In an impact momentum is conserved, but kinetic energy only in the perfectly elastic limit. Calculate how much kinetic energy is lost in a straight central impact and converted into deformation, heat and sound. In forming, hammering and pile driving this very loss is the actually usable work.
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In an impact momentum is conserved, but kinetic energy only in the perfectly elastic limit. Calculate how much kinetic energy is lost in a straight central impact and converted into deformation, heat and sound. In forming, hammering and pile driving this very loss is the actually usable work.
A body with m₁ = 2 kg and v₁ = 5 m/s strikes an approaching one with m₂ = 3 kg and v₂ = −1 m/s. The closing velocity is v₁ − v₂ = 6 m/s and the reduced mass m₁m₂/(m₁+m₂) = 6/5 = 1.2 kg. At k = 0.6 this gives ΔT = ½·(1−0.36)·1.2 kg·36 m²/s² = ½·0.64·43.2 J = 13.824 J. In the perfectly plastic limit k = 0 it would be 21.6 J, the full amount, and in the perfectly elastic case k = 1 exactly zero. The total kinetic energy before impact is ½·2·25 + ½·3·1 = 26.5 J – so at k = 0.6 a good half of it is lost.
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. The coefficient of restitution is assumed known and constant; in reality it falls as impact speed rises and is not a material constant. The calculator says how much energy is lost but not where it goes in detail: the split between plastic deformation, heat, sound and vibration depends on material and geometry and is not part of the model. The oblique and eccentric impact, rotation and angular momentum, friction in the contact area, the time history and magnitude of the impact force, and stress waves in the material are likewise excluded.
Momentum is conserved in every impact – kinetic energy only in the ideal elastic limit. This calculator determines how much kinetic energy is lost in a straight central impact, and can be rearranged to infer the coefficient of restitution from a measured loss.
Kinetic energy is the energy of motion, usually written T in mechanics; the symbol Δ (delta) denotes a difference. ΔT is therefore the difference between the kinetic energies before and after the impact. From conservation of momentum and the coefficient of restitution k the closed-form expression follows: ΔT = ½(1−k²)·m₁m₂/(m₁+m₂)·(v₁−v₂)². Two quantities in it deserve their own names: v₁ − v₂ is the closing velocity at which the bodies approach each other before impact, and m₁m₂/(m₁+m₂) is the reduced mass – the effective substitute mass of the two-body system for the impact.
A hammer blow on a glowing workpiece deforms it; the same blow on a steel ball makes the ball bounce back. The hammer's energy is the same in both cases – it is just converted almost entirely into deformation in the first and returned almost entirely in the second. The energy loss is therefore not a defect but, in forming, precisely the goal.
ΔT = ½ · (1 − k²) · m₁·m₂/(m₁+m₂) · (v₁ − v₂)²
Energy loss: ΔT = ½·(1−k²)·m₁m₂/(m₁+m₂)·(v₁−v₂)²Reduced mass: mred = m₁m₂/(m₁+m₂)Closing velocity: Δv = v₁ − v₂Perfectly plastic (k = 0): ΔT = ½·mred·Δv², the largest possible lossPerfectly elastic (k = 1): ΔT = 0Coefficient of restitution from a measured loss: k = √(1 − 2ΔT/(mred·Δv²))Coefficient of restitution from a rebound test: k = √(h/H)| Symbol / input | Meaning |
|---|---|
| Energy loss ΔT | The kinetic energy lost in the impact, i.e. the difference between the kinetic energies before and after it. The symbol Δ (delta) denotes a difference and T is the usual symbol for kinetic energy in mechanics. This energy does not disappear but is converted into plastic deformation, heat, sound and vibration. In forging, riveting and pile driving it is the desired useful part; in buffers and stops it is the intended damping; in a vehicle crash it is the measure of deformation. |
| Mass m₁ | The mass of the first body, for example the hammer, the pile driver or the striking vehicle. Only the product and the sum of both masses enter; the expression m₁m₂/(m₁+m₂) is called the reduced mass and describes the effective substitute mass of the two-body system for the impact. |
| Mass m₂ | The mass of the second body, for example the workpiece, the pile or the struck vehicle. When a body hits a far larger mass at rest such as a wall or a foundation, the reduced mass approaches m₁ – the energy loss is then governed by the smaller mass alone. |
| Velocity v₁ before impact | The velocity of the first body immediately before impact, signed with respect to a previously chosen positive direction. Only the difference v₁ − v₂ enters – the closing velocity – and it enters squared: twice the closing velocity means four times the energy loss. |
| 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. Because only the square of the difference enters, the sign of the difference itself does not matter – its magnitude very much does. |
| Coefficient of restitution k | The coefficient of restitution states what fraction of the closing velocity returns as separation velocity, with 0 ≤ k ≤ 1. k = 0 is the perfectly plastic impact with the largest possible energy loss, k = 1 the perfectly elastic impact with none at all. Source: 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 and modelling clay close to 0. Because (1−k²) enters, even k = 0.7 destroys about half the available impact energy. |
m₁ and m₂ are the two masses, for example hammer and workpiece or pile driver and pile. v₁ and v₂ are the velocities immediately before impact, each signed with respect to a previously chosen positive direction; a body at rest has velocity zero. k is the coefficient of restitution between 0 and 1, usually from a rebound test via k = √(h/H) with drop height H and rebound height h.
