amazon.co.ukSUVAT equations, calculus kinematics, Newton's laws, connected particles, friction, inclined planes and projectiles - A-Level Maths mechanics (Edexcel 9MA0 Topics 17-18, AQA 7357 P-R) revised with worked examples and mark-scheme logic.
When acceleration is constant, five formulae connect the five quantities s (displacement), u (initial velocity), v (final velocity), a (acceleration) and t (time) (Edexcel 9MA0 Topic 17; AQA 7357 Q1-Q2; OCR A H640 22.01; WJEC/Eduqas and CCEA applied papers): v = u + at, s = ut + 1/2 at^2, s = 1/2 (u + v)t, v^2 = u^2 + 2as and s = vt - 1/2 at^2. Each formula misses one quantity, so the selection routine is: list what you know, list what you want, and use the equation that avoids the quantity you do not have. They are valid only for constant acceleration - a question switching to variable acceleration needs calculus, and using SUVAT there is a method error.
Two discipline points: choose a positive direction and stick to it (deceleration is a negative acceleration once a direction is fixed), and convert units before substituting (km/h to m/s by dividing by 3.6). Velocity-time graphs support the work: the gradient is the acceleration and the area under the graph is the displacement - useful checks and a source of graph questions.
A car travelling at 25 m/s brakes uniformly and comes to rest in 50 m. Find the deceleration and the braking time.
Line 1: u = 25, v = 0, s = 50, a unknown, t unknown. v^2 = u^2 + 2as avoids t (M1: correct selection).
Line 2: 0 = 25^2 + 2a(50) = 625 + 100a, so a = -6.25 m/s^2 (A1: deceleration 6.25 m/s^2).
Line 3: v = u + at: 0 = 25 - 6.25t, so t = 4 s (A1).
Note the sign: the acceleration is negative because it opposes the motion - an answer of 6.25 m/s^2 without the direction comment can lose the interpretation mark when the question asks for deceleration explicitly.
When acceleration is not constant, calculus takes over (Edexcel 9MA0 Topic 17; AQA 7357 Q3; OCR A H640 22.01): v = ds/dt and a = dv/dt differentiate downwards, while s = integral of v dt and v = integral of a dt integrate upwards, with initial conditions fixing the constants of integration. The chain r(t) -> v(t) -> a(t) runs downhill by differentiation and uphill by integration - know which direction the question demands.
The subtlety exam questions love: displacement versus distance. Displacement is the integral of v dt; distance is the integral of |v| dt - when the particle changes direction (v = 0), you must split the integral at the turning points and add the absolute values. Find where v = 0 first in any distance question.
A particle moves with velocity v(t) = 3t^2 - 12t + 9 m/s. Find its displacement and the distance travelled from t = 0 to t = 3.
Line 1: s(t) = t^3 - 6t^2 + 9t (integral of v, s(0) = 0) (M1 A1).
Line 2: Displacement = s(3) - s(0) = 27 - 54 + 27 = 0 m (A1).
Line 3: v(t) = 3(t - 1)(t - 3), so v = 0 at t = 1 and t = 3 - the particle changes direction (M1: direction changes found).
Line 4: s(1) = 1 - 6 + 9 = 4. Distance = |s(1) - s(0)| + |s(3) - s(1)| = 4 + 4 = 8 m (A1: split at turning points).
Displacement zero with distance 8 m is the classic contrast: the particle returns to its start but travelled 8 m doing it. Always find v = 0 before answering a distance question, and state both quantities when the question asks for distance - quoting displacement for distance is a standard mark loss.
Newton's second law: the resultant force on a body is F = ma (in newtons, kg and m/s^2) (Edexcel 9MA0 Topic 18; AQA 7357 R1-R3; OCR A H640 23.01/24.01). The forces to model: weight mg vertically down, the normal reaction R perpendicular to the surface, tension T along light inextensible strings, and applied forces. Draw a force diagram first in every dynamics question - it is the model, and the equations come from resolving.
Newton's third law: forces between two interacting bodies are equal, opposite and of the same type - the reason the tension is the same throughout a light string over a smooth pulley. Connected particles: write F = ma for each mass separately and solve simultaneously; the whole-system shortcut (treating the system as one mass with the external resultant force) finds the acceleration fast, but the individual equations are needed for the tension.
Particles of mass 3 kg and 5 kg are connected by a light inextensible string over a smooth pulley. Find the acceleration and the tension (g = 9.8 m/s^2).
