A Level Maths 9709 · Paper 4 Mechanics · 4.3 · ~4 min read
Momentum
Momentum as mass × velocity, the principle of conservation of momentum for direct collisions in one dimension, particles that coalesce, and deciding whether further collisions happen.
MomentumConservation of momentumDirect collisionsCoalescing particlesSigns & directionsLoss of kinetic energyFurther collisions
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end
In 30 seconds
- Momentum = mv, in N s or kg m s⁻¹. It is a vector, so direction matters: pick a positive direction.
- Conservation: in a collision, total momentum before = total momentum after (no external forces along the line of motion).
- Coalescing particles stick together and move with one common velocity.
- Further collisions happen if a particle behind is moving faster (in the same direction) than the one in front.
By the end you can
- Use momentum as mass × velocity, taking account of direction.
- Use conservation of momentum for direct impacts in one dimension, including coalescing particles.
- Decide whether further collisions take place.
1Conservation of momentum
Before = after
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
velocities with signs; one positive direction for all- Draw before-and-after diagrams with arrows and masses. Mark unknown velocities in the positive direction; a negative answer means it moves the other way.
Worked example
A (3 kg) moves at 4 m s⁻¹ towards B (2 kg), which moves at 1 m s⁻¹ towards A. They coalesce. Find their common velocity.
- positive in A's direction: 3 × 4 + 2 × (−1) = 5v M1 A1
- v = 2 m s⁻¹ in A's original direction A1
Worked example
P (0.5 kg) moving at 6 m s⁻¹ hits Q (0.3 kg) at rest. After the collision P moves at 2 m s⁻¹ in the same direction. Find Q's speed and the loss of kinetic energy.
- 0.5 × 6 = 0.5 × 2 + 0.3v ⇒ v = 20/3 ≈ 6.67 m s⁻¹ M1 A1
- KE before = ½ × 0.5 × 36 = 9 J; after = 1 + ½ × 0.3 × (20/3)² = 7.67 J M1
- loss = 1.33 J A1
Common mistakeSigns dropped: a particle moving left must enter as −u. The 2024 Paper 4 report: most errors in momentum were sign errors.
2Further collisions
Chains of impacts
- Compare velocities after each impact. If the one behind moves faster in the same direction, they collide again.
Worked example
P (2 kg) moving at 5 m s⁻¹ hits Q (1 kg) at rest; afterwards P moves at 2 m s⁻¹ in the same direction. Q then hits R (3 kg) at rest, and Q and R coalesce. Will P collide with the combined particle?
- first impact: 10 = 4 + vQ ⇒ vQ = 6 m s⁻¹ M1 A1
- second impact: 1 × 6 = 4v ⇒ v = 1.5 m s⁻¹ M1 A1
- P moves at 2 > 1.5 in the same direction, so yes, P catches up and collides again A1
Common mistakeAssuming kinetic energy is conserved. Momentum is conserved in every collision; kinetic energy usually isn't.
Linking questionChanges in kinetic energy link to Energy, work and power; momentum, impulse and elastic collisions in Physics are in Dynamics.
Test yourself — cover the answers first
Quick check
- Find the momentum of a 0.2 kg ball moving at 15 m s⁻¹.
- Two particles coalesce. How many velocities are there after?
- A 2 kg particle at 3 m s⁻¹ hits a 1 kg particle at rest and they coalesce. Find v.
- Is kinetic energy conserved when particles coalesce?
- A velocity comes out negative. What does it mean?
Exam-style
- A truck of mass 4000 kg moving at 3 m s⁻¹ couples to a stationary truck of mass m kg; they move off at 2 m s⁻¹. Find m.
- A (4 kg, 5 m s⁻¹) and B (2 kg, −4 m s⁻¹) collide; A's velocity after is −1 m s⁻¹. Find B's velocity.
- For the collision above, find the loss of kinetic energy.
Show answers
- 3 N s
- one common velocity
- 2 m s⁻¹
- no, some is lost
- it moves in the opposite direction to the one chosen as positive
- 12000 = 2(4000 + m) ⇒ m = 2000
- 20 − 8 = −4 + 2v ⇒ v = 8 m s⁻¹
- before 50 + 16 = 66 J; after 2 + 64 = 66 J; no loss (perfectly elastic)
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