Flashcards · 9660 · Topic 7

Unit M2: Mechanics (International A-level) flashcards

38 cards on Topic 7, subtopics M2.1 to M2.7. Pick a subtopic or study them all.

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M2.1 Mathematical modelling

The mathematical modelling cycle: 1. Simplify and state assumptions. 2. Write the mathematics and solve. 3. Interpret in context. 4. Compare with evidence and judge validity. 5. Refine the assumptions if needed.
Modelling a body as a particle: Its size and rotation are ignored and all forces act at one point. A large body can be a point mass when its size is small compared with the distances involved.
Modelling words: light, inextensible, smooth, and motion under gravity: Light: negligible mass, same tension all along a string. Inextensible: fixed length, so connected objects share speed and acceleration. Smooth: no friction. Motion under gravity: only gravity acts, g constant, no air resistance.
How do you comment on the validity of a model?: Name the assumption, say which way the prediction shifts, and suggest an improvement. Example: ignoring air resistance makes a predicted fall time too short and the impact speed too large.

M2.2 Kinematics

Vector kinematics: links between r, v and a: v = dr/dt and a = dv/dt, component by component. r = ∫ v dt and v = ∫ a dt, each with a constant vector. Speed = |v|.
a = 6t i − 2j and v = 4i when t = 0. Find v and the speed at t = 1.: v = (3t² + 4)i − 2t j (the j constant is 0). At t = 1: v = 7i − 2j, speed √53 ≈ 7.28.
Conditions for "at rest" and for "moving parallel to i": At rest: every component of v is zero at the same time. Parallel to i: only the other component(s) of v are zero.
How do you show that two particles collide?: Equate the i components and the j components of their position vectors. They collide only if one value of t satisfies every component; then substitute to give the position.
Direction of motion when v = 3i − 4j: tan⁻¹(4/3) = 53.1° below the direction of i. State the axis the angle is measured from.

M2.3 Statics and forces

Friction in equilibrium, and the condition to rest on a rough slope with no other force: F ≤ μR; F = μR only when on the point of slipping (limiting). A particle rests on a slope at angle α with no other force if μ ≥ tan α.
A 6 kg particle is on a rough slope at 30° with μ = 0.2. Find the range of a force P up the slope that keeps it at rest.: R = 6g cos 30° = 50.92, so maximum friction 10.18; weight component 6g sin 30° = 29.4. About to slide down: P = 29.4 − 10.18 = 19.2. About to slide up: P = 29.4 + 10.18 = 39.6. 19.2 N ≤ P ≤ 39.6 N.
Moment of a force, including a force at angle θ to a rod: Moment = force × perpendicular distance from the point to its line of action (N m). Force F at angle θ to a rod, a distance d along it: moment F d sin θ.
Conditions for a rigid body in equilibrium, and the tilting condition for a beam: Resultant force zero and resultant moment zero about any point. A beam about to tilt about one support has zero reaction at the other support.
Centre of mass of particles and of a composite lamina with a hole: X Σm = Σ mx and Y Σm = Σ my. Uniform lamina: use areas in place of masses; a removed piece has negative area: X = (A_whole x_whole − A_hole x_hole)/(A_whole − A_hole).
A lamina has centre of mass (3.5, 1.5) and hangs from the origin. Find the angle between the x-axis edge and the vertical.: The line from the origin to G is vertical. tan α = 1.5/3.5, so α = 23.2° (3 s.f.) between the edge and the vertical.

M2.4 Newton's Law of Motion

Acceleration of a particle sliding down a smooth slope and down a rough slope at angle θ: Smooth: a = g sin θ, independent of mass. Rough, sliding down: a = g(sin θ − μ cos θ). R = mg cos θ when no other force acts perpendicular to the slope.
Particle projected up a rough slope: deceleration going up, acceleration coming back down: Going up, friction acts down the slope: deceleration g(sin θ + μ cos θ). Coming down, friction acts up the slope: a = g(sin θ − μ cos θ). It slides back only if tan θ > μ.
A force F = 6t N acts on a 3 kg particle starting from rest. Find v and r.: a = F/m = 2t. v = ∫ 2t dt = t² (constant 0, since it starts from rest). r = t³/3 + constant.
r = (t³ − t)i + 3t² j for a particle of mass 3 kg. Find the force.: a = d²r/dt² = 6t i + 6j. F = ma = 18t i + 18j.
A 6 kg particle on a rough slope at 20° (μ = 0.3) is pushed up the slope by 50 N parallel to it. Find a.: R = 6(9.8) cos 20° = 55.25 N; friction 0.3R = 16.58 N down; weight component 20.11 N down. 50 − 20.11 − 16.58 = 6a, so a = 2.22 m s⁻² up the slope.

