Kinematics describes motion without asking what causes it. In the 2025 SL course you need to tell distance from displacement and average from instantaneous values, use the four equations of uniformly accelerated motion, and read velocity and acceleration from the gradients and areas of motion graphs.
Projectiles are analysed by splitting the motion into independent horizontal and vertical components, with a constant g and no air resistance. Fluid resistance is tested qualitatively only: its effect on the range, the time of flight, the shape of the path and terminal speed.
31 questions
196 marks
Paper 1A: 15
Paper 1B: 7
Paper 2: 9
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31 practice questions on A.1 Kinematics
1A-1A-01
Distance, displacement & averages·A.1 Kinematics
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksState
An athlete runs once around a circular track of circumference 400 m, returning to the starting point after 50 s.
What are the magnitude of the average velocity and the average speed of the athlete?
Average velocity / m s−1Average speed / m s−1
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Marking point
Mark
Notes
Step 1Displacement is the change in position. After one complete lap the athlete is back at the start, so the displacement is 0 and the average velocity is 0.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Distance is the total path length, 400 m, so the average speed is 400 / 50 = 8.0 m s−1.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the displacement after one complete lap is zero, so the average velocity is zero; the distance travelled is 400 m, so the average speed is 8.0 m s⁻¹.
BThe average speed cannot be zero: the athlete covers a distance of 400 m, and average speed is distance divided by time.
CThis reverses the two quantities. Velocity uses displacement, which is zero here; speed uses distance, which is not.
DThis treats displacement as equal to distance. They are equal only for motion in a straight line without reversal.
Syllabus understandingA.1 — the difference between distance and displacement; the difference between instantaneous and average values of velocity, speed and acceleration Command term: State
2A-1A-02
Equations of motion·A.1 Kinematics
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A stone is thrown vertically upwards with speed u from the edge of a cliff. It later lands in the sea, a vertical distance h below the point of release. Air resistance is negligible.
What is the speed of the stone as it reaches the sea?
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Marking point
Mark
Notes
Step 1Take upwards as positive. Over the whole flight the displacement is s = −h and the acceleration is a = −g; the path up and back down does not need to be split.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2v² = u² + 2as = u² + 2(−g)(−h) = u² + 2gh, so the speed is √(u² + 2gh).
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis is the speed of a stone dropped from rest. The stone returns to the cliff edge moving downwards at u, so its initial speed cannot be ignored.
BThis takes the displacement as +h while keeping a = −g: a sign error. The stone ends up below the point of release, so s and a have the same sign.
CThis adds the two speeds as numbers. The contributions combine through v² = u² + 2as, not by addition.
DCorrect: with s = −h and a = −g, v² = u² + 2gh. The same result follows from energy conservation, ½mv² = ½mu² + mgh.
Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by v² = u² + 2as; the change in position is the displacement Command term: Deduce
3A-1A-03
Projectile motion·A.1 Kinematics
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Two identical stones are projected from level ground with the same speed. Stone P is projected at 30° above the horizontal and stone Q at 60° above the horizontal. Air resistance is negligible.
What are the ratios of the range and of the maximum height of Q to those of P?
Range of Q ÷ range of PMaximum height of Q ÷ maximum height of P
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Notes
Step 1Range = (u cos θ)(2u sin θ/g) ∝ sin θ cos θ, and sin 60° cos 60° = sin 30° cos 30°, so the ranges are equal: ratio 1.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Maximum height = (u sin θ)²/2g ∝ sin²θ, so the ratio is (sin 60°/sin 30°)² = (√3)² = 3.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe range is right, but the height ratio uses sin 60°/sin 30° without squaring. The height depends on the square of the vertical component, (u sin θ)²/2g.
BCorrect: Q stays in the air √3 times longer but moves horizontally √3 times more slowly, so the ranges are equal; the heights are in the ratio sin²60°/sin²30° = 3.
CThe height ratio is right, but the range is taken to depend only on the time of flight. Q's horizontal component, u cos 60°, is smaller by the same factor √3.
DThis scales both quantities with the vertical component of velocity alone, forgetting the smaller horizontal component for the range and the square for the height.
