Further Mechanics Problem Solving

This final lesson brings together everything you have learned in this course. Exam questions in further mechanics often combine multiple topics — impulse with 2D momentum, circular motion with energy conservation, banked curves with friction. The key to solving these multi-step problems is systematic working and knowing which principles to apply.

Problem-Solving Strategy

flowchart TD
    A["Read the question\nIdentify the physical scenario"] --> B["Draw a clear diagram\nLabel all forces, velocities, angles"]
    B --> C{"What type of problem?"}
    C -->|"Collision / Explosion"| D["Use conservation of momentum\n(2D: resolve into x and y)"]
    C -->|"Circular motion"| E["Identify the centripetal force\nApply F = mv²/r or F = mω²r"]
    C -->|"Vertical circle"| F["Newton's 2nd law at each position\n+ energy conservation for speeds"]
    C -->|"Rotation"| G["Use τ = Iα\nor conservation of L = Iω"]
    D --> H["Check: is KE conserved?\nClassify collision type"]
    E --> I["Solve for unknown\n(speed, radius, force, angle)"]
    F --> I
    G --> I
    H --> I
    I --> J["Check answer is\nphysically reasonable"]

For any further mechanics problem, follow these steps:

Draw a clear diagram showing all forces, velocities, and angles
Identify the physics — which principles apply? (momentum conservation, energy conservation, centripetal force, impulse, etc.)
Define your coordinate system and sign convention
Write the equations using the appropriate formulae
Solve step by step, keeping units consistent
Check your answer — is it physically reasonable?

Multi-Step Problem 1: 2D Collision with Energy Analysis

A 0.15 kg snooker ball travelling at 6.0 m s⁻¹ hits a stationary ball of the same mass. After the collision, the cue ball moves at 3.5 m s⁻¹ at 35° to its original direction. Find the velocity of the object ball and determine whether the collision is elastic.

Step 1: Momentum Conservation

x-direction: $0.15 \times 6.0 = 0.15 \times 3.5\cos 35° + 0.15 \times v_x$ $6.0 = 2.867 + v_x$ $v_x = 3.133 \text{ m s}^{-1}$

y-direction: $0 = 0.15 \times 3.5\sin 35° + 0.15 \times v_y$ $v_y = -2.008 \text{ m s}^{-1}$

Object ball speed: $v = \sqrt{3.133^2 + 2.008^2} = \sqrt{9.82 + 4.03} = \sqrt{13.85} = 3.72 \text{ m s}^{-1}$

Direction: $\theta = \tan^{-1}\left(\frac{2.008}{3.133}\right) = 32.7° \text{ below original direction}$

Step 2: Energy Check

KE before: ½ × 0.15 × 6.0² = 2.70 J

KE after: ½ × 0.15 × 3.5² + ½ × 0.15 × 3.72² = 0.919 + 1.038 = 1.957 J

KE lost: 2.70 − 1.96 = 0.74 J (27% lost)

The collision is inelastic — 27% of the kinetic energy was lost to heat and sound.

Multi-Step Problem 2: Banked Curve with Friction

A car of mass 1400 kg travels around a banked curve of radius 80 m, banked at 20°. The coefficient of friction is 0.30. Find the maximum speed the car can take the curve without skidding.

Analysis

At maximum speed, friction acts down the slope (towards the centre), adding to the centripetal force component of the normal force.

Perpendicular to the slope: $N\cos 20° - mg - \mu N\sin 20° = 0$ $N(\cos 20° - \mu\sin 20°) = mg$ $N \times 0.837 = mg$ $N = \frac{mg}{0.837}$

Along the slope (towards centre): $N\sin 20° + \mu N\cos 20° = \frac{mv^2}{r}$ $N(\sin 20° + \mu\cos 20°) = \frac{mv^2}{r}$ $N \times 0.624 = \frac{mv^2}{r}$

Substituting N = mg/0.837: $\frac{mg \times 0.624}{0.837} = \frac{mv^2}{r}$ $v^2 = \frac{0.624 \times 9.81 \times 80}{0.837} = 585$ $v = 24.2 \text{ m s}^{-1} \approx 87 \text{ km/h}$

Without friction (μ = 0), the design speed would be 16.9 m s⁻¹. Friction allows a significantly higher maximum speed.

Multi-Step Problem 3: Vertical Circle with Energy

A ball of mass 0.30 kg is attached to a string of length 0.80 m and swung in a vertical circle. At the bottom of the circle, the ball has speed 7.0 m s⁻¹.

(a) What is the speed at the top?

