Calculate the velocity v₁' of the first body after a straight central impact from both masses, both initial velocities and the coefficient of restitution k. Conservation of momentum together with the definition of the coefficient of restitution yields both final velocities; that of the second body follows by swapping the two mass and velocity pairs.
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Calculate the velocity v₁' of the first body after a straight central impact from both masses, both initial velocities and the coefficient of restitution k. Conservation of momentum together with the definition of the coefficient of restitution yields both final velocities; that of the second body follows by swapping the two mass and velocity pairs.
A body with m₁ = 2 kg and v₁ = 5 m/s strikes an approaching body with m₂ = 3 kg and v₂ = −1 m/s at a coefficient of restitution k = 0.6. This gives v₁' = [2·5 + 3·(−1) + 0.6·3·(−1−5)]/5 = (10 − 3 − 10.8)/5 = −0.76 m/s – the first body therefore rebounds. Swapping the pairs gives v₂' = +2.84 m/s. Momentum check: 2·(−0.76) + 3·2.84 = −1.52 + 8.52 = 7.0 kg·m/s, exactly as before at 2·5 + 3·(−1) = 7.0 kg·m/s. The limiting cases with the same masses: at k = 0 both continue at 1.4 m/s, at k = 1 v₁' = −2.2 m/s and v₂' = +3.8 m/s.
Straight central impact: the centres of mass of both bodies move along one common line perpendicular to the plane of contact, so the motion stays one-dimensional. External forces are neglected during the short impact, which is why total momentum is conserved; the impact forces are internal to the two-body system. The oblique and the eccentric impact (which additionally transfers angular momentum), rotation and angular momentum of the bodies, friction in the contact area, the deformation and the time history of the impact force, and stress waves in the material are all excluded. The coefficient of restitution is assumed known and constant – it is not a material constant and must be determined for the speed range and geometry considered. Except in the perfectly elastic impact there is always a loss of kinetic energy, which leaves as deformation, heat and sound energy.
When two bodies collide, momentum is conserved but kinetic energy usually is not. The coefficient of restitution k closes that gap: with it, one formula covers the perfectly plastic impact, the perfectly elastic one, and every real impact in between.
A straight central impact occurs when the centres of mass of both bodies move along one common line perpendicular to the plane of contact – the motion therefore stays one-dimensional. Because the impact forces are internal to the two-body system and external forces may be neglected during the short impact, conservation of momentum always holds: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'. That is one equation for two unknowns. The second is supplied by the coefficient of restitution k = (v₂'−v₁')/(v₁−v₂): the ratio of separation to closing velocity. Together they determine both final velocities uniquely.
Billiard balls bounce apart almost unchanged – they are nearly elastic, k close to 1. Two balls of modelling clay stick together and move on as one – perfectly plastic, k = 0. Almost all real impacts lie in between: part of the kinetic energy returns, the rest stays in the material as deformation and heat.
v₁' = [m₁v₁ + m₂v₂ + k·m₂·(v₂ − v₁)] / (m₁ + m₂)
Momentum conservation: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'Coefficient of restitution: k = (v₂' − v₁')/(v₁ − v₂), 0 ≤ k ≤ 1First body: v₁' = [m₁v₁ + m₂v₂ + k·m₂·(v₂ − v₁)]/(m₁ + m₂)Second body: v₂' = [m₁v₁ + m₂v₂ + k·m₁·(v₁ − v₂)]/(m₁ + m₂)Perfectly plastic (k = 0): v₁' = v₂' = (m₁v₁ + m₂v₂)/(m₁ + m₂)Energy loss: ΔT = ½·(1−k²)·m₁m₂/(m₁+m₂)·(v₁−v₂)²Coefficient of restitution from a rebound test: k = √(h/H)| Symbol / input | Meaning |
|---|---|
| Velocity v₁' after impact | The velocity of the first body immediately after impact, signed with respect to the same positive direction as all other velocities. A negative result means the body rebounds, i.e. has reversed its direction of travel. Obtain the velocity of the second body by swapping m₁ with m₂ and v₁ with v₂ and calculating again. |
| Mass m₁ | The mass of the first body. Only the mass ratio affects the result, not the absolute size. Important for an impact against a wall, a foundation or the ground: there m₂ is far larger than m₁, so in practice m₂ → ∞ is used, i.e. a very large number. |
| Mass m₂ | The mass of the second body, in the same system of units as m₁. Equal masses give two notable special cases: in a perfectly elastic impact the bodies exchange their velocities, and in a perfectly plastic impact they continue at the arithmetic mean. |
| Velocity v₁ before impact | The velocity of the first body immediately before impact, with sign. Define a positive direction first and keep it for all four velocities – that is the single most important rule of the whole calculation. For the bodies to collide at all, v₁ must be greater than v₂. |
| Velocity v₂ before impact | The velocity of the second body immediately before impact, signed in the same positive direction. A body at rest has v₂ = 0, an approaching one a negative sign. The difference v₁ − v₂ is the closing velocity and must be positive. |
| Coefficient of restitution k | The coefficient of restitution states what fraction of the closing velocity returns as separation velocity: k = (v₂'−v₁')/(v₁−v₂), with 0 ≤ k ≤ 1. k = 0 means perfectly plastic (the bodies stay together), k = 1 perfectly elastic (no loss of kinetic energy). Source: a rebound test, k = √(h/H) with drop height H and rebound height h. Typical values: steel on steel ≈ 0.6…0.8, wood ≈ 0.5, lead or modelling clay close to 0. Important: k is not a material constant but depends on both impact partners, the speed and the geometry. |
m₁ and m₂ are the two masses; only their ratio affects the result. v₁ and v₂ are the velocities immediately before impact, each signed with respect to a previously chosen positive direction. k is the coefficient of restitution between 0 and 1, obtained from a rebound test via k = √(h/H) with drop height H and rebound height h. Typical values: steel on steel about 0.6 to 0.8, wood about 0.5, lead or modelling clay close to 0.