Syllabus understandingA.1 — the behaviour of projectiles in the absence of fluid resistance, and the application of the equations of motion resolved into vertical and horizontal components Command term: Determine
4A-1A-04
Equations of motion·A.1 Kinematics
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A driver sees a hazard and, after a fixed reaction time, brakes with a constant deceleration until the car stops. The reaction time and the deceleration do not depend on the speed of the car.
The initial speed of the car is doubled. What happens to the distance travelled during the reaction time and to the braking distance?
Distance during reaction timeBraking distance
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Mark
Notes
Step 1During the reaction time the speed is constant, so that distance is ut, proportional to u: × 2.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2While braking, 0 = u² − 2as gives s = u²/2a, proportional to u²: × 4.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the reaction distance ut doubles, and the braking distance u²/2a increases by 2² = 4.
BThis assumes the braking time stays the same. At twice the speed the car needs twice as long to stop (t = u/a) at twice the average speed, so the braking distance increases four times.
CThis applies the square law to both distances. During the reaction time there is no acceleration, so that distance is simply ut.
DThis swaps the two dependences: it is the braking distance that goes as u² and the reaction distance as u.
Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion; how the analysis of motion can be used to solve real-life problems Command term: Determine
5A-1B-01
Uncertainties & processing data·A.1 Kinematics
Paper 1BEasy10 marks
Data-based question10 steps to full marksDetermine
A student measures her reaction time with a ruler-drop test. A partner holds a 30 cm ruler vertically, with its zero mark level with the top of the student's open thumb and finger, and releases it without warning. The student catches the ruler as quickly as possible and the distance d through which it has fallen is read at the top of the thumb. The ruler is graduated in millimetres.
Ten trials are made with the student's dominant hand. The ruler falls freely from rest.
Trial
1
2
3
4
5
6
7
8
9
10
d / cm
19.4
21.0
18.2
20.6
17.9
19.8
22.1
18.8
20.3
19.5
(a)
State the resolution of the ruler.
(1)
(b)
(i)
Calculate the mean value of d.
(1)
(ii)
Estimate the absolute uncertainty in d, using half the range of the readings.
(1)
(c)
Show that the mean reaction time t of the student is about 0.20 s.
(1)
(d)
Determine the reaction time together with its absolute uncertainty.
(3)
(e)
The test is repeated with the non-dominant hand, giving t = 0.213 ± 0.010 s. The student concludes that her reaction time is longer with her non-dominant hand. Discuss whether the data support this conclusion.
(2)
(f)
Suggest one improvement to the procedure that would make the comparison in (e) more reliable.
(1)
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Marking point
Mark
Notes
Part (a)
0.1 cm OR 1 mm
✓ 1
Accept ±0.05 cm OR ±0.5 mm. Do not accept 0.1 m or 1 cm.
Part (b)(i)
mean d = 19.76 ≈ 19.8 cm
✓ 1
Accept 19.76 cm.
Part (b)(ii)
½ × (22.1 − 17.9) = 2.1 cm
✓ 1
Accept 2 cm.
Part (c)
t = √(2d/g) = √(2 × 0.1976 / 9.81) = 0.201 s
✓ 1
Must see full substitution OR answer to at least 3 s.f. Use of d in cm (giving 2.0 s) scores 0.
Part (d)
fractional uncertainty in d = 2.1 / 19.8 = 0.106 OR 10.6 %
✓ 1
Allow ECF from (b). Accept 0.11 OR 11 %.
t ∝ d1/2, so fractional uncertainty in t = ½ × 0.106 = 0.053 OR 5.3 %
✓ 1
A candidate who doubles instead of halving (21 %) loses MP2 but may score MP3 by ECF.
t = 0.201 ± 0.011 s OR t = 0.20 ± 0.01 s
✓ 1
MP3 is for matching the precision of value and uncertainty; unit required. Accept Δt = 0.010–0.011 s.
Part (e)
ranges are 0.190–0.212 s «dominant» and 0.203–0.223 s «non-dominant», so the ranges overlap
✓ 1
Accept the equivalent statement that the difference in the means, 0.012 s, is smaller than the combined uncertainty, 0.021 s. Allow ECF from (d).
so the difference is not significant and the data do not support the conclusion
✓ 1
OWTTE. Do not award for 'the non-dominant time is larger, so the conclusion is correct'.
