NCEA Level 2 Physics 91171 Mechanics: past papers, practice and marking
- NCEA Level 2
- Physics
- AS91171
- 6 credits
- External exam
About 91171
AS91171, Demonstrate understanding of mechanics. Motion, projectiles, circular motion, forces and torques, momentum, springs and energy: the 6-credit external, in the six question types it keeps reusing.
What 91171 covers
- KinematicsEquations of motion vf = vi + at, d = vit + ½at², d = (vi + vf)t/2, vf² = vi² + 2ad, negative acceleration, free fall at 9.8 m s⁻², unbalanced force and acceleration
- Projectile motionComponents v cos θ and v sin θ, maximum height, time of flight, range, launch height, clearing an obstacle, angle of the final velocity, explaining the path with force, acceleration and velocity
- Circular motionFc = mv²/r and ac = v²/r, what provides the centripetal force, accelerating at constant speed, moving off at a tangent
- Forces and torqueForce diagrams, vector addition, force components, equilibrium, τ = Fd, support forces and tensions, Newton’s third law
- MomentumMomentum p = mv as a vector, impulse = change in momentum = FΔt, safety features, conservation of momentum in collisions, coupling and explosions
- EnergyF = −kx, spring constant from data, springs combined, Ep = ½kx², W = Fd, ΔEp = mgΔh, Ek = ½mv², P = W/t, conservation of energy
91171 past papers
Open a 91171 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 2 Physics past papers
Practice questions
Question 1: Kinematics
A cyclist speeds up uniformly from 4.0 m s⁻¹ to 10.0 m s⁻¹ over 42 m of straight, flat road.
- Calculate the acceleration of the cyclist.(up to Achieved)
- Calculate the time taken.(up to Achieved)
Check your answer
Part a: what gets the marks
- No time is given, so use vf² = vi² + 2ad: 10.0² = 4.0² + 2 × a × 42 a = 84/84 = 1.0 m s⁻²
A full answer: a = 1.0 m s⁻²
Part b: what gets the marks
- Use the acceleration: t = (vf − vi)/a = (10.0 − 4.0)/1.0 = 6.0 s
A full answer: t = 6.0 s
What gets each grade
Question 2: Kinematics
A bus pulls away from rest at a stop and accelerates uniformly at 1.6 m s⁻² for 12.5 s along a straight road.
- Calculate the speed of the bus after 12.5 s.(up to Achieved)
- Calculate the distance the bus travels in this time.(up to Achieved)
Check your answer
Part a: what gets the marks
- Constant acceleration from rest: vf = vi + at = 0 + 1.6 × 12.5 = 20 m s⁻¹
A full answer: vf = 20 m s⁻¹
Part b: what gets the marks
- Displacement from rest: d = vit + ½at² = 0 + ½ × 1.6 × 12.5² = 125 m
A full answer: d = 125 m
What gets each grade
Question 3: Kinematics
A bus travelling at 20 m s⁻¹ along a straight road brakes uniformly and stops in 50.0 m.
- Calculate the acceleration of the bus and the time it takes to stop, and explain the sign of the acceleration.(up to Merit)
- The driver brakes with the same deceleration from 10 m s⁻¹ instead. Discuss how the stopping distance compares, and what this means for a driver who goes faster.(up to Excellence)
Check your answer
Part a: what gets the marks
- Acceleration from the stopping distance: vf² = vi² + 2ad 0 = 20² + 2 × a × 50.0 a = −400/100 = −4.0 m s⁻²
- Time to stop: d = (vi + vf)t/2 50.0 = (20 + 0) × t/2 t = 5.0 s
- With forward as positive, the negative acceleration (−4.0 m s⁻²) points backwards, opposite to the velocity, so the bus slows down.
A full answer: a = −4.0 m s⁻² and t = 5.0 s. The acceleration is negative because it acts opposite to the direction of motion: the velocity is forwards (positive) and decreasing, so the change in velocity is backwards. Assumption: the bus moves in a straight line with constant acceleration.
Part b: what gets the marks
- New stopping distance: d = (vf² − vi²)/2a = (0 − 10²)/(2 × −4.0) = 12.5 m
- Half the speed gives a quarter of the stopping distance (12.5 m against 50.0 m), because d = v²/2a at the same deceleration, so a driver going twice as fast needs four times the distance to stop.
A full answer: From 10 m s⁻¹ the bus stops in 12.5 m, a quarter of 50.0 m, even though the speed is only halved. With the same deceleration, vf² = vi² + 2ad gives d = vi²/2a, so the stopping distance is proportional to the square of the speed. A driver who doubles their speed needs four times the braking distance, which is why a small increase in speed makes a big difference to stopping safely.