First define a positive direction and refer all four velocities to it consistently – an approaching body gets a negative sign. Then enter masses, initial velocities and the coefficient of restitution; the result is v₁'. For the velocity of the second body, swap m₁ with m₂ and v₁ with v₂ and calculate again. Then check by comparing momentum before and after: m₁v₁ + m₂v₂ must equal m₁v₁' + m₂v₂'. Conversely, if a final velocity has been measured and k is unknown, select k as the target.
A body with m₁ = 2 kg and v₁ = 5 m/s strikes an approaching one with m₂ = 3 kg and v₂ = −1 m/s at k = 0.6. Then v₁' = [2·5 + 3·(−1) + 0.6·3·(−1−5)]/5 = (10 − 3 − 10.8)/5 = −0.76 m/s: the lighter body rebounds. Swapping the pairs gives v₂' = +2.84 m/s. Momentum check: 2·(−0.76) + 3·2.84 = 7.0 kg·m/s, exactly as before. The limiting cases with the same data: at k = 0 both continue together at 1.4 m/s, at k = 1 v₁' = −2.2 m/s and v₂' = +3.8 m/s. The energy loss follows from ΔT = ½(1−k²)·m₁m₂/(m₁+m₂)·(v₁−v₂)² and is about 13.8 J at k = 0.6, the full 21.6 J at k = 0, and exactly zero at k = 1.
The coefficient of restitution governs the energy loss alone, not the momentum: momentum is conserved exactly for every k. The loss scales with (1−k²), so even k = 0.7 destroys roughly half the available impact energy. Two special cases are particularly vivid: with equal masses an elastic impact exchanges the velocities completely, while a plastic impact leaves both moving at the arithmetic mean. And when a body strikes a far larger mass at rest – a wall, a foundation, the ground – it rebounds at −k·v₁ while the large mass stays practically still.
The calculation runs in kilograms, metres and seconds internally. Velocities may be entered in m/s, km/h, ft/s, mph, m/min or mm/min, masses in mg, g, kg, t, oz or lb. Since only the mass ratio enters, the result is independent of the mass unit system – but both masses must use the same unit. The coefficient of restitution is dimensionless and entered as a decimal.
Hammers and striking tools, where the most plastic impact possible maximises the forming work; pile drivers and hammer drills; buffers, stops and end dampers in machine design; designing catch devices and impact protection; estimating vehicle and transport collisions; rebound tests to determine the coefficient of restitution; sports and games technology; assessing component damage from impact loading.
Valid for the straight central impact, i.e. one-dimensional motion along the common line of centres perpendicular to the plane of contact. The oblique impact (velocities not on that line) and the eccentric impact, which additionally transfers angular momentum and sets the bodies rotating, are not covered. Friction in the contact area, rotation and angular momentum, the time history and magnitude of the impact force, the deformation itself and stress waves in the material also lie outside the model. The coefficient of restitution is assumed known and constant; in reality it decreases as impact speed rises.
Common mistake: By far the most common error is inconsistent signs: all four velocities must refer to the same positive direction, and an approaching body carries a negative sign. Second, k is often treated as a material constant – but it depends on both impact partners, the speed and the geometry, and is therefore transferable only to comparable conditions. Third, kinetic energy is expected to be conserved in a plastic impact; only momentum is, and the energy loss is maximal there. And fourth, a high k is not automatically better: in hammering or pile driving a low k is exactly what is wanted, because only the lost energy goes into forming work.
Hammers and striking tools, where the most plastic impact possible maximises the forming work; pile drivers and hammer drills; buffers, stops and end dampers in machine design; designing catch devices and impact protection; estimating vehicle and transport collisions; rebound tests to determine the coefficient of restitution; sports and games technology; assessing component damage from impact loading.
m₁ and m₂ are the two masses; only their ratio affects the result. v₁ and v₂ are the velocities immediately before impact, each signed with respect to a previously chosen positive direction. k is the coefficient of restitution between 0 and 1, obtained from a rebound test via k = √(h/H) with drop height H and rebound height h. Typical values: steel on steel about 0.6 to 0.8, wood about 0.5, lead or modelling clay close to 0.
Valid for the straight central impact, i.e. one-dimensional motion along the common line of centres perpendicular to the plane of contact. The oblique impact (velocities not on that line) and the eccentric impact, which additionally transfers angular momentum and sets the bodies rotating, are not covered. Friction in the contact area, rotation and angular momentum, the time history and magnitude of the impact force, the deformation itself and stress waves in the material also lie outside the model. The coefficient of restitution is assumed known and constant; in reality it decreases as impact speed rises.