Part (f)
any one specific improvement, eg: many more trials with each hand to reduce the random uncertainty in each mean; alternate the hands between trials to remove a learning or fatigue effect; vary the delay before release so the student cannot anticipate it; keep the same partner and starting position for every trial
✓ 1
Do not accept 'repeat the experiment' or 'be more careful' without a purpose.
Answers: (b)(i) 19.8 cm · (b)(ii) ±2.1 cm · (c) 0.201 s · (d) 0.201 ± 0.011 s (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by s = ut + ½at²; Tool 3 — propagate uncertainties in processed data in calculations involving raising to a power Command term: Determine
6A-1B-02
Distance, displacement & averages·A.1 Kinematics
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
A trolley is released from rest at the same point on a straight inclined track for every run. A light gate is fixed at a point P further down the track. Cards of different length L are mounted on the trolley in turn, always with their leading edge in the same place, so the leading edge of every card reaches P with the same instantaneous speed u. The timer records the time Δt for which each card blocks the beam and the average speed vavg = L/Δt is calculated.
The uncertainty in each value of vavg is ±0.010 m s−1. The graph shows vavg against Δt with error bars.
L / m
Δt / s
vavg / m s−1
0.025
0.0384
0.651
0.050
0.0747
0.669
0.075
0.1075
0.100
0.1400
0.714
0.150
0.1971
0.200
0.2517
0.795
Graph drawn to scale
(a)
The trolley has a constant acceleration a. Show that vavg = u + ½aΔt.
(1)
(b)
Complete the table by calculating the two missing values of vavg.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
(i)
Determine u, the instantaneous speed of the trolley at P.
(1)
(ii)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in u.
(2)
(e)
Determine the acceleration of the trolley.
(2)
(f)
(i)
The light-gate software reports L/Δt as “the speed at P”. Explain why this value is always greater than u in this experiment.
(1)
(ii)
A student suggests using a card only 5 mm long so that L/Δt is closer to u. Suggest a disadvantage of this.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
s = ut + ½at² with s = L and t = Δt, so L/Δt = u + ½aΔt
✓ 1
Accept the argument that for uniform acceleration the average velocity equals the velocity at the mid-time, u + a(Δt/2).
Part (b)
0.075/0.1075 = 0.698 «m s−1»
✓ 1
0.150/0.1971 = 0.761 «m s−1»
✓ 1
Both values to 3 s.f. for the second mark. Award [1] for one correct value.
Part (c)
single straight line passing through all error bars
✓ 1
Do not accept a line joined point to point or a line forced through the origin.
Part (d)(i)
u = intercept on the vavg axis = 0.621 «m s−1»
✓ 1
Accept 0.612–0.630 m s−1.
Part (d)(ii)
lines of maximum and minimum gradient drawn through all error bars, extrapolated to Δt = 0
✓ 1
Δu = ½ × difference of their intercepts ≈ 0.011 m s−1, so u = 0.62 ± 0.01 m s−1
✓ 1
Accept 0.005–0.015 m s−1. Allow ECF from their lines.
Part (e)
gradient = ½a; gradient read from line ≈ 0.69 m s−2
✓ 1
a = 2 × gradient = 1.38 m s−2
✓ 1
Accept 1.30–1.46 m s−2. Award [1] for the gradient alone.
Part (f)(i)
the trolley speeds up while the card passes P, so the average speed over Δt exceeds the speed of the leading edge at P
✓ 1
OWTTE. Accept a numerical illustration, eg the 0.100 m card gives 0.714 m s−1, about 15 % more than u.
Part (f)(ii)
the percentage uncertainty in L «and in Δt» becomes very large, eg 0.5 mm in 5 mm is 10 %
✓ 1
OWTTE.
Answers: (b) 0.698 and 0.761 m s−1 · (d)(i) 0.62 m s−1 · (d)(ii) ±0.01 m s−1 · (e) 1.4 m s−2(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them; Tool 3 — determining the uncertainty in gradients and intercepts Command term: Determine
7A-2-01
Motion graphs·A.1 Kinematics
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksSketch
An electric scooter and its rider, of total mass 95 kg, travel along a straight, level cycle path. Figure 1 shows how the velocity v of the scooter varies with time t.
