Mechanics Problem Solving

This final lesson brings together all the topics covered in this course — scalars and vectors, kinematics, projectile motion, Newton’s laws, forces, moments, work-energy-power, momentum, and impulse. Real exam questions rarely test a single concept in isolation. Instead, they combine multiple ideas in multi-step problems that require careful analysis and a clear strategy. This lesson works through several extended problems and highlights the techniques that lead to success.

Problem-Solving Strategy

flowchart TD
    A["1. Read the question\nIdentify knowns and unknowns"] --> B["2. Draw a diagram\nFree-body diagram or sketch"]
    B --> C["3. Choose your approach"]
    C --> D["Forces & Newton's 2nd law\n(F = ma)"]
    C --> E["Energy conservation\n(KE + GPE = const)"]
    C --> F["Momentum conservation\n(p before = p after)"]
    C --> G["Impulse\n(FΔt = Δp)"]
    D --> H["4. Define sign convention\n5. Write equations\n6. Solve and check"]
    E --> H
    F --> H
    G --> H

Detailed Steps

Read the question carefully. Identify what quantities are given and what you are asked to find.
Draw a diagram. A clear free-body diagram or sketch of the situation saves time and prevents errors.
Choose your approach. Decide whether to use Newton’s second law (forces), conservation of energy, conservation of momentum, or a combination.
Define a sign convention. Choose positive directions and stick to them throughout.
Write equations. Apply the relevant principles and solve algebraically before substituting numbers.
Check your answer. Does it have the right units? Is the magnitude reasonable? Does the direction make sense?

When to Use Each Method

Approach	Best For	Key Equation
SUVAT	Constant acceleration, finding s, u, v, a, or t	v² = u² + 2as, etc.
F = ma	Finding acceleration, tension, or forces	F_net = ma
Energy conservation	Height/speed changes, frictionless or with known friction	KE + GPE = const
Work-energy theorem	Relating force, distance, and speed change	W_net = ΔKE
Momentum conservation	Collisions and explosions	p_before = p_after
Impulse	Relating force, time, and momentum change	FΔt = Δp
Moments	Beams, levers, equilibrium of extended objects	Sum CW = Sum ACW

Problem 1: Inclined Plane with Friction

A 5.0 kg block is placed on a rough plane inclined at 35° to the horizontal. The coefficient of friction is 0.25. The block is released from rest.

(a) Show that the block accelerates down the slope.

Force down slope = mg sin(35°) = 5.0 x 9.81 x 0.5736 = 28.1 N Normal reaction R = mg cos(35°) = 5.0 x 9.81 x 0.8192 = 40.2 N Maximum friction = mu R = 0.25 x 40.2 = 10.1 N

Since 28.1 N > 10.1 N, the gravitational component exceeds maximum friction, so the block accelerates.

(b) Find the acceleration.

Net force down slope = 28.1 - 10.1 = 18.0 N a = F/m = 18.0 / 5.0 = 3.6 m/s²

(c) The block slides 2.0 m down the slope. Find its speed at the bottom using both SUVAT and energy methods.

SUVAT method: v² = u² + 2as = 0 + 2(3.6)(2.0) = 14.4 v = sqrt(14.4) = 3.8 m/s

Energy method: Height dropped: h = 2.0 sin(35°) = 2.0 x 0.5736 = 1.15 m GPE lost = mgh = 5.0 x 9.81 x 1.15 = 56.4 J Work done against friction = friction x distance = 10.1 x 2.0 = 20.2 J KE gained = GPE lost - work against friction = 56.4 - 20.2 = 36.2 J (1/2)mv² = 36.2 v² = 2 x 36.2 / 5.0 = 14.5 v = sqrt(14.5) = 3.8 m/s (consistent with SUVAT)

(d) The block then slides onto a rough horizontal surface with the same coefficient of friction. How far does it travel before stopping?

On the horizontal: friction = mu mg = 0.25 x 5.0 x 9.81 = 12.3 N Deceleration: a = F/m = 12.3/5.0 = 2.45 m/s²

Using v² = u² + 2as with v = 0, u = 3.8 m/s: 0 = 14.4 - 2(2.45)s s = 14.4/4.90 = 2.9 m

Problem 2: Connected Bodies (Atwood-Style)

Two blocks are connected by a light, inextensible string over a smooth pulley. Block A (mass 4.0 kg) sits on a smooth horizontal table. Block B (mass 3.0 kg) hangs vertically. The system is released from rest.

