amazon.co.ukA-Level Physics mechanics revision: SUVAT equations with derivations, projectile motion, Newton's laws, momentum conservation, work-energy-power, moments and the free-fall required practical - mapped to AQA 7408 3.4, OCR A H556 Module 3, Edexcel 9PH0 Topic 2, WJEC/Eduqas and CCEA.
Kinematics is the description of motion: displacement s, velocity v and acceleration a. On a displacement-time graph the gradient is velocity; on a velocity-time graph the gradient is acceleration and the area under the graph is displacement (AQA 7408 3.4.1.3; OCR A H556 3.1; Edexcel 9PH0 Topic 2; WJEC/Eduqas and CCEA AS mechanics). The five SUVAT equations connect s, u, v, a and t when acceleration is constant - and you should be able to derive them, not just quote them.
Derivation 1: v = u + at. Acceleration is defined as a = (v - u)/t for uniform acceleration. Multiply both sides by t: at = v - u, so v = u + at.
Derivation 2: s = ut + ½at^2. Displacement is the area under the v-t graph: a rectangle of height u and width t (area ut) plus a triangle of base t and height at (area ½ × t × at). So s = ut + ½at^2. The same result follows by integrating: s = ∫v dt = ∫(u + at) dt = ut + ½at^2 with s(0) = 0.
Derivation 3: s = ½(u + v)t. For a straight-line v-t graph the area is the mean velocity times time: s = ((u + v)/2)t.
Derivation 4: v^2 = u^2 + 2as. Rearrange derivation 1 for t = (v - u)/a and substitute into derivation 2: s = u(v - u)/a + ½a(v - u)^2/a^2. Expand: s = (uv - u^2)/a + (v^2 - 2uv + u^2)/2a. Put over a common denominator: 2as = 2uv - 2u^2 + v^2 - 2uv + u^2 = v^2 - u^2, so v^2 = u^2 + 2as.
Selection routine: list knowns, list the unknown, use the equation that omits the quantity you do not need. Choose a positive direction and keep signs consistent; convert km/h to m/s by dividing by 3.6. Free fall is the standard application: with u = 0 downward, s = ½gt^2.
Drop a ball bearing through a measured height s past a light gate (or use two gates, or a video with a metre rule in shot). Timing starts when the ball is released (u = 0) and stops when it breaks the gate, so human reaction time - roughly 0.2 s, a large systematic error in stopwatch methods - is eliminated. Rearrange s = ½gt^2 to g = 2s/t^2, or better, plot s against t^2: a straight line through the origin with gradient g/2, so g = 2 × gradient. Using many heights and a best-fit line averages random error. Uncertainty in the gradient comes from the scatter and the resolution of the rule (±1 mm on a metre rule read to 1 mm gives about 0.1% per metre). Percentage difference from the accepted value 9.81 m/s^2 concludes the evaluation.
A projectile moves under gravity alone, so the two components are independent (AQA 7408 3.4.1.4): horizontal velocity stays constant at u cos θ (no horizontal force), and vertical velocity starts at u sin θ with acceleration -g. Resolve first, then apply SUVAT to each component separately; the link between them is time.
Derivation: range on level ground. At maximum height the vertical velocity is zero, so from v = u + at: 0 = u sin θ - gt_half, giving t_half = u sin θ/g. By symmetry the total flight time is T = 2u sin θ/g. The range is the constant horizontal velocity times the flight time: R = u cos θ × 2u sin θ/g. Since 2 sin θ cos θ = sin 2θ, R = u^2 sin 2θ/g. Maximum range on level ground occurs at θ = 45 degrees because sin 2θ peaks at 1 when 2θ = 90 degrees - a favourite one-mark question.
Maximum height: using v^2 = u^2 + 2as vertically with v = 0: 0 = u^2 sin^2 θ - 2gh, so h = u^2 sin^2 θ/(2g).
Trajectory shape: eliminating t between x = u cos θ × t and y = u sin θ × t - ½gt^2 gives y = x tan θ - gx^2/(2u^2 cos^2 θ) - a parabola (y is quadratic in x). The path is symmetric only when launch and landing are at the same height; with air resistance the path shortens and steepens after the peak.
A ball is projected at 20 m/s at 30 degrees above the horizontal from level ground (g = 9.8 m/s^2).
Line 1: u_x = 20 cos 30 = 17.3 m/s, u_y = 20 sin 30 = 10.0 m/s (M1: resolve into components).
Line 2: Vertical, at max height v = 0: 0 = 10.0^2 - 2(9.8)h, so h = 100/19.6 = 5.10 m (M1 A1).
Line 3: T = 2u_y/g = 2 × 10.0/9.8 = 2.04 s (A1: symmetry).
Line 4: R = u_x T = 17.3 × 2.04 = 35.3 m (A1).
Typical follow-ups: state the modelling assumptions (no air resistance so horizontal velocity constant; the ball is a particle; g uniform) and describe the effect of removing one - naming the assumption and its consequence earns the mark.
Newton's first law: a body remains at rest or in uniform motion unless acted on by a resultant force - equilibrium means ΣF = 0. Second law: the resultant force equals the rate of change of momentum; for constant mass this reduces to F = ma. Third law: if A exerts a force on B, B exerts an equal and opposite force on A of the same type (AQA 7408 3.4.1.5; OCR A H556 3.5; Edexcel 9PH0 Topic 2).
Derivation: F = ma from F = Δp/Δt. Momentum p = mv. If the mass is constant and velocity changes from u to v in time t: F = Δp/Δt = (mv - mu)/t = m(v - u)/t = ma. For changing mass (a rocket) or changing velocity you must use the full F = Δp/Δt form.
