum \tfrac12 m_i v_i^2 = \text{const}$ (kinetic energy)
Pendulum period: $T = 2\pi\sqrt{\dfrac{L}{g}}$
KE loss per collision: $\propto (1-e^2)$
Check: with N = 5, lifting k = 1 sends out just one ball (the others stay still), k = 2 sends out two and k = 3 sends out three from the other side (with e = 1, p and KE are fully conserved).
What Is Newton's Cradle?
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Newton's cradle is that desk toy where you pull back one ball at the end and let go, and just one ball flies out on the other side, right? Why do the middle balls stay still while only the end balls move?
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Roughly speaking, it is because both momentum and kinetic energy are conserved. When balls of equal mass collide perfectly elastically (coefficient of restitution e = 1), the ball that hits stops, and its momentum and energy are passed on "exactly as they are" to the next ball. Keep the "Coefficient of restitution" slider at 1 in the simulator and pull back one ball. You will clearly see the momentum and kinetic-energy bars return to the reference line at every collision (= conserved across the collision).
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Really! So what happens if I make the coefficient of restitution smaller than 1? What does this "e" represent?
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In practice perfectly elastic collisions with e = 1 hardly exist; 0.95–0.99 is common. e is the ratio of relative speeds after and before the collision, and below 1 the collision is inelastic. In other words, each collision turns some kinetic energy into heat, sound and deformation, and it is lost. Lower e to about 0.8 in the simulator. You should see that the momentum does not change from just before to just after a collision (p is conserved), yet the height the kinetic-energy bar reaches drops with every collision until everything stops.
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I see! When I set "Balls lifted k" to 2, two balls really do come out of the other side. Is that also decided by the conservation laws?
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Exactly. k balls in, k balls out: this is the only solution that satisfies momentum conservation and kinetic-energy conservation at the same time. If you assume "only one ball flies out at k times the speed", the momentum matches but the kinetic energy does not. If instead you assume "all the balls move slowly at the same speed", the energy matches but the momentum does not. Only "the same number of balls leaving at the same speed" satisfies both. These conservation laws are the very basics of vibration and impact analysis in CAE too.
Frequently Asked Questions
Drag the ball up to a sufficient height and release the mouse. If the pull-back angle is too small, the chain of collisions is weak. With a coefficient of restitution below 1, energy decays and the chain breaks off more easily. Set the coefficient of restitution to 1.0 and check again.
The coefficient of restitution e sets the speed ratio after a collision: e = 1 is perfectly elastic and e = 0 perfectly inelastic (the bodies stick together). In CAE it is used in designing impact-absorbing materials (a small e to absorb energy) and in evaluating crash-safety components.
This tool fixes the string length at L = 1.0 m and the gravitational acceleration at g = 9.81 m/s². The period is T = 2π√(L/g): a longer L lengthens the period and spaces out the collisions, while a larger g shortens the period. In either case the outcome of the collisions, "k balls in, k balls out", does not change.
This tool shows simplified elastic collisions of equal masses, whereas LS-DYNA and Abaqus/Explicit analyse contact between complex shapes and materials with the finite element method. The idea of the coefficient of restitution and the law of momentum conservation are shared, so it gives an intuitive grasp of the basic concepts of impact analysis.
Real-World Applications
Crash-safety design (cars and railways):In vehicle crash simulation (LS-DYNA and others), the coefficient of restitution e is an important parameter for modelling the behaviour of impact-absorbing materials. The value of e determines how much energy is absorbed (that is, the impact on the occupants).
Manufacturing processes (ball mills, stamping):In grinding mills and impact processes in metalworking, collisions between the media (balls or hammers) govern product quality. The efficiency of energy transfer in a collision is analysed with exactly the physics learned in this tool.
Sports equipment development:In collisions such as a golf club head and ball or a tennis racket and ball, the coefficient of restitution (COR) is the key to performance. For some equipment the regulations set an upper limit.
Space engineering (docking, sample return):Spacecraft docking and the impacts during sample collection by asteroid probes take place at extremely low relative speeds. Designs that dissipate kinetic energy through inelastic collisions (e < 1) are important.
Common Misconceptions and Points to Note
First,are you assuming that "because there are five balls, pulling one end ball makes just one ball move on the other side"? In fact that only holds in the ideal case where the initial conditions are perfectly symmetric and the collision is perfectly elastic (e = 1). For example, if you pull back two balls together and let go, two balls fly out of the other side. This follows necessarily from conservation of momentum and energy. Check it in the simulator. Next,a pitfall in setting parameters. If you increase the "pull-back angle", the pendulum period becomes longer than the small-amplitude value 2π√(L/g) (about 4% longer at 45° and about 7% at 60°). The period shown is the small-amplitude approximation, so remember that at large angles it differs from the actual interval between collisions. Finally,the misconception that "the coefficient of restitution e is determined by the material alone". In reality it also changes with impact speed, temperature and surface condition. In CAE impact analysis, rather than treating e as a constant, a speed-dependent model is often used. The trick is to observe the change in behaviour when you vary e in this tool not simply as "damping" but in connection with "the fraction of energy lost".
Set "Number of balls N" between 2 and 9 to change the size of the cradle
Set "Balls lifted k" between 1 and N−1 and confirm that k balls in means k balls out on the other side
Change the coefficient of restitution e between 0.60 and 1.00 to compare perfectly elastic (e = 1.0) and inelastic collisions (with e = 1, the momentum and kinetic-energy bars return to the reference line at every collision and the energy loss stays at 0%)
Adjust the pull-back angle θ and the playback speed (0.25× / 1× / 2×) to watch the chain of collisions in slow motion or fast forward. For elastic collisions of equal masses, momentum conservation Σm·v = constant always holds
Worked Example
In this tool the string length is fixed at L = 1.0 m and g = 9.81 m/s², so the small-amplitude period is T = 2π√(L/g) ≈ 2.01 s (independent of mass; at a 45° pull-back angle the actual period is about 4% longer). Each ball is treated as 1 kg and energies are shown in J/kg. Pulling a ball back by 45° raises it by h = L(1 − cos 45°) ≈ 0.293 m, giving a potential energy of g·h ≈ 2.87 J/kg, which becomes the same amount of kinetic energy at the bottom, at a speed of √(2gh) ≈ 2.40 m/s. With k = 2 the total energy doubles to about 5.75 J/kg. With a coefficient of restitution e < 1, one collision with a ball at rest loses (1 − e²)/2 of the kinetic energy (about 4.9% for e = 0.95). With five balls in a row the collisions chain within each swing, so the energy falls even faster.
Practical Notes
This tool resolves instantaneous collisions between neighbouring balls one after another. Real balls transmit force while deforming slightly, so even with e close to 1 the balls on the far side can gradually start to move, departing from the ideal "k balls in, k balls out"
The string length is fixed at 1.0 m. The period is proportional to √L, so making the string 1.5 times longer lengthens the period by about 1.22 times (independent of mass and the number of balls)
Real steel balls have a coefficient of restitution of about 0.95–0.99, and air resistance and friction in the strings and pivots also act, so a real cradle stops within tens of seconds to a few minutes. This tool does not include air resistance, so with e = 1 it keeps moving