Flashcards · YMA01 · Topic 6

Unit M2: Mechanics 2 flashcards

29 cards on Topic 6, subtopics M2.1 to M2.5. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 29 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

Projectile launched at speed U, angle α above the horizontal: velocity and displacement components: Horizontal: velocity U cos α, displacement (U cos α)t. Vertical (up positive): velocity U sin α − gt, displacement (U sin α)t − ½gt².
Time of flight, range and greatest height on level ground: Time of flight T = 2U sin α/g. Range = (U cos α)T. Greatest height H = (U sin α)²/(2g), reached at T/2.
A particle is launched horizontally at 15 m s⁻¹ from 19.6 m above level ground. Find the time and the horizontal distance (g = 9.8).: 19.6 = 4.9t², so t = 2 s. Horizontal distance = 15 × 2 = 30 m.
Variable acceleration: how are x, v and a linked by calculus?: v = dx/dt, a = dv/dt. x = ∫ v dt and v = ∫ a dt, each with a constant found from a given condition. suvat does not apply. Velocity is greatest or least when a = 0.
How do you find the distance travelled when v changes sign?: Solve v = 0 to find the turning times, find x at the start, at each turning time and at the end, and add the sizes of the changes. A single definite integral of v gives only displacement.
r = 2t³ i − 5t j. Find v, a and the speed at t = 1.: v = 6t² i − 5j, a = 12t i. At t = 1: v = 6i − 5j, speed √(36 + 25) = √61 = 7.81 (3 s.f.).

Centre of mass of a system of particles: x̄ = Σ(mx)/Σm and ȳ = Σ(my)/Σm.
Centre of mass of a composite uniform lamina: Use areas in place of masses: x̄ = Σ(Ax)/ΣA. A hole counts as a negative area. Measure every coordinate from the same origin.
Centre of mass of a uniform triangular lamina and of a uniform semicircular lamina: Triangle: on a median, one third of the way from the side to the opposite vertex. Semicircle of radius r: on the axis of symmetry, 4r/(3π) from the diameter.
Centre of mass of a uniform sector of radius r and half-angle α: On the axis of symmetry, 2r sin α/(3α) from the centre (α in radians).
A lamina hangs freely from a point. Where is its centre of mass?: Directly below the point of suspension. Use the right-angled triangle formed by the suspension point, G and an edge to find the angle asked for.
Body on a rough slope: conditions for sliding and for toppling: Slips when tan α > μ. Topples when the vertical through G passes beyond the lowest point of contact, at tan α = (half the base)/(height of G). Whichever angle is smaller happens first.

Kinetic energy and gravitational potential energy: KE = ½mv² (J). GPE = mgh (J), where h is the vertical height above a chosen zero level. Moving a distance d up a slope at angle α raises it by d sin α.
Work done by a constant force: Work = Fd cos θ, where θ is the angle between the force and the direction of motion. A force at right angles to the motion (such as a normal reaction) does no work.
The work-energy principle: Initial energy + work done by driving forces − work done against resistances = final energy.
Power: Power is the rate of doing work, in watts. For a vehicle, P = Fv, where F is the driving force and v the speed.
A lorry with engine power 30 kW has resistance 1500 N on level ground. Find its maximum speed.: At maximum speed a = 0, so driving force = resistance = 1500 N. v = P/F = 30 000/1500 = 20 m s⁻¹.
A 1200 kg car at 20 m s⁻¹ coasts against a constant 400 N resistance. How far does it go before stopping?: Work-energy: ½ × 1200 × 20² = 400d, so 240 000 = 400d and d = 600 m.

Newton's law of restitution: Speed of separation = e × speed of approach, with 0 ≤ e ≤ 1. For particle 1 behind particle 2: v₂ − v₁ = e(u₁ − u₂).
What do e = 1 and e = 0 mean?: e = 1: no loss of mechanical energy (perfectly elastic). e = 0: the particles move on together after impact. 0 < e < 1: some kinetic energy is lost.
A (1 kg, 6 m s⁻¹) hits B (2 kg, at rest) with e = 0.5. Find the speeds after.: Momentum: v_A + 2v_B = 6. Restitution: v_B − v_A = 0.5 × 6 = 3. Adding: 3v_B = 9, so v_B = 3 m s⁻¹ and v_A = 0.
Direct impact with a fixed wall: Speed after = e × speed before, with the direction reversed. Momentum of the particle is not conserved, because the wall gives an impulse. A ball dropped from height h rebounds to e²h.
Kinetic energy and momentum in a collision between two particles: Momentum is conserved whatever e is (no external impulse along the line). Kinetic energy is conserved only when e = 1. Loss of KE = KE before − KE after, always ≥ 0.
Successive impacts: how do you decide whether there is another collision?: After each impact update every velocity. If the particle behind is moving faster in the same direction as the one in front (or they are moving towards each other), they collide again.

Conditions for a rigid body to be in equilibrium: The resultant force is zero (resolve in two directions) and the sum of the moments about any point is zero.
Moment of a force F at angle θ to a rod, acting a distance d along the rod from the pivot: Moment = F d sin θ (force × perpendicular distance).
Ladder resting against a smooth wall on rough ground: method: Forces: weight at the centre of mass, wall reaction S (horizontal), ground reaction R (vertical), friction F at the foot. Resolve: F = S and R = total weight. Take moments about the foot. Then F ≤ μR.
A uniform ladder rests at 50° to rough ground against a smooth wall. Find the least μ.: Moments about the foot: S L sin 50° = W (L/2) cos 50°, so S = W/(2 tan 50°) = 0.420W. F = S and R = W, so μ ≥ 0.420 (3 s.f.).
Smooth wall compared with rough wall: A smooth wall gives only a reaction perpendicular to the wall. A rough wall also gives friction along the wall, which must be included in the vertical equation.