M2.5 Projectiles

Projectile equations for launch speed V at angle α (origin at the launch point, y up): x = (V cos α)t y = (V sin α)t − ½gt² Horizontal velocity V cos α; vertical velocity V sin α − gt.
Time of flight, greatest height and range on level ground, and how to get them: Derive them in answers, do not quote: y = 0 gives time of flight 2V sin α/g. Vertical velocity 0 gives greatest height (V sin α)²/(2g). x at landing gives range V² sin 2α/g.
Which launch angle gives the greatest range, and which angles give equal ranges?: The range is greatest at 45°. Angles α and 90° − α give the same range on level ground, with different heights and times.
Equation of the trajectory: Eliminate t using t = x/(V cos α): y = x tan α − gx²/(2V² cos²α) = x tan α − gx²(1 + tan²α)/(2V²). The second form gives a quadratic in tan α when the angle is unknown.
V = 15 m/s at 60° on level ground. Find the time of flight, greatest height and range.: Components 7.5 and 12.99 m/s. Time = 2 × 12.99/9.8 = 2.65 s. Height = 12.99²/19.6 = 8.61 m. Range = 7.5 × 2.651 = 19.9 m.
Assumptions in the projectile model, and the effect of air resistance: Particle (size and spin ignored), no air resistance or wind, constant g downwards, motion in a vertical plane. Air resistance would normally make the real range and greatest height smaller.

M2.6 Work and energy

Work done by a constant force: Work = Fd cos θ (J), where θ is the angle between the force and the motion. A normal reaction does no work; friction and drag do negative work.
Kinetic energy, gravitational potential energy, and the work-energy principle: KE = ½mv². GPE = mgh, using vertical height only (h = s sin α on a slope). Total work done = final KE − initial KE.
A bead slides from rest down a smooth wire, dropping 1.8 m. Find its speed.: Only gravity does work, so KE + GPE is constant: v² = 2 × 9.8 × 1.8 = 35.28, v = 5.94 m s⁻¹ (3 s.f.).
Power: Power is the rate of doing work: P = work/time = Fv, where F is the driving force. 1 W = 1 J s⁻¹; convert kW to W.
A 1500 kg van has power 45 kW and resistance 1000 N. Find its maximum speed on level ground and its acceleration at 20 m s⁻¹.: Maximum speed: 45 000/1000 = 45 m s⁻¹. At 20 m s⁻¹: driving force 45 000/20 = 2250 N, so a = (2250 − 1000)/1500 = 0.833 m s⁻².
A 5000 kg lorry climbs at constant speed where sin α = 1/25, resistance 1500 N, power 60 kW. Find its speed.: F = 1500 + 5000 × 9.8/25 = 1500 + 1960 = 3460 N. v = 60 000/3460 = 17.3 m s⁻¹ (3 s.f.).

M2.7 Uniform circular motion

Formulae for uniform circular motion: v = rω. Acceleration towards the centre a = rω² = v²/r. ω = 2π/T for a period T. ω in rad s⁻¹.
Convert n rev/min to rad s⁻¹, and 30 rev/min: ω = n × 2π/60. 30 rev/min = π rad s⁻¹.
Is there a separate "centripetal force" to add to a force diagram?: No. Draw only the real forces (tension, friction, normal reaction, weight, gravity). Their resultant points to the centre and equals ma, with a = v²/r or rω².
Conical pendulum: the two resolving equations and the tension: Vertically: T cos θ = mg. Towards the centre: T sin θ = mrω² with r = L sin θ. Dividing by sin θ gives T = mLω².
Conical pendulum with L = 0.75 m and θ = 40°. Find the period.: Depth h = L cos θ = 0.5745 m and ω² = g/h. Period = 2π√(h/g) = 2π√(0.5745/9.8) = 1.52 s (3 s.f.).
Vector form of circular motion with centre O, radius r, angular speed ω: r = r cos(ωt) i + r sin(ωt) j v = −rω sin(ωt) i + rω cos(ωt) j a = −ω² r (towards the centre); |v| = rω and v.r = 0.