Using energy conservation (height gain = 2r = 1.60 m): $v_{\text{top}}^2 = v_{\text{bottom}}^2 - 4gr = 49 - 4 \times 9.81 \times 0.80 = 49 - 31.4 = 17.6$ $v_{\text{top}} = 4.20 \text{ m s}^{-1}$

Since 4.20 > √(gr) = √(9.81 × 0.80) = 2.80 m s⁻¹, the ball completes the circle.

(b) What is the tension at the top? $T + mg = \frac{mv_{\text{top}}^2}{r}$ $T = \frac{0.30 \times 17.6}{0.80} - 0.30 \times 9.81 = 6.6 - 2.94 = 3.66 \text{ N}$

(c) What is the tension at the bottom? $T - mg = \frac{mv_{\text{bottom}}^2}{r}$ $T = \frac{0.30 \times 49}{0.80} + 0.30 \times 9.81 = 18.4 + 2.94 = 21.3 \text{ N}$

The tension at the bottom (21.3 N) is nearly 6 times greater than at the top (3.66 N).

(d) What is the tension at the 3 o'clock position?

Using energy conservation (height gain = r = 0.80 m): $v_{\text{side}}^2 = 49 - 2 \times 9.81 \times 0.80 = 49 - 15.7 = 33.3$

At the side, weight is perpendicular to the radius: $T = \frac{mv_{\text{side}}^2}{r} = \frac{0.30 \times 33.3}{0.80} = 12.5 \text{ N}$

Multi-Step Problem 4: Impulse Leading to 2D Collision

A 0.15 kg ball A moving east at 8.0 m s⁻¹ receives an impulse of 0.60 N s directed north (from a glancing blow with ball B of mass 0.15 kg initially at rest). Find the final velocities of both balls.

Ball A after impulse:

p_x(A) = 0.15 × 8.0 = 1.20 N s (east, unchanged)
p_y(A) = 0 + 0.60 = 0.60 N s (north, from impulse)
v_x(A) = 8.0 m s⁻¹, v_y(A) = 0.60/0.15 = 4.0 m s⁻¹
Speed(A) = √(64 + 16) = √80 = 8.94 m s⁻¹
Direction: tan⁻¹(4.0/8.0) = 26.6° north of east

Ball B (by Newton's third law, receives opposite impulse):

p_x(B) = 0 (no x-impulse)
p_y(B) = 0 − 0.60 = −0.60 N s (south)
v_y(B) = −0.60/0.15 = −4.0 m s⁻¹
Speed(B) = 4.0 m s⁻¹ due south

Verification (total momentum):

x: 0.15 × 8.0 + 0 = 1.20 N s before; 1.20 + 0 = 1.20 N s after ✓
y: 0 before; 0.60 + (−0.60) = 0 after ✓

Multi-Step Problem 5: Satellite Speed and Period

A satellite orbits a planet of mass 4.0 × 10²³ kg at an altitude of 500 km above the surface. The planet's radius is 3.0 × 10⁶ m.

(a) Orbital speed: $r = 3.0 \times 10^6 + 500{,}000 = 3.5 \times 10^6 \text{ m}$ $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 4.0 \times 10^{23}}{3.5 \times 10^6}} = \sqrt{7.62 \times 10^6} = 2760 \text{ m s}^{-1}$

(b) Orbital period: $T = \frac{2\pi r}{v} = \frac{2\pi \times 3.5 \times 10^6}{2760} = 7970 \text{ s} = 133 \text{ minutes}$

Quick Reference: Key Formulae

Topic	Formula	When to Use
Impulse	FΔt = Δp = m(v − u)	Force, time, momentum change
Force–time graph	Area = impulse	Variable force over time
2D momentum (x)	Σp_x(before) = Σp_x(after)	2D collisions/explosions
2D momentum (y)	Σp_y(before) = Σp_y(after)	2D collisions/explosions
KE check	KE = ½mv²	Classify collision type
Coefficient of restitution	e = (v₂ − v₁)/(u₁ − u₂)	Bounciness of collision
Angular velocity	ω = 2πf = 2π/T	Circular motion
Linear ↔ angular speed	v = rω	Connecting v and ω
Centripetal acceleration	a = v²/r = ω²r	Circular motion
Centripetal force	F = mv²/r = mω²r	Circular motion
Flat curve max speed	v_max = √(μgr)	Car on flat bend
Banked curve design speed	v = √(rg tan θ)	Banked curve, no friction
Vertical circle (top)	T + mg = mv²/r	Ball on string at top
Vertical circle (bottom)	T − mg = mv²/r	Ball on string at bottom
Min speed at top	v_min = √(gr)	Complete vertical circle
Energy (vertical circle)	v²_bottom = v²_top + 4gr	Relating speeds at top/bottom
Orbital speed	v = √(GM/r)	Satellites
Kepler's third law	T² ∝ r³	Orbital periods
Torque	τ = Fd or Fr sin θ	Rotation problems
Newton's 2nd (rotation)	τ = Iα	Angular acceleration
Angular momentum	L = Iω	Rotation problems
Conservation of L	I₁ω₁ = I₂ω₂	No external torque

