9231 Further Mathematics — Paper 3: Mechanics
m₁ m₂ u₁ u₂ line of centres Newton's Law: e = sep/approach Impulse: J = m(v − u)
9231 Further MathsPaper 3: Mechanics › Lesson 06

Impulse & Momentum

Conservation of momentum, Newton's experimental law, coefficient of restitution, direct and oblique impact — the physics of collisions.

🎯 Cambridge 9231 P3 📐 Syllabus Section 3.6 ⭐ Oblique Impact Key ⏱ ~100 min
06Lesson

1. Impulse and Momentum

The linear momentum of a particle is the product of its mass and velocity. It is a vector quantity.

Momentum p= mvkg m/s (vector)
Impulse J= Ft = m(v − u) = change in momentumN s
Impulse-Mom.J = Δpimpulse = change in momentum
💡 Impulse from a Variable Force

When force varies with time, impulse = ∫F dt. For a constant force over time t: impulse = Ft. Either way, impulse = change in momentum.

Conservation of Linear Momentum

When no external forces act on a system, the total linear momentum is conserved. This is the most important principle in collision problems.

⭐ Conservation of Momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Always holds — regardless of the type of collision (elastic, inelastic, or perfectly inelastic).

2. Newton's Experimental Law — Coefficient of Restitution

Newton's experimental law states that the relative speed of separation after impact equals e times the relative speed of approach before impact, where e is the coefficient of restitution.

⭐ Newton's Law of Restitution
e = (speed of separation) / (speed of approach)
= (v₂ − v₁) / (u₁ − u₂)   [along line of centres]

Always measured along the line of centres (the line joining the centres of the two spheres at impact).

The Coefficient of Restitution e

0 0.5 1
e=0: Perfectly inelastic
(bodies coalesce)
0<e<1: Inelastic
(KE lost)
e=1: Perfectly elastic
(KE conserved)
⚠ Sign Convention — Critical

Always define a positive direction before writing equations. Velocities in the opposite direction are negative. The restitution equation uses:

Approach speed= u₁ − u₂ (must be positive for collision to occur)
Separation speed= v₂ − v₁ (must be positive after collision)

3. Direct Impact — Two Spheres

A direct impact (head-on collision) occurs when the velocities of both particles are along the line of centres. We have two unknowns (v₁ and v₂) and two equations.

BEFORE A B u₁→ ←u₂ AFTER v₁ v₂

Direct collision: velocities along line of centres.

Two equations, two unknowns:

Eq 1 — Conservation of momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Eq 2 — Newton's law of restitution
v₂ − v₁ = e(u₁ − u₂)

Solve simultaneously. Always check: 0 ≤ e ≤ 1 and no sphere passes through the other (v₁ ≤ v₂ if A was overtaking B).

4. Impact with a Fixed Surface

When a sphere hits a fixed smooth surface, the component of velocity perpendicular to the surface is reversed and scaled by e. The component parallel to the surface is unchanged (smooth surface).

⭐ Sphere Hitting a Fixed Surface
Perpendicularv_⊥ = e × u_⊥ (reversed in direction)
Parallelv_∥ = u_∥ (unchanged)

For a direct impact on a wall (velocity perpendicular to wall): v = eu. The wall exerts an impulse J = m(v + u) = m(e+1)u on the sphere.

5. Oblique Impact — The Key 9231 Topic

In an oblique impact, the initial velocities are not along the line of centres. This is the most frequently examined and most challenging part of the momentum topic.

🔑 Method for Oblique Impact of Two Smooth Spheres
  • Step 1 — Set up axes: Take the x-axis along the line of centres, y-axis perpendicular to it.
  • Step 2 — Resolve velocities of each sphere into x and y components before impact.
  • Step 3 — Along line of centres (x): Apply conservation of momentum AND Newton's restitution law → find v₁ₓ and v₂ₓ.
  • Step 4 — Perpendicular to line of centres (y): Smooth spheres exert no force in this direction → each sphere's y-component is unchanged.
  • Step 5 — Reconstruct final velocity vectors and find speed/angle if required.
x (line of centres) y A u₁ α u₁cosα u₁sinα B After: v₂ v₁ stays same y-comp

Oblique impact: resolve into x (line of centres) and y (perpendicular). Only x-components change.

6. Energy Loss in Collisions

Kinetic energy is conserved only in a perfectly elastic collision (e = 1). For all other collisions (0 ≤ e < 1), KE is lost as heat, sound, and deformation.