(a) Find the acceleration of the system.

For the system, the driving force is the weight of block B: W_B = 3.0 x 9.81 = 29.4 N The total mass being accelerated = 4.0 + 3.0 = 7.0 kg

a = F_net / m_total = 29.4 / 7.0 = 4.2 m/s²

(b) Find the tension in the string.

Consider block A on the table: T = m_A x a = 4.0 x 4.2 = 16.8 N

Check with block B: W_B - T = m_B x a 29.4 - 16.8 = 12.6 N, and m_B x a = 3.0 x 4.2 = 12.6 N. Consistent.

(c) After falling 1.5 m, block B hits the ground and stops. Block A continues sliding. How far does block A travel after block B stops? (Table is smooth.)

First, find the speed when B hits the ground: v² = u² + 2as = 0 + 2(4.2)(1.5) = 12.6 v = sqrt(12.6) = 3.55 m/s

After B stops, there is no tension and no friction (smooth table), so block A continues at 3.55 m/s with no force to stop it. On a smooth table, block A would continue indefinitely.

If the table had friction (say mu = 0.20): Friction = mu x m_A x g = 0.20 x 4.0 x 9.81 = 7.85 N Deceleration = 7.85 / 4.0 = 1.96 m/s² Using v² = u² + 2as with v = 0: 0 = 3.55² - 2(1.96)s s = 12.6 / 3.92 = 3.2 m

Problem 3: Projectile and Energy Combined

A ball of mass 0.20 kg is launched from ground level at 25 m/s at 40° above the horizontal.

(a) Find the speed of the ball at its maximum height. (Ignore air resistance.)

At maximum height, the vertical component of velocity is zero. The horizontal component remains constant:

v_x = 25 cos(40°) = 25 x 0.766 = 19.1 m/s

This is the speed at the top.

(b) Verify the maximum height using both SUVAT and energy.

SUVAT method: u_y = 25 sin(40°) = 25 x 0.643 = 16.1 m/s At max height, v_y = 0: v_y² = u_y² - 2gH 0 = 16.1² - 2(9.81)H H = 259.2 / 19.62 = 13.2 m

Energy method: At launch: KE = (1/2)(0.20)(25²) = 62.5 J, GPE = 0 At max height: KE = (1/2)(0.20)(19.1²) = 36.5 J, GPE = mgh = (0.20)(9.81)(h)

Energy conservation: 62.5 = 36.5 + (0.20)(9.81)h 26.0 = 1.962 h h = 26.0 / 1.962 = 13.3 m (consistent, small difference due to rounding)

(c) Find the speed of the ball when it is at half the maximum height (h = 6.6 m).

Using energy: (1/2)(0.20)v² = (1/2)(0.20)(25²) - (0.20)(9.81)(6.6) 0.10 v² = 62.5 - 12.9 = 49.6 v² = 496 v = 22.3 m/s

Problem 4: Collision and Momentum

A 2.0 kg trolley A moves at 4.0 m/s and collides with a stationary 3.0 kg trolley B on a smooth surface. After the collision, trolley A moves at 1.0 m/s in the same direction.

(a) Find the velocity of trolley B after the collision.

Conservation of momentum: (2.0)(4.0) + (3.0)(0) = (2.0)(1.0) + (3.0)(v_B) 8.0 = 2.0 + 3.0 v_B 3.0 v_B = 6.0 v_B = 2.0 m/s (same direction as A’s initial motion)

(b) Is the collision elastic?

KE before = (1/2)(2.0)(4.0²) = 16.0 J KE after = (1/2)(2.0)(1.0²) + (1/2)(3.0)(2.0²) = 1.0 + 6.0 = 7.0 J

KE before (16.0 J) does not equal KE after (7.0 J), so the collision is inelastic. 9.0 J of kinetic energy was lost to other forms (heat, sound, deformation).

(c) What was the impulse on trolley B?

Impulse = change in momentum of B = (3.0)(2.0) - (3.0)(0) = 6.0 N s

(d) If the collision lasted 0.15 s, what was the average force between the trolleys?

F = impulse / time = 6.0 / 0.15 = 40 N

Problem 5: Power and Motion on a Hill

A car of mass 1400 kg drives up a hill inclined at 5.0° to the horizontal at a constant speed of 20 m/s. The resistive forces total 600 N.

(a) Find the driving force.

