PHYSICS · LESSON 07 OF 9

Impulse and momentum

Momentum p = mv is a vector. A force acting for a time changes it (impulse). In a collision with no external force, total momentum is conserved.

What this lesson explains

Momentum is the natural quantity for collisions, impacts and explosions — crash safety, pile drivers, jet and rocket propulsion, the recoil of a gun. During a collision the forces are huge, brief and hard to measure, but momentum lets you predict the outcome without knowing them. Impulse explains why airbags, crumple zones and bent knees reduce injuries.

Before you begin

Signs in one dimension

Choose a positive direction and give every velocity a sign. A rebound reverses the sign of velocity.

Newton’s third law

Forces between two objects are equal in size and opposite in direction, and act on different objects.

The idea, made visible

Momentum p = mv combines mass and velocity. It is a vector, so direction matters: with right positive, a 2 kg ball at 3 m/s to the left has p = −6 kg·m/s.

Newton’s second law can be written ΣF = Δp / Δt. Rearranged, impulse J = FΔt = Δp: a force acting for a time changes momentum. To stop a moving object you need a definite impulse (its momentum). You can deliver it with a large force for a short time or a small force for a long time. Airbags and crumple zones lengthen the stopping time, so the force on the passenger is smaller.

Momentum before (6 kg·m/s) equals momentum after (6 kg·m/s). Kinetic energy is not conserved: it drops from 9 J to 6 J.

In a collision between two objects, each pushes the other with equal and opposite forces for the same time (Newton’s third law). So the impulses are equal and opposite and the momentum lost by one is gained by the other. If no external net force acts on the system, the total momentum is conserved: Σp_before = Σp_after. This is true for every collision, bouncy or sticky.

Kinetic energy is a different story. In an elastic collision kinetic energy is also conserved (ideal billiard balls). In an inelastic collision some kinetic energy becomes heat, sound and deformation. In a perfectly inelastic collision the objects stick together, and the loss of kinetic energy is as large as momentum conservation allows.

The centre of mass of a system moves as if all the mass were there and all external forces acted there. With no external force, the centre of mass moves at constant velocity, even while the parts collide or fly apart.

Momentum and kinetic energy in the collision shown
	momentum (kg·m/s)	kinetic energy (J)
before	2 × 3 + 1 × 0 = 6	½ × 2 × 3² = 9
after	3 × 2 = 6	½ × 3 × 2² = 6
change	0 (conserved)	−3 (lost to heat, sound, deformation)

Always conserve momentum in collisions; conserve kinetic energy only if the collision is stated to be elastic.

Key terms

Momentum p: p = mv, a vector in the direction of the velocity. Unit kg·m/s.
Impulse J: J = F_avgΔt = Δp; equals the area under a force–time graph. Unit N·s (= kg·m/s).
Isolated system: A system on which the net external force is zero (or negligible during a brief collision).
Elastic collision: Both momentum and kinetic energy are conserved.
Inelastic collision: Momentum is conserved; kinetic energy is not. Perfectly inelastic: the objects stick together.
Centre of mass: x_cm = m₁x₁ + m₂x₂ + … / m₁ + m₂ + …, the mass-weighted average position.

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
p	momentum	kg·m/s
J	impulse	N·s
F_avg	average force during contact	N
Δt	contact time	s
v_i, v_f	velocities before and after (signed)	m/s

Impulse–momentum theorem

J = F_avgΔt = Δp = mv_f − mv_i

Conditions and limits: Always; F_avg is the time-averaged net force. Use signed velocities.

Conservation of momentum

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

Conditions and limits: No net external force on the system during the interaction (or the collision is so brief that external impulses are negligible). In 2D, conserve each component separately.

Perfectly inelastic

v_f = m₁v_1i + m₂v_2i / m₁ + m₂

Conditions and limits: Objects stick together after the collision.

Elastic, target at rest (1D)

v_1f = m₁ − m₂ / m₁ + m₂v_1i; v_2f = 2m₁ / m₁ + m₂v_1i

Conditions and limits: Head-on elastic collision with m₂ initially at rest.

Why it works: Why momentum is conserved in a collision

Two carts collide. During contact, cart 1 pushes cart 2 with force F and cart 2 pushes cart 1 with −F (third law), for the same time Δt.

Impulse on cart 2: Δp₂ = FΔt.
Impulse–momentum theorem.
Impulse on cart 1: Δp₁ = −FΔt.
Equal and opposite force, same time.
Δp₁ + Δp₂ = 0.
Add them.

The total momentum does not change. External forces (friction with the track) would spoil this, but during a brief collision their impulse is usually negligible.

A first worked example

Momentum problems

Define the system (usually all colliding objects) and check that external forces are zero or negligible during the interaction.
Choose a positive direction; write every velocity with its sign.
Write total momentum before = total momentum after (component by component in 2D).
If the collision is elastic, add conservation of kinetic energy; if the objects stick, use one final velocity.
For forces during impact, use impulse: F_avg = Δp/Δt.
Check signs (does a rebound come out negative?) and compare kinetic energies before and after (KE cannot increase unless stored energy is released).