Kinetic energy lost in direct impact
ΔKE= ½m₁u₁² + ½m₂u₂² − (½m₁v₁² + ½m₂v₂²)
ΔKE= [m₁m₂(1−e²)(u₁−u₂)²] / [2(m₁+m₂)]standard result
✓ Key Result

When e = 1, ΔKE = 0 (elastic). When e = 0, ΔKE is maximum (perfectly inelastic — maximum energy loss). Energy lost ∝ (1 − e²) × (approach speed)².

Example 1 Direct Impact — Finding Post-Collision Velocities ★☆☆ Standard
A sphere A of mass 3 kg moving at 8 m/s collides directly with a sphere B of mass 2 kg moving at 2 m/s in the same direction. The coefficient of restitution is 0.5. Find the velocities after impact and the loss in kinetic energy.
1
Set up — positive direction →, u₁=8, u₂=2
CLM:3(8) + 2(2) = 3v₁ + 2v₂ → 3v₁ + 2v₂ = 28
NEL:v₂ − v₁ = e(u₁ − u₂) = 0.5(8−2) = 3
2
Solve simultaneously
From NEL:v₂ = v₁ + 3
Sub into CLM:3v₁ + 2(v₁+3) = 28 → 5v₁ = 22 → v₁ = 4.4 m/s
v₂= 4.4 + 3 = 7.4 m/s
v₁= 4.4 m/s

Check: v₂ > v₁ ✓ (B faster than A — they don't overlap)

3
KE loss
KE before= ½(3)(64) + ½(2)(4) = 96 + 4 = 100 J
KE after= ½(3)(19.36) + ½(2)(54.76) = 29.04 + 54.76 = 83.8 J
ΔKE= 100 − 83.8 = 16.2 J
Example 2 Sphere Bouncing Off a Fixed Wall ★☆☆ Standard
A ball of mass 0.3 kg hits a smooth fixed vertical wall at speed 10 m/s, making an angle of 30° with the wall. The coefficient of restitution with the wall is 0.6. Find the speed and direction of the ball after impact.
1
Resolve into components (normal to wall = perpendicular, along wall = parallel)

Angle to wall = 30°, so angle to normal = 60°.

u_⊥= 10 sin30° = 5 m/s (towards wall)
u_∥= 10 cos30° = 5√3 m/s (along wall)
2
Apply restitution and smooth wall conditions
v_⊥= e × u_⊥ = 0.6 × 5 = 3 m/s (away from wall)
v_∥= u_∥ = 5√3 m/s (unchanged)
3
Reconstruct final velocity
Speed= √(3² + (5√3)²) = √(9 + 75) = √84 = 2√21 ≈ 9.17 m/s
Angle to wall= arctan(v_⊥/v_∥) = arctan(3/5√3) = arctan(√3/5) = 19.1°

The angle to the wall decreases from 30° to 19.1° because the normal component is reduced by e while the tangential component is unchanged.

Example 3 Oblique Impact of Two Smooth Spheres ★★★ A* Level
Two smooth spheres A (mass 2 kg) and B (mass 3 kg) of equal radius collide. Before impact, A moves at 6 m/s at 30° to the line of centres, and B moves at 4 m/s in the opposite direction along the line of centres. The coefficient of restitution is 0.5. Find the velocities of A and B after impact.
1
Resolve initial velocities (x = line of centres →, y = perpendicular)
A: u_Ax= 6 cos30° = 3√3 m/s
A: u_Ay= 6 sin30° = 3 m/s
B: u_Bx= −4 m/s (opposite to positive x)
B: u_By= 0
2
Along line of centres (x): CLM + NEL
CLM(x):2(3√3) + 3(−4) = 2v_Ax + 3v_Bx
6√3 − 12 = 2v_Ax + 3v_Bx   … (i)
NEL(x):v_Bx − v_Ax = 0.5(3√3 − (−4)) = 0.5(3√3 + 4)   … (ii)
3
Solve (i) and (ii)
3√3 ≈ 5.196→ 6√3 ≈ 10.39
From (i):2v_Ax + 3v_Bx = 10.39 − 12 = −1.61
From (ii):v_Bx − v_Ax = 0.5(5.196 + 4) = 4.598
Solving:v_Ax = (−1.61 − 3×4.598)/5 = (−1.61−13.79)/5 = −3.08 m/s
v_Bx = −3.08 + 4.598 = 1.52 m/s
4
Perpendicular components (y) — unchanged
v_Ay= u_Ay = 3 m/s
v_By= u_By = 0 m/s
Final speeds
Speed of A= √(3.08² + 3²) = √(9.49 + 9) = √18.49 = 4.30 m/s
Speed of B= √(1.52² + 0²) = 1.52 m/s
Example 4 Successive Impacts — Finding e ★★☆ Challenging
A ball is dropped from rest at height H above a hard floor. It bounces and reaches height h₁ after the first bounce, then h₂ after the second. Show that e² = h₁/H and e⁴ = h₂/H, and find e if H = 1.6 m, h₁ = 0.9 m.
1
Speed just before first impact
u₁= √(2gH)from free fall
2
Speed just after first bounce and height reached
v₁= e × u₁ = e√(2gH)restitution with floor
h₁= v₁²/(2g) = e²(2gH)/(2g) = e²H → e² = h₁/H ∎
3
Second bounce
u₂ = √(2gh₁)= e√(2gH)
v₂= e × u₂ = e²√(2gH)
h₂= v₂²/(2g) = e⁴H → e⁴ = h₂/H ∎
e= √(h₁/H) = √(0.9/1.6) = √(0.5625) = 0.75

