NCEA Level 3 Physics 91524 Mechanical Systems: past papers, practice and marking
- NCEA Level 3
- Physics
- AS91524
- 6 credits
- External exam
About 91524
AS91524, Demonstrate understanding of mechanical systems. Circular motion, gravitation, rotation, momentum in two dimensions and simple harmonic motion: the 6-credit external, taught as the handful of question types it actually contains.
What 91524 covers
- Banked cornerBanked curves and level turns: horizontal net force, tan θ = v²/(rg), reaction force
- Vertical loopVertical circles: reaction force at the top and bottom, feeling heavier or lighter, F_R = 0, energy
- Rotational kinematicsConstant angular acceleration equations, rpm and revolutions, v = rω, a = rα
- Rotational inertiaTorque τ = Fr (perpendicular component), τ = Iα, I = Σmr², tension torque, rolling energy
- Angular momentumL = Iω and L = mvr; conserved with no external torque; combining I of two parts
- SHM equationsConditions for SHM, T = 2π√(m/k) and 2π√(l/g), y, v and a equations in radians, reference circle
- SHM graphsPhase of y, v and a; damping; energy transfer; resonance
- Gradient of a graphGraphs of a against y (gradient −ω²), ω against t (gradient α, area θ), damped and SHM displacement
- MomentumCentre of mass in one and two dimensions, planet-moon systems; momentum and impulse in two dimensions
- OrbitsF = GMm/r² as the centripetal force, g = GM/r², orbital speed and period, fields between two bodies, apparent weightlessness
91524 past papers
Open a 91524 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 3 Physics past papers
Practice questions
Question 1: Banked corner
An aircraft flying at 200 m s⁻¹ banks at 35.0° to make a level turn. In level flight the lift force is perpendicular to the wings, exactly as the reaction force on a banked road is perpendicular to the road surface.
- Calculate the radius of the turn.(up to Achieved)
- The pilot wants a tighter turn at the same speed. Explain what must change, and what the consequence is for the lift force.(up to Excellence)
Check your answer
Part a: what gets the marks
- Recognise this as the banked-corner problem with lift in place of the reaction force: Vertical: L cos θ = mg Horizontal: L sin θ = m v² / r
- Divide, exactly as before: tan θ = v² / (r g) → r = v² / (g tan θ)
- Substitute: r = 200² / (9.81 × tan 35.0°) = 40000 / (9.81 × 0.7002) = 40000 / 6.869 = 5.82 × 10³ m
A full answer: r = 5.82 × 10³ m (about 5.8 km).
Part b: what gets the marks
- r = v²/(g tan θ). At fixed v, a smaller r requires a larger tan θ, so a steeper bank angle.
- Follow the consequence through the vertical equation: L = mg / cos θ
- As θ increases, cos θ falls, so L = mg/cos θ rises steeply, faster than the angle does.
A full answer: From r = v²/(g tan θ), at fixed v a smaller r needs a larger tan θ: the pilot must bank more steeply.
The vertical component of lift must still equal the weight while L sin θ grows:
L cos θ = mg, so L = mg / cos θ
As θ increases, cos θ falls, so the lift rises sharply, not linearly (at 35°, L = 1.22 mg; at 60°, L = 2.00 mg).
What gets each grade
Question 2: Banked corner
Two students disagree. Ari says a heavily loaded truck needs a lower design speed on a banked corner than an empty car, because it is heavier. Bree says both vehicles have the same design speed.
- State who is correct, and justify your answer with the physics.(up to Excellence)
- The same corner is to be rebuilt with a radius of 60.0 m and a design speed of 11.2 m s⁻¹. Calculate the bank angle required.(up to Achieved)
Check your answer
Part a: what gets the marks
- Write both equations with the mass left in: Vertical: R cos θ = m g Horizontal: R sin θ = m v² / r
- Divide: tan θ = v² / (r g)
- The mass appears on both sides and cancels completely. It is not that mass has a small effect: it has no effect at all.
- A heavier vehicle needs a proportionally larger centripetal force, and being heavier it also presses into the road harder, so the reaction force, and therefore the horizontal component available, grows in exactly the same proportion. The two effects cancel.
A full answer: Bree is correct.
Vertically: R cos θ = mg
Horizontally: R sin θ = mv²/r
Dividing: tan θ = v²/(rg), so v = √(rg tan θ)
The mass appears on both sides and cancels completely, so it has no effect on the design speed.
Physically: the heavier truck needs a proportionally larger centripetal force, but it also presses harder into the road, so R and its horizontal component grow in the same proportion. The two effects cancel.
Part b: what gets the marks
- Rearrange for the angle: tan θ = v²/(rg) = 11.2² / (60.0 × 9.81) = 125.4 / 588.6 = 0.2131
- Inverse tangent: θ = 12.0°
A full answer: θ = 12.0°