Define a positive direction and refer both initial velocities to it consistently. Enter masses and the coefficient of restitution; the result is the kinetic energy lost. For the inverse – the coefficient of restitution from a measured energy loss, such as the deformation work of an upsetting test – select k as the target. For context it is worth comparing with the total kinetic energy before impact, ½m₁v₁² + ½m₂v₂²: the loss as a fraction of that shows how much of the energy input actually goes into the process.
m₁ = 2 kg at v₁ = 5 m/s strikes m₂ = 3 kg at v₂ = −1 m/s, i.e. an approaching body. The closing velocity is v₁ − v₂ = 6 m/s and the reduced mass 2·3/(2+3) = 1.2 kg. At k = 0.6 this gives ΔT = ½·(1 − 0.6²)·1.2·6² = ½·0.64·1.2·36 = 13.824 J. The limiting cases with the same numbers: perfectly plastic (k = 0) loses 21.6 J, perfectly elastic (k = 1) exactly zero. The total kinetic energy before impact is ½·2·5² + ½·3·1² = 26.5 J – so at k = 0.6 about 52 per cent of it goes into deformation, heat and sound.
Three dependencies matter. First, the closing velocity enters squared: twice the speed means four times the loss – which is why striking speed matters far more than hammer mass in hammering. Second, the coefficient of restitution appears as (1−k²), i.e. not linearly: even k = 0.7 destroys about half the impact energy, and only close to k = 1 does the loss become genuinely small. Third, the reduced mass caps the loss: it is always smaller than the lesser of the two masses, which is why a light hammer on a heavy workpiece can never convert more energy than its own kinetic energy.
The calculation runs in joules, kilograms, metres and seconds internally. Energy may be entered in J, kJ, MJ, mJ, µJ, N·mm, Wh, kWh or ft·lbf, masses in mg, g, kg, t, oz or lb, and velocities in m/s, km/h, ft/s, mph, m/min or mm/min. Both masses must use the same unit. The coefficient of restitution is dimensionless and entered as a decimal, not a percentage.
Forging and forming hammers, where the lowest possible k maximises the forming work; pile drivers and hammer drills; riveting and upsetting; designing buffers, stops and end dampers where the loss is the intended damping; estimating the deformation energy in vehicle and transport collisions; assessing how much energy an impact guard must absorb.
Valid for the straight central impact, i.e. one-dimensional motion along the common line of centres perpendicular to the plane of contact. The coefficient of restitution is assumed known and constant, although in reality it falls as impact speed rises and is not a material constant. The calculator says how much energy is lost but not where it goes: the split between plastic deformation, heat, sound and vibration depends on material and geometry. The oblique and eccentric impact, rotation and angular momentum, friction in the contact area, the magnitude and time history of the impact force, and stress waves in the material are not covered.
Common mistake: The most common error is inconsistent signs in the initial velocities; since the difference is squared, a sign error does not stand out in the result but distorts it considerably. Second, the energy loss is often confused with the total kinetic energy: only the part associated with the relative motion is lost – the common centre-of-mass motion is always retained, which is why ΔT never reaches the full initial energy. Third, a high loss is reflexively judged as bad; in forming, pile driving and damping it is the goal. And fourth, the formula holds only for the straight central impact – in an oblique or eccentric collision additional energy goes into rotation.
Forging and forming hammers, where the lowest possible k maximises the forming work; pile drivers and hammer drills; riveting and upsetting; designing buffers, stops and end dampers where the loss is the intended damping; estimating the deformation energy in vehicle and transport collisions; assessing how much energy an impact guard must absorb.
m₁ and m₂ are the two masses, for example hammer and workpiece or pile driver and pile. v₁ and v₂ are the velocities immediately before impact, each signed with respect to a previously chosen positive direction; a body at rest has velocity zero. k is the coefficient of restitution between 0 and 1, usually from a rebound test via k = √(h/H) with drop height H and rebound height h.
Valid for the straight central impact, i.e. one-dimensional motion along the common line of centres perpendicular to the plane of contact. The coefficient of restitution is assumed known and constant, although in reality it falls as impact speed rises and is not a material constant. The calculator says how much energy is lost but not where it goes: the split between plastic deformation, heat, sound and vibration depends on material and geometry. The oblique and eccentric impact, rotation and angular momentum, friction in the contact area, the magnitude and time history of the impact force, and stress waves in the material are not covered.