Impulse, Momentum & Collision Notes

Last reviewed 16 Sept 2026 · 5 min read

Momentum and impulse

Linear momentum $p = m v$ (kg·m/s) — a vector along the velocity.

Impulse of a force $= \int F d t$ ; for a constant force, $F t$ . Unit N·s (= kg·m/s).

FormulaImpulse–momentum principle

\int_{t_{1}}^{t_{2}} F d t = m v_{2} - m v_{1}

The impulse of the resultant force equals the change in momentum. Average force during an impact $= \frac{Δ ( m v )}{Δ t}$ .

An impulsive force is very large and acts for a very short time (hammer blow, pile driving, collision). Its impulse is finite even though the force itself is hard to measure.

Exam TipWhy helmets and crumple zones work

For a given change of momentum, stretching the impact time reduces the average force: $F_{a v g} = Δ p /Δ t$ . Sand beds for long jumpers, airbags and fender piles on jetties use the same idea.

Conservation of linear momentum

If the net external impulse on a system is zero, its total momentum is conserved:

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

Applications: collisions, explosions, recoil.

Recoil of a gun

A gun of mass $M$ fires a bullet of mass $m$ at velocity $v$ . Initially at rest:

M V = m v \Rightarrow V = \frac{m v}{M}

The kinetic energy is shared inversely as the masses: the bullet gets far more energy than the gun.

Collision of elastic bodies

When two bodies collide:

Period of compression — they deform until they move with a common velocity.
Period of restitution — they partly or fully recover their shape and separate.

Direct (central) impact: velocities along the line joining centres. Oblique impact: velocities inclined to that line.

Newton's law of restitution

FormulaCoefficient of restitution

e = \frac{velocity of separation}{velocity of approach} = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}

Perfectly elastic impact: $e = 1$ (no loss of kinetic energy).
Perfectly plastic (inelastic) impact: $e = 0$ (bodies move together).
Real impacts: $0 < e < 1$ (e.g. glass on glass ≈ 0.95, steel ≈ 0.6, lead ≈ 0.2).

Velocities after direct impact

Combining momentum conservation with restitution:

v_{1} = \frac{m _{1} u _{1} + m _{2} u _{2} - m _{2} e ( u _{1} - u _{2} )}{m _{1} + m _{2}}, v_{2} = \frac{m _{1} u _{1} + m _{2} u _{2} + m _{1} e ( u _{1} - u _{2} )}{m _{1} + m _{2}}

Special results (perfectly elastic, $e = 1$ , equal masses): the bodies exchange velocities.

Loss of kinetic energy

Δ K E = \frac{m _{1} m _{2}}{2 ( m _{1} + m _{2} )} (u_{1} - u_{2})^{2} (1 - e^{2})

Zero for $e = 1$ ; maximum for $e = 0$ .

Ball dropped on a floor

Dropped from height $h$ onto a fixed floor: velocity of approach $2 g h$ ; rebound velocity $e 2 g h$ ; rebound height $h_{1} = e^{2} h$ . After $n$ bounces $h_{n} = e^{2 n} h$ . Total distance travelled before coming to rest $= h \frac{1 + e ^{2}}{1 - e ^{2}}$ .

Oblique impact of smooth spheres

Components along the line of centres obey momentum conservation and restitution.
Components perpendicular to it are unchanged for each sphere (smooth surfaces transmit no tangential impulse).

A ball striking a smooth fixed surface at angle $α$ to the normal rebounds at $β$ with $tan β = \frac{tan α}{e}$ .

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Momentum and impulse

Conservation of linear momentum

Recoil of a gun

Collision of elastic bodies

Newton's law of restitution

Velocities after direct impact

Loss of kinetic energy

Ball dropped on a floor

Oblique impact of smooth spheres

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