Derivation: conservation of momentum from Newton's third law. Consider two bodies A and B interacting. By the third law, F_AB = -F_AB (action on B equals minus reaction on A), and they act for the same time Δt. Impulse on A: Δp_A = F_AB Δt; impulse on B: Δp_B = -F_AB Δt. Adding: Δp_A + Δp_B = 0, so the total momentum change is zero and Σp_before = Σp_after - momentum is conserved provided no external resultant force acts.
Impulse: J = FΔt = Δp, measured in N s; it equals the area under a force-time graph. This is the physics of safety design: crumple zones and air bags increase the collision time Δt, so for a fixed Δp the force F = Δp/Δt is smaller.
Collisions: momentum is always conserved, but kinetic energy only in elastic collisions. In an inelastic collision KE decreases (transferred to deformation, heat, sound); an explosion converts stored energy into KE so total KE increases while momentum stays zero overall.
A 1200 kg car at 10 m/s hits a stationary 800 kg car and they couple together. Find the common speed and the KE lost.
Line 1: 1200 × 10 + 800 × 0 = 2000v (M1: conservation of momentum).
Line 2: v = 12000/2000 = 6.0 m/s (A1).
Line 3: KE before = ½ × 1200 × 10^2 = 6.0 × 10^4 J; KE after = ½ × 2000 × 6^2 = 3.6 × 10^4 J (M1).
Line 4: KE lost = 2.4 × 10^4 J to deformation and heat (A1: inelastic).
Work is energy transferred by a force: W = Fs cos θ, where θ is the angle between force and displacement - only the component of force along the displacement does work (AQA 7408 3.4.1.1/3.4.1.1 energy; OCR A H556 3.3).
Derivation: KE = ½mv^2. Work-energy theorem: the work done by the resultant force equals the gain in KE. From Newton's second law F = ma and the constant-acceleration equation v^2 = u^2 + 2as with u = 0: a = v^2/2s. So W = Fs = m(v^2/2s)s = ½mv^2. Starting from rest, the work done on the body is W = ½mv^2, which is therefore the kinetic energy.
Gravitational PE near Earth's surface is GPE = mgh (work done lifting mass m through height h against weight mg). Conservation of energy: in the absence of friction, ΣE is constant - falling converts GPE to KE: mgh = ½mv^2, giving v = √(2gh) independent of mass. With friction, the 'missing' mechanical energy equals work done against friction (heat).
Power is the rate of energy transfer: P = W/t, and since W = Fs, P = Fv. A car at steady top speed has engine driving force equal to total resistance, and P = Fv sets the limit. Efficiency = useful output/total input (× 100%).
Moments (AQA 7408 3.4.1.2): the moment of a force about a point is M = Fd, where d is the perpendicular distance from the line of action to the point. A couple is a pair of equal, opposite, parallel forces; its moment is one force × separation and it produces rotation only. Equilibrium of a rigid body requires both ΣF = 0 and ΣM = 0 about any point. The centre of mass is the point where the weight appears to act; for a uniform body it is at the geometric centre. Take moments about a support to eliminate its (unknown) reaction.
A uniform 4.0 m beam of mass 20 kg rests on supports at 0.5 m and 3.0 m from its left end (g = 9.8 m/s^2).
Line 1: Weight 196 N acts at the centre, 2.0 m from the left end (B1: uniform beam).
Line 2: Moments about the left support (at 0.5 m): weight is 1.5 m to the right of it, right support is 2.5 m to the right of it: R_2 × 2.5 = 196 × 1.5 (M1: principle of moments).
Line 3: R_2 = 294/2.5 = 117.6 N (A1).
Line 4: Vertical equilibrium: R_1 + R_2 = 196, so R_1 = 78.4 N (M1 A1).
Beyond the free-fall practical (AQA RP3, OCR PAG1, Edexcel CP1), mechanics practicals measure g with a simple pendulum (T = 2π√(L/g), so g = 4π^2L/T^2 from the gradient of T^2 against L - the pendulum belongs properly to oscillations but is a standard g method), terminal velocity with falling filters or ball bearings (weight equals drag at terminal velocity, so the graph of velocity against time plateaus), and force-extension work with springs (elastic PE = ½kx^2 = area under the F-x graph).
Uncertainty rules (AQA 7408 3.1.2; Edexcel 9PH0 Topic 1; OCR Module 1): the uncertainty in a single reading is at least half the resolution (±0.5 mm on a mm rule); for repeated readings use half the range of the repeats. Combining: adding/subtracting quantities - add absolute uncertainties; multiplying/dividing - add percentage uncertainties; raising to a power n - multiply the percentage uncertainty by n. Percentage error against an accepted value: |experimental - accepted|/accepted × 100%.
A ball falls s = 1.200 ± 0.002 m in t = 0.495 ± 0.005 s. Find g and its uncertainty.
Line 1: g = 2s/t^2 = 2 × 1.200/0.495^2 = 9.80 m/s^2 (M1 A1).
Line 2: Percentage uncertainty: s contributes 0.002/1.200 = 0.17%; t^2 contributes 2 × (0.005/0.495) = 2.0% (M1: power rule for t^2).
Line 3: Total: 0.17% + 2.0% = 2.2%, so Δg = 0.022 × 9.80 = 0.22 m/s^2 (A1).
Line 4: g = 9.8 ± 0.2 m/s^2 - the timing uncertainty dominates, so improving the timing (light gates, video analysis) is the evaluation point (A1: conclusion).
Exam evaluation points for mechanics: reaction time (eliminate with light gates), parallax in length measurement (read at eye level, use a set square to mark heights), air resistance on falling objects (limits the accuracy of g), and repeating measurements to average random error. State which uncertainty is largest and how to reduce it - that comparison earns the top evaluation mark.