Exam Tips for Multi-Step Problems

Show all working — method marks are available even if the final answer is wrong
State the principles you are using (e.g. "By conservation of momentum in the x-direction...")
Draw diagrams for every problem — they help you and they earn marks
Check units at every stage — if units do not match, you have made an error
Sanity check — is a car speed of 500 m s⁻¹ reasonable? (No — check your work)
Keep extra significant figures in intermediate steps to avoid rounding errors

Common Multi-Step Errors

Error	How to Avoid It
Forgetting to convert rpm to rad s⁻¹	Always: ω = 2π × (rpm/60)
Forgetting direction reversal in impulse	Always define +/− and use signed velocities
Confusing radius with diameter	Read the question: "radius 0.50 m" vs "diameter 1.0 m"
Not adding planet radius to altitude	r_orbital = R_planet + h_altitude
Using wrong force equation at top/bottom of vertical circle	Top: T + mg = mv²/r. Bottom: T − mg = mv²/r
Assuming KE is conserved in all collisions	Only elastic collisions conserve KE; always check
Rounding too early	Keep 4+ s.f. in intermediate steps; round at the end

Summary

Multi-step problems require combining impulse, 2D momentum, circular motion, and energy conservation
Always draw a diagram, define coordinates, and state the principles used
Banked curve problems: resolve forces perpendicular to and along the slope
Vertical circle problems: use energy conservation for speeds and Newton's second law for forces at each position
Satellite problems: equate gravitational force with centripetal force, remember orbital radius = planet radius + altitude
Show all working and check answers are physically reasonable

A-Level Deep Dive: Further Mechanics Problem Solving

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics is the synoptic capstone of the Year 2 mechanics strand: it draws together impulse and momentum (Topic 2 carried through), circular motion, centripetal acceleration and force, and links forward to gravitational fields (Topic 8) and oscillations (Topic 7). The spec requires candidates to apply $\Delta p = F \Delta t$ , conserve momentum in two dimensions for collisions and explosions, distinguish elastic from inelastic processes, and analyse motion in circular paths using $a = v^2/r = \omega^2 r$ and $F = mv^2/r$ (refer to the official specification document for exact wording). Problem-solving questions in this topic are deliberately multi-step: a single stem may demand momentum conservation, then energy bookkeeping, then circular-motion analysis, then a comment on assumptions. The Edexcel Physics formula booklet provides $F = mv^2/r$ , $\omega = v/r$ , and the gravitational expressions, but candidates must select and combine them; no single equation suffices.

Worked example with full mark scheme

Question (12 marks): A satellite of mass $m = 1500\,\text{kg}$ is to be placed in a circular orbit of radius $r = 7.40 \times 10^6\,\text{m}$ around the Earth ( $M_E = 5.97 \times 10^{24}\,\text{kg}$ , $G = 6.67 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}$ , Earth radius $R_E = 6.37 \times 10^6\,\text{m}$ ).

(a) Show that the orbital speed is approximately $7.34\,\text{km s}^{-1}$ . (3)

(b) Calculate the minimum kinetic energy that must be supplied at the Earth's surface to lift the satellite to this orbit and give it the orbital speed in (a). Ignore atmospheric drag and the Earth's rotation. (6)

(c) Discuss two physical effects that would increase the actual launch energy required compared with your answer to (b). (3)

Solution with mark scheme:

(a) Step 1 — equate gravitational pull with centripetal requirement.

$\dfrac{GM_E m}{r^2} = \dfrac{mv^2}{r} \implies v = \sqrt{\dfrac{GM_E}{r}}$

M1 — correct equation, mass $m$ cancels.

Step 2 — substitute.

$v = \sqrt{\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.40 \times 10^6}} = \sqrt{5.38 \times 10^7} \approx 7.34 \times 10^3\,\text{m s}^{-1}$

M1 — correct substitution into a correct rearrangement. A1 — $7.34\,\text{km s}^{-1}$ to 3 s.f., agreeing with the printed value.