Line 1: The heavier particle moves down: for the 5 kg: 5g - T = 5a (M1: F = ma resolved).
Line 2: For the 3 kg: T - 3g = 3a (M1).
Line 3: Adding: 2g = 8a, so a = g/4 = 2.45 m/s^2 (A1).
Line 4: T = 3g + 3a = 3(9.8) + 3(2.45) = 36.75 N (A1: tension from one individual equation).
Check the tension by substituting into the other equation: 5g - 5a = 49 - 12.25 = 36.75 N - both equations agree. Modelling assumptions matter here: light (massless) string and smooth (frictionless) pulley give equal tension throughout; an extensible string would not.
Friction opposes relative motion and has magnitude at most F = mu R (Edexcel 9MA0 Topic 18; AQA 7357 R4-R5; OCR A H640 25.01), where mu is the coefficient of friction and R the normal reaction. Friction takes any value up to that maximum and reaches the maximum only in limiting equilibrium or when sliding - a body at rest on a shallow slope has friction exactly balancing the weight component, not necessarily mu R. Questions asking whether the body slides compare the required friction with mu R.
On an incline of angle theta, resolve parallel and perpendicular to the slope: the normal reaction is R = mg cos(theta) and the weight component along the slope is mg sin(theta). Resolving along the slope (not horizontally and vertically) is the method - diagrams with the angle marked earn the setup marks.
A mass of 4 kg slides down a slope inclined at 30 degrees to the horizontal, with mu = 0.2. Find its acceleration (g = 9.8 m/s^2).
Line 1: Resolve perpendicular: R = mg cos(30 degrees) = 4(9.8)(sqrt(3)/2) = 33.9 N (M1 A1).
Line 2: Friction (opposing the downward motion): F = mu R = 0.2 x 33.9 = 6.79 N (M1 A1).
Line 3: Weight component down the slope: mg sin(30 degrees) = 4(9.8)(0.5) = 19.6 N (B1).
Line 4: Resultant: 19.6 - 6.79 = 12.8 N down the slope (M1).
Line 5: a = F/m = 12.8/4 = 3.20 m/s^2 (A1).
Common errors: using sin and cos the wrong way round (R uses cos - the reaction is perpendicular to the slope), forgetting that friction acts up the slope when the body slides down, and using F = mu R when the body is not sliding. State the direction of friction in your working - it is part of the model.
A projectile under gravity alone has two independent motions: horizontal velocity constant at u cos(theta) (no horizontal force) and vertical velocity u sin(theta) with acceleration -g (Edexcel 9MA0 Topic 17; AQA 7357 Q4; OCR A H640 24.01). At maximum height the vertical velocity is zero. The three standard quantities: maximum height (from the vertical SUVAT with v = 0), time of flight (twice the time to maximum height, for level ground), and range (horizontal velocity x time of flight).
A ball is projected from ground level with speed 20 m/s at 30 degrees above the horizontal. Find the maximum height, the time of flight and the range (g = 9.8 m/s^2).
Line 1: Resolve: u_x = 20cos(30 degrees) = 17.3 m/s, u_y = 20sin(30 degrees) = 10 m/s (M1: components).
Line 2: Vertical, at max height v = 0: 0 = 10^2 - 2(9.8)s, so s = 100/19.6 = 5.10 m (M1 A1).
Line 3: Time to max height: 0 = 10 - 9.8t, so t = 1.04 s; flight time 2t = 2.04 s (A1: symmetry).
Line 4: Range = u_x x time of flight = 17.3 x 2.04 = 35.4 m (A1).
Modelling assumptions are examinable: no air resistance (so horizontal velocity is constant), the projectile is a particle (no spin or size), and gravity is uniform at 9.8 m/s^2. Questions ask which assumption is being made or what its effect would be - name it and state the consequence in context.
Exam technique across the topic: (1) draw the force diagram or resolve into components before writing equations - the diagram earns the setup marks; (2) check whether acceleration is constant (SUVAT) or variable (calculus) before selecting a method; (3) find v = 0 before any distance question; (4) state the positive direction and keep signs consistent; (5) in connected-particle questions, give the tension from an individual equation, not the whole-system equation; (6) convert units before substituting. Board coverage: Edexcel 9MA0 Topics 17-18, AQA 7357 P1-R5, OCR A H640 22.01-25.01, WJEC/Eduqas AS/A2 mechanics, CCEA AS2/A22 or AS3/A23 - identical techniques, differing wording and g conventions (9.8 or 9.81 as your board's formula booklet states).