Impulse and a change of direction

Problem. A 2 kg body changes velocity from +5 m/s to +2 m/s. Find the impulse. Then find the impulse if instead it rebounds at −2 m/s.

J = m(v_f − v_i) = 2(2 − 5) = −6 N·s.
Slowing down in the positive direction: negative impulse.
Rebound: J = 2(−2 − 5) = −14 N·s.
The velocity change is 7 m/s, not 3 m/s.

Result: −6 N·s; with a rebound, −14 N·s.

What it means: Bouncing requires a larger impulse than stopping: the wall must first stop the body and then push it back.

A different case

Average force in an impact

Problem. A 0.15 kg ball moving at 40 m/s is caught and stopped in 0.02 s. Find the average force. What if the hand moves back so that stopping takes 0.10 s?

Δp = 0.15(0 − 40) = −6.0 kg·m/s.
Change in momentum.
F_avg = Δp / Δt = −6.0 / 0.02 = −300 N.
Impulse–momentum theorem.
With Δt = 0.10 s: F_avg = −60 N.
Same impulse over five times the time.

Result: 300 N, reduced to 60 N by extending the stopping time.

What it means: This is the principle of airbags, crumple zones and bending your knees on landing.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Perfectly inelastic collision

Problem. A 2 kg cart at 3 m/s hits a 1 kg cart at rest and they stick together. Find their common speed and the kinetic energy lost.

2(3) + 1(0) = (2 + 1)v_f.
Conservation of momentum.
v_f = 2 m/s.
Solve.
K before 9 J; K after 1 / 2(3)(2)² = 6 J. Lost: 3 J.
Compare kinetic energies.

Result: v_f = 2 m/s; 3 J of kinetic energy lost.

What it means: Momentum is conserved while one-third of the kinetic energy is not.

Elastic collision of equal masses

Problem. A 1 kg ball at 4 m/s hits an identical ball at rest, head-on and elastically. Find both final velocities.

m₁ = m₂, so v_1f = 0 / 2 × 4 = 0 and v_2f = 2 / 2 × 4 = 4 m/s.
Elastic formulas with the target at rest.
Check momentum: 4 = 0 + 4 ✓. Kinetic energy: 8 J = 0 + 8 J ✓.
Both conserved.

Result: The first ball stops; the second moves off at 4 m/s.

What it means: This is what you see in a Newton’s cradle.

Recoil (an “explosion”)

Problem. A 60 kg skater at rest throws a 3 kg ball forward at 10 m/s. Find the skater’s recoil velocity.

Total momentum before = 0.
Everything starts at rest.
0 = 3(10) + 60v ⇒ v = −0.5 m/s.
Momentum after must also total zero.

Result: The skater moves backwards at 0.5 m/s.

What it means: Kinetic energy increased (from 0 to 157.5 J), supplied by the skater’s muscles. Momentum is still conserved.

Locating the centre of mass

Problem. A 2 kg mass is at x = 0 and a 1 kg mass is at x = 6 m. Find the centre of mass.

x_cm = m₁x₁ + m₂x₂ / m₁ + m₂ = 2(0) + 1(6) / 2 + 1.
Weight each position by its mass.
x_cm = 6 / 3 = 2 m.
Divide by the total mass.

Result: x_cm = 2 m, one third of the way from the heavier mass.

What it means: The centre of mass lies closer to the larger mass; a plain average of the positions (3 m) is correct only for equal masses. In the recoil example, the centre of mass of skater and ball stays at rest.

Common misunderstandings

Misunderstanding: Ignoring the signs of velocities.
Correct idea: Momentum is a vector. Opposite directions need opposite signs.
Misunderstanding: Conserving kinetic energy in every collision.
Correct idea: Only in elastic collisions. Momentum is conserved in all isolated collisions.
Misunderstanding: Using conservation of momentum when an external force acts over a long time.
Correct idea: Check the system is isolated, or the interaction brief.
Misunderstanding: Treating impulse as a force.
Correct idea: Impulse is force × time, with units N·s.

Keep in mind

J = F_avgΔt = Δp = mv_f − mv_iImpulse–momentum theorem
m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2fConservation of momentum
v_f = m₁v_1i + m₂v_2i / m₁ + m₂Perfectly inelastic
v_1f = m₁ − m₂ / m₁ + m₂v_1i; v_2f = 2m₁ / m₁ + m₂v_1iElastic, target at rest (1D)

Scope of this lesson

Collisions in one dimension are treated fully; two-dimensional collisions use the same idea with components.
Variable-mass systems (rockets) are beyond this topic.

Next: Circular motion. An object moving in a circle at constant speed is accelerating towards the centre, a = v²/r. Some real force (tension, friction, gravity, normal force) must supply the net inward force mv²/r.

Original study text. Sources and credits.

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