Interactive Collision Laboratory

Set masses, initial velocities and the coefficient of restitution, then watch the direct collision play out. See post-collision velocities and energy loss computed in real time.

Direct Collision Simulator
3.0 kg
8.0 m/s →
2.0 kg
2.0 m/s →
0.50
v₁ after (m/s)
v₂ after (m/s)
KE loss (J)
Collision Type

Practice Questions

Question 1 — Direct Collision
Sphere A (mass 4 kg, velocity 6 m/s) collides directly with sphere B (mass 2 kg, velocity 1 m/s, same direction). The coefficient of restitution is 0.4. Find the velocities after impact and verify that A does not pass through B.
CLM: 4(6)+2(1) = 4v₁+2v₂. NEL: v₂−v₁ = 0.4(6−1). Solve simultaneously. Then check v₁ ≤ v₂.
✓ Solution
CLM:26 = 4v₁ + 2v₂ → 2v₁ + v₂ = 13 ... (i)
NEL:v₂ − v₁ = 0.4(5) = 2 → v₂ = v₁ + 2 ... (ii)
Sub (ii)→(i):2v₁ + v₁ + 2 = 13 → 3v₁ = 11 → v₁ = 11/3 ≈ 3.67 m/s
v₂= 11/3 + 2 = 17/3 ≈ 5.67 m/s
Check:v₂ = 5.67 > v₁ = 3.67 ✓ (A no longer catching B)
Question 2 — Ball and Wall
A ball hits a smooth fixed wall at speed 8 m/s, at angle 45° to the wall. The coefficient of restitution is 0.7. Find (a) the speed after impact, (b) the angle it makes with the wall after impact, (c) the impulse exerted by the wall on the ball (mass 0.2 kg).
Perpendicular to wall: v_⊥ = e × u_⊥ = 0.7 × 8sin45°. Parallel: v_∥ = 8cos45°. Speed = √(v_⊥² + v_∥²). Angle to wall = arctan(v_⊥/v_∥). Impulse = m(v_⊥ + u_⊥) in perpendicular direction.
✓ Solution
u_⊥= 8sin45° = 4√2, u_∥ = 8cos45° = 4√2
v_⊥= 0.7 × 4√2 = 2.8√2
v_∥= 4√2 (unchanged)
(a) Speed= √((2.8√2)² + (4√2)²) = √(15.68+32) = √47.68 = 6.91 m/s
(b) Angle= arctan(2.8√2/4√2) = arctan(0.7) = 34.99° ≈ 35.0°
(c) Impulse= 0.2(v_⊥ + u_⊥) = 0.2(2.8√2 + 4√2) = 0.2×6.8√2 = 1.36√2 ≈ 1.92 N s
Question 3 — Conservation + Restitution (find e)
A particle A (mass 3 kg, speed 5 m/s) collides directly with a stationary particle B (mass 2 kg). After impact, A has speed 1 m/s in the same direction. Find (a) the speed of B after impact, (b) the coefficient of restitution, (c) the kinetic energy lost.
Use CLM to find v_B. Then e = (v_B − v_A)/(u_A − u_B) = (v_B − 1)/(5 − 0). KE lost = KE before − KE after.
✓ Solution
CLM:3(5) + 0 = 3(1) + 2v_B → 2v_B = 12 → v_B = 6 m/s
(b) e= (6−1)/(5−0) = 5/5 = 1.0
(c) KE before= ½(3)(25) = 37.5 J
KE after= ½(3)(1) + ½(2)(36) = 1.5 + 36 = 37.5 J
ΔKE= 0 J — perfectly elastic!
Question 4 — Oblique Impact (A* Level)
A smooth sphere A (mass m) moves with speed u at angle 60° to the line of centres when it strikes a smooth stationary sphere B (also mass m). The coefficient of restitution is e. Show that the speed of A after impact is u√((1 + 3e²)/4) and find the direction of A's motion relative to the line of centres.
u_Ax = u cos60° = u/2, u_Ay = u sin60° = u√3/2. B is stationary. CLM(x): m(u/2) = mv_Ax + mv_Bx. NEL: v_Bx − v_Ax = e(u/2). Solve for v_Ax and v_Bx. v_Ay = u_Ay = u√3/2 (unchanged). Then |v_A| = √(v_Ax² + v_Ay²).
✓ Solution
u_Ax = u/2u_Ay = u√3/2, u_Bx = u_By = 0
CLM(x):m(u/2) = mv_Ax + mv_Bx → v_Ax + v_Bx = u/2 ... (i)
NEL(x):v_Bx − v_Ax = e(u/2) ... (ii)
Add (i)+(ii):2v_Bx = u/2(1+e) → v_Bx = u(1+e)/4
v_Ax= u/2 − v_Bx = u/2 − u(1+e)/4 = u(1−e)/4
v_Ay= u√3/2 (unchanged)
|v_A|²= [u(1−e)/4]² + [u√3/2]² = u²(1−e)²/16 + 3u²/4
= u²[(1−e)² + 12]/16 = u²[1−2e+e²+12]/16 = u²(13−2e+e²)/16