At constant velocity, net force = 0: Driving force = component of weight down the hill + resistive forces = mg sin(5.0°) + 600 = (1400)(9.81)(0.08716) + 600 = 1197 + 600 = 1797 N (approximately 1800 N)

(b) Find the power output of the engine.

P = Fv = 1797 x 20 = 35 940 W = 35.9 kW

(c) The car reaches the top and drives on a flat road at the same power. What is its new top speed?

On the flat, at constant velocity: Driving force = resistive force = 600 N P = Fv: 35 940 = 600 x v v = 35 940 / 600 = 59.9 m/s

(d) The car then starts climbing a steeper hill (10°) at the same power. What is its constant speed on this hill?

At constant velocity on the 10° hill: Driving force = mg sin(10°) + 600 = 1400 x 9.81 x 0.1736 + 600 = 2384 + 600 = 2984 N

P = Fv: 35 940 = 2984 x v v = 35 940 / 2984 = 12.0 m/s

The steeper the hill, the lower the speed at a given power output.

Problem 6: Moments and Beams

A non-uniform plank of weight 180 N and length 5.0 m is supported at two points: A at the left end and B at 3.5 m from A. When a 60 N weight is hung from the right end, the plank is on the point of tipping about B.

(a) Find the position of the centre of gravity of the plank.

On the point of tipping about B, the reaction at A is zero. All the clockwise moment (from the plank weight and the 60 N load) about B must be balanced.

Taking moments about B: The 60 N weight is 5.0 - 3.5 = 1.5 m to the right of B. Let the centre of gravity be at distance d from A (i.e. d - 3.5 m from B).

For the plank to be on the point of tipping, the plank weight must create an anticlockwise moment about B: 180 x (3.5 - d) = 60 x 1.5 180(3.5 - d) = 90 3.5 - d = 0.50 d = 3.0 m from A

(b) With the 60 N weight removed, what is the reaction at each support?

Taking moments about A: R_B x 3.5 = 180 x 3.0 R_B = 540/3.5 = 154 N

R_A = 180 - 154 = 26 N

Problem 7: Multi-Concept Problem

A 0.50 kg ball is thrown horizontally at 8.0 m/s from the top of a 20 m cliff. It hits the ground and bounces, reaching a maximum height of 5.0 m vertically.

(a) Find the speed of the ball just before it hits the ground.

Horizontal: v_x = 8.0 m/s (constant) Vertical: v_y² = 0 + 2(9.81)(20) = 392.4, v_y = 19.8 m/s

Speed at impact = sqrt(8.0² + 19.8²) = sqrt(64 + 392) = sqrt(456) = 21.4 m/s

(b) Find the speed of the ball just after the bounce (assuming it bounces vertically upward).

Using energy: (1/2)mv² = mgh v = sqrt(2 x 9.81 x 5.0) = sqrt(98.1) = 9.9 m/s

(c) How much energy is dissipated during the bounce?

KE before bounce = (1/2)(0.50)(21.4²) = (1/2)(0.50)(456) = 114 J KE after bounce = (1/2)(0.50)(9.9²) = (1/2)(0.50)(98.1) = 24.5 J

Energy dissipated = 114 - 24.5 = 89.5 J

(d) If the bounce lasts 0.020 s, estimate the average force on the ball during the bounce.

This is complex because the ball changes direction. Before bounce: velocity is 21.4 m/s downward (approximately, dominated by vertical component of 19.8 m/s). After bounce: 9.9 m/s upward.

Taking upward as positive: Impulse = m(v - u) = 0.50(9.9 - (-19.8)) = 0.50(29.7) = 14.85 N s

(Note: we use the vertical component before impact, 19.8 m/s downward, and 9.9 m/s upward after.)

Average force = 14.85/0.020 = 742 N (upward, net of weight)

Common Mistakes Reference

Mistake	How to Avoid It
Using the wrong SUVAT equation	List knowns first, identify which variable is missing
Forgetting friction on slopes	Always check: is the surface "smooth" or "rough"?
Wrong sign convention	State your positive direction explicitly
Not resolving forces on slopes	Use parallel and perpendicular to slope, not H and V
Confusing speed and velocity	Speed = magnitude only; velocity includes direction
KE proportional to v, not v²	KE = (1/2)mv² — doubling speed quadruples KE
Momentum is a scalar	No! Momentum is a vector; assign +/- signs to directions
Ignoring beam weight in moments	Uniform beam weight acts at its centre