(b) Step 1 — energy bookkeeping. The minimum supplied kinetic energy equals the change in mechanical energy from the surface (at rest) to the orbit (moving at $v$ ):

$E_\text{supplied} = \Delta E_k + \Delta E_p = \tfrac{1}{2}mv^2 + \left(-\dfrac{GM_E m}{r}\right) - \left(-\dfrac{GM_E m}{R_E}\right)$

M1 — recognising both kinetic and gravitational PE changes are needed; using $E_p = -GMm/r$ (not $mgh$ , which is invalid over this range).

Step 2 — kinetic term.

$\tfrac{1}{2}mv^2 = \tfrac{1}{2}(1500)(7.34 \times 10^3)^2 \approx 4.04 \times 10^{10}\,\text{J}$

A1 — correct kinetic term to 3 s.f.

Step 3 — potential term.

$\Delta E_p = GM_E m\left(\dfrac{1}{R_E} - \dfrac{1}{r}\right) = (6.67 \times 10^{-11})(5.97 \times 10^{24})(1500)\left(\dfrac{1}{6.37 \times 10^6} - \dfrac{1}{7.40 \times 10^6}\right)$

The bracket evaluates to $(1.5699 - 1.3514) \times 10^{-7} = 2.185 \times 10^{-8}\,\text{m}^{-1}$ . So $\Delta E_p \approx (5.97 \times 10^{17})(2.185 \times 10^{-8}) \approx 1.30 \times 10^{10}\,\text{J}$ .

M1 — correct expression for $\Delta E_p$ between two finite radii. A1 — value to within rounding tolerance.

Step 4 — total.

$E_\text{supplied} \approx 4.04 \times 10^{10} + 1.30 \times 10^{10} \approx 5.3 \times 10^{10}\,\text{J}$

A1 — final answer to appropriate precision.

(c) Any two from: atmospheric drag dissipates energy as heat during ascent (so more chemical energy must be supplied); the rocket equation means a substantial fraction of fuel is spent accelerating unburnt fuel (the "tyranny of the rocket equation"); the orbit insertion burn requires careful thrust-vector control and gravity losses during ascent; the Earth's rotation can be exploited for an eastward launch but only reduces the requirement, not increases it. B1 B1 for two valid effects, B1 for a brief justification linking the effect to extra energy.

Total: 12 marks (M3 A4 + M1 A1 + B3, structure as above).

Specimen question modelled on the Edexcel 9PH0 Paper 3 format

Question (10 marks): A puck of mass $0.40\,\text{kg}$ travelling at $3.0\,\text{m s}^{-1}$ on a frictionless horizontal surface collides with a stationary puck of mass $0.60\,\text{kg}$ . After the collision, the $0.40\,\text{kg}$ puck moves at $2.0\,\text{m s}^{-1}$ at $30°$ to its original direction.

(a) Calculate the speed and direction of the $0.60\,\text{kg}$ puck after the collision. (5)

(b) Determine whether the collision is elastic. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2) — applying conservation of momentum in two perpendicular directions (along and perpendicular to the original line of motion).
M1 (AO1) — $x$ -component: $(0.40)(3.0) = (0.40)(2.0)\cos 30° + (0.60)v_x$ giving $v_x \approx 0.847\,\text{m s}^{-1}$ .
M1 (AO1) — $y$ -component: $0 = (0.40)(2.0)\sin 30° + (0.60)v_y$ giving $v_y \approx -0.667\,\text{m s}^{-1}$ (opposite sense to the first puck's deflection).
A1 (AO2) — speed $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \approx 1.08\,\text{m s}^{-1}$ .
A1 (AO2) — direction $\theta = \arctan(|v_y|/v_x) \approx 38°$ on the opposite side of the original line.

(b)

M1 (AO1) — $E_k$ before $= \tfrac{1}{2}(0.40)(3.0)^2 = 1.80\,\text{J}$ .
M1 (AO1) — $E_k$ after $= \tfrac{1}{2}(0.40)(2.0)^2 + \tfrac{1}{2}(0.60)(1.08)^2 \approx 0.80 + 0.35 = 1.15\,\text{J}$ .
A1 (AO3) — kinetic energy is not conserved ( $1.15 < 1.80\,\text{J}$ ), so the collision is inelastic.

(c) Frictionless surface; pucks are point particles; no rotation or spin transfers energy; collision is instantaneous so no external impulse during contact. B1 B1 for any two clearly stated.