Hmm — let's recheck. Using (1+3e²)/4: |v_A|² = u²(1+3e²)/4 requires v_Ax = u(1−e)/4, v_Ay = u√3/2:

|v_A|²= u²(1−e)²/16 + 3u²/4 = u²[(1−e)²+12]/16
For claim to hold:[(1−e)²+12]/16 = (1+3e²)/4
→ (1−e)²+12= 4(1+3e²) = 4+12e²
→ 1−2e+e²+12= 4+12e² → 9=2e+11e²

This identity only holds for specific e. The correct expression for speed of A is u√[(13−2e+e²)/16]. Direction: tan θ = v_Ay/v_Ax = (u√3/2)/[u(1−e)/4] = 2√3/(1−e) from the line of centres.

Formula Reference Sheet

Complete reference for Momentum — Cambridge 9231 P3, Section 3.6.

Momentum and Impulse
p = mvmomentum (vector)
J = Ft = m(v−u)impulse
J = Δpimpulse-momentum theorem
Conservation of Momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Always conservedall collision types
Newton's Law of Restitution
e = sep / approach
v₂−v₁ = e(u₁−u₂)along line of centres
0 ≤ e ≤ 1
e=0: coalescee=1: elastic
Impact with Fixed Surface
v_⊥ = e × u_⊥reversed
v_∥ = u_∥unchanged (smooth)
J = m(1+e)u_⊥impulse from wall
Oblique Impact — Method
x-axis: line of centres
x: CLM + NEL → v_x
y: v_y = u_y (unchanged)
Reconstruct speed + angle
KE Loss in Collision
ΔKE = KE_before − KE_after
= m₁m₂(1−e²)(u₁−u₂)²/ 2(m₁+m₂)
e=1: ΔKE=0 (elastic)
e=0: ΔKE maximum
📋 Cambridge Exam Strategy — Momentum
  • Always define positive direction at the start and mark it on your diagram. All velocities must be consistent with this choice.
  • Write out both CLM and NEL clearly before solving — examiners expect to see both equations stated.
  • After solving, check the physics: if two objects approach (A overtaking B), ensure v₁ ≤ v₂ after (A no longer overtaking). If not, something is wrong.
  • For oblique impacts, resolve along and perpendicular to the line of centres — the y-component is unchanged for smooth spheres. This is the key exam distinction.
  • For a ball bouncing on a floor, the coefficient of restitution gives e = √(h₁/H) — a quick check available in successive bounce questions.
  • KE loss questions: always compute KE before and after separately. Don't use the formula unless you've derived it — show working.