Exam Tips for Edexcel Physics

Show all working. Method marks are available even if you get the final answer wrong.
State principles. Write “by conservation of momentum” or “using Newton’s second law” to show the examiner which physics you are applying.
Units matter. Include units in every answer. Convert to SI if necessary.
Significant figures. Give your answer to the same number of significant figures as the data in the question (usually 2 or 3). Do not round intermediate calculations.
Check directions. Sign errors are the most common mistake. Define positive direction clearly and be consistent.
Re-read the question. Make sure you have answered what was actually asked.
Use energy methods as a check. If you solved a problem with SUVAT, verify it with energy. If the answers agree, you can be confident.
For multi-part questions, use answers from earlier parts. But if you could not get part (a), use the given value (if provided) to attempt part (b).
When stuck, draw a bigger diagram. Label everything. Many students lose marks because their diagram was too small or unclear.
Time management. In the Edexcel exam, aim for roughly 1 mark per minute. If a question is worth 6 marks, spend about 6 minutes on it.

Key Points

Multi-step problems require combining SUVAT, Newton’s laws, energy, and momentum.
Draw diagrams and define sign conventions before starting calculations.
For inclined planes: resolve forces parallel and perpendicular to the slope.
For connected bodies: consider the system as a whole for acceleration, then individual objects for tension.
Energy methods often provide an alternative (and sometimes simpler) route to the same answer as SUVAT or force methods.
Always verify answers are physically reasonable.
Use two different methods to check important results: if SUVAT and energy give the same answer, your working is likely correct.

A-Level Deep Dive: Mechanics Problem Solving

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, when read as a whole rather than as isolated sub-topics, is examined as a synoptic capstone (refer to the official Pearson Edexcel specification document for exact wording). The mechanics problem-solving lesson is not a discrete spec point — it is the integration of every prior sub-topic: 2.1 motion (kinematics, projectiles), 2.2 forces and Newton's laws, 2.3 work, energy and power, and the momentum strands developed within 2.2 and extended in Topic 6 (Further mechanics). Synoptic mechanics items appear most heavily on Paper 3 — General and Practical Principles in Physics, where AO3 (analyse, interpret and evaluate) carries roughly half the assessment weight, but multi-method mechanics problems also surface on Papers 1 and 2 wherever a question chains two or more techniques (e.g. impulse then energy, or projectile then collision). The Edexcel data, formulae and relationships booklet provides the kinematic, energy, momentum and oscillation equations, but does not tell candidates which equation to use first — the AO3 mark for "selecting an appropriate model" is exactly the strategic choice this lesson trains.

Worked example with full mark scheme

Question (12 marks):

A car of mass 1200 kg brakes and skids along a horizontal road before colliding with a stationary 4800 kg truck. The two vehicles coalesce as a wreckage. Pre-impact skid marks (car only): 18.0 m. Post-impact skid marks (wreckage): 4.5 m. Coefficient of friction between tyres and road: 0.70. Take $g = 9.81$ m s⁻².

(a) Calculate the speed of the wreckage immediately after the collision. (3)

(b) Hence calculate the speed of the car at impact. (3)

(d) Calculate the fraction of pre-collision KE dissipated in the collision. (2)

Solution with mark scheme: Take rightward (car's direction) positive throughout.

(a) Post-impact wreckage decelerated by friction $f = \mu m_\text{w} g$ over 4.5 m. Work-energy: $\tfrac{1}{2} m_\text{w} v_\text{w}^2 = \mu m_\text{w} g d_2$ — mass cancels.

M1 identifies friction as decelerating force on combined mass $m_\text{w} = 6000$ kg. M1 equates KE to work done against friction (equivalent: $v^2 = u^2 - 2\mu g s$ ).

$v_\text{w} = \sqrt{2 \times 0.70 \times 9.81 \times 4.5} = 7.86$ m s⁻¹. A1 to 3 sf with units.

(b) Conservation of linear momentum across the brief collision (no significant external horizontal impulse during contact):

$m_\text{c} v_\text{c} = m_\text{w} v_\text{w} \implies v_\text{c} = (6000/1200) \times 7.86 = 39.3$ m s⁻¹.

M1 invokes momentum conservation by name. M1 correct substitution using (a)'s $v_\text{w}$ . A1 $v_\text{c} = 39.3$ m s⁻¹.