Momentum: Revision Notes

Condensed for the final weeks. For the full explanation, use the Momentum study guide.

Definitions and equations

momentum   p = m v          kg m/s   -- a VECTOR
impulse    F t = m v - m u  N s      -- change in momentum
force      F = (m v - m u) / t       -- Newton's second law, full form

Momentum is a vector: direction matters. Assign one direction positive and stick to it throughout — the commonest source of error in this topic.

Conservation of momentum

In a closed system with no external forces, total momentum before = total momentum after.

The principle assumes no significant external force during the event — it cannot be applied, for example, to a trolley slowed by friction over a long time, since friction is an external force acting throughout.

m1 u1  +  m2 u2   =   m1 v1  +  m2 v2

Applies to all collisions and explosions, elastic or not.

The two collision types

	Elastic	Inelastic
Momentum	Conserved	Conserved
Kinetic energy	Conserved	Not conserved
Objects	Bounce apart	Often stick together

Momentum is conserved in both. Only kinetic energy distinguishes them — students routinely claim momentum is lost in an inelastic collision, which is wrong.

Explosions

Total momentum before = zero (both at rest), so afterwards the momenta must be equal and opposite:

0 = m1 v1 + m2 v2      ->    m1 v1 = -m2 v2

This is recoil: a rifle and bullet, a rocket and exhaust gases, two skaters pushing apart.

Worked example. A 5 kg object at rest explodes into two parts. A 2 kg piece moves off at 6 m/s. Find the velocity of the remaining 3 kg piece.

momentum before = 0
momentum after = (2 x 6) + (3 x v) = 0
12 + 3v = 0
v = -4 m/s   (4 m/s in the opposite direction to the 2 kg piece)

The negative sign is doing real work: the two pieces must move in opposite directions for the total to remain zero.

Worked examples

Collision, sticking together: A 2 kg trolley at 3 m/s hits a stationary 4 kg trolley and they move off together.

before:  (2 x 3) + (4 x 0) = 6 kg m/s
after:   (2 + 4) x v = 6v

6v = 6   ->   v = 1 m/s

Impulse: A 0.15 kg ball hits a wall at 12 m/s and rebounds at 8 m/s. Find the change in momentum.

Take towards the wall as positive:
initial = 0.15 x 12  = +1.8
final   = 0.15 x -8  = -1.2

change = -1.2 - 1.8 = -3.0 kg m/s   (magnitude 3.0 N s)

Note the sign change — forgetting it gives 0.6 instead of 3.0, and is the classic trap.

Safety applications

Crumple zones, airbags, seatbelts, crash mats and gloves all increase the time over which momentum changes. Since F = Δp / t, a longer time means a smaller force, and therefore less injury.

F = Δp/Δt is equivalent to F = ma when mass is constant, since Δp = Δ(mv) = mΔv, giving F = mΔv/Δt = ma — the same law, just written in a form that also applies where mass itself changes.

Exam traps

• Assign and keep a positive direction; a rebound reverses the sign.
• Momentum is conserved in inelastic collisions.
• Units: momentum kg m/s, impulse N s — numerically the same.
• “Objects stick together” means they share a common final velocity.
• In an explosion, total momentum before is zero, not “each object has zero momentum after”.
• Forgetting the negative sign shows the two exploding pieces move in opposite directions.
• Applying conservation of momentum where a significant external force (such as friction over a long time) acts during the event.

Self-test

1. State the principle of conservation of momentum.
2. Is momentum conserved in an inelastic collision?
3. A 1 kg ball at 5 m/s hits a stationary 4 kg ball; they move off together. Find their speed.
4. Why do airbags reduce injury?
5. A 0.2 kg ball hits a wall at 10 m/s and rebounds at 6 m/s. Find the impulse.
6. A 5 kg object at rest explodes into a 2 kg piece moving at 6 m/s and a 3 kg piece. Find the 3 kg piece’s velocity.
7. Show that F = Δp/Δt is equivalent to F = ma for constant mass.

Answers: 1. In a closed system with no external forces, total momentum before a collision equals total momentum after. 2. Yes — momentum is always conserved; only kinetic energy is not. 3. (1×5) = 5 = (1+4)v → v = 1 m/s. 4. They increase the time over which the passenger’s momentum changes, so by F = Δp/t the force on the body is reduced. 5. Taking towards the wall as positive: (0.2×−6) − (0.2×10) = −1.2 − 2.0 = −3.2 kg m/s, magnitude 3.2 N s. 6. Momentum before = 0, so (2×6) + (3×v) = 0 → 12 + 3v = 0 → v = −4 m/s (4 m/s opposite to the 2 kg piece). 7. Δp = Δ(mv) = mΔv for constant m, so F = Δp/Δt = mΔv/Δt = ma.