Figure 2 is a set of axes for part (c).
Figure 1 — graph drawn to scaleFigure 2 — axes for (c)
(a)
(i)
State what is represented by the gradient of a velocity–time graph.
(1)
(ii)
Calculate the acceleration of the scooter during the first 4.0 s.
(1)
(b)
(i)
Show that the scooter travels about 70 m in the 15 s shown.
(1)
(ii)
Calculate the average speed of the scooter for the 15 s.
(1)
(c)
Sketch, on Figure 2, a graph to show how the acceleration a of the scooter varies with time from t = 0 to t = 15 s.
(3)
(d)
(i)
Calculate the magnitude of the resultant force on the scooter and rider between t = 12 s and t = 15 s.
(1)
(ii)
The rider brakes between t = 12 s and t = 15 s. Explain why the force exerted by the brakes is smaller than your answer to (d)(i).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
acceleration
✓ 1
Do not accept “speed” or “change in velocity”.
Part (a)(ii)
a = 6.0 / 4.0 = 1.5 m s−2
✓ 1
Part (b)(i)
area under the graph = ½ × 4.0 × 6.0 + 8.0 × 6.0 + ½ × 3.0 × 6.0 = 12 + 48 + 9 «= 69 m»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept the trapezium ½(8.0 + 15) × 6.0.
Part (b)(ii)
69 / 15 = 4.6 m s−1
✓ 1
Allow ECF from (b)(i). Accept 4.7 m s−1 from 70 m.
Part (c)
horizontal line at a = +1.5 m s−2 from 0 to 4.0 s
✓ 1
Allow ECF from (a)(ii).
a = 0 «line along the time axis» from 4.0 s to 12 s
✓ 1
horizontal line at a = −2.0 m s−2 from 12 s to 15 s, below the time axis
✓ 1
Vertical joining lines may be shown or omitted. Do not award MP3 for a positive value.
Part (d)(i)
F = ma = 95 × 2.0 = 190 N
✓ 1
Accept 1.9 × 102 N.
Part (d)(ii)
air resistance/rolling friction also act on the scooter opposite to its velocity «in the same direction as the braking force»
✓ 1
the resultant force is the sum of the braking force and these resistive forces, so the brakes provide less than 190 N
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Answers: (a)(ii) 1.5 m s−2 · (b)(i) 69 m · (b)(ii) 4.6 m s−1 · (d)(i) 190 N (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — velocity is the rate of change of position, and acceleration is the rate of change of velocity; the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them; motion with uniform and non-uniform acceleration Command term: Sketch
8A-2-02
Projectile motion·A.1 Kinematics
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksCalculate
A garden sprinkler at ground level sends out a jet of water at a speed of 13 m s−1 at an angle of 32° above the horizontal. The ground is level. Each small mass of water in the jet may be treated as a projectile, and air resistance is negligible.
(a)
(i)
Show that the vertical component of the initial velocity of the water is about 7 m s−1.
(1)
(ii)
Calculate the horizontal component of the initial velocity of the water.
(1)
(b)
Calculate the maximum height reached by the water.
(2)
(c)
A flower bed, in line with the jet, extends from 14.0 m to 17.0 m from the sprinkler. Determine whether the water lands in the flower bed.
(3)
(d)
State and explain the speed of the water as it reaches the ground.
(2)
(e)
Calculate the height of the water above the ground at a horizontal distance of 6.0 m from the sprinkler.
(2)
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Marking point
Mark
Notes
Part (a)(i)
uy = 13 sin 32° «= 6.89 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
ux = 13 cos 32° = 11.0 m s−1
✓ 1
Accept 11 m s−1.
Part (b)
v² = u² − 2gh with v = 0 at the top: h = 6.89² / (2 × 9.81)
✓ 1
Accept h = uyt − ½gt² with t = uy/g.
h = 2.42 m
✓ 1
Accept 2.4–2.5 m «using 7 m s−1». Award [2] for CNA.
Part (c)
time of flight = 2 × 6.89 / 9.81 = 1.40 s
✓ 1
Allow use of symmetry: twice the time to the highest point.
horizontal distance = 11.02 × 1.404 = 15.5 m
✓ 1
Accept 15.4–15.7 m. Allow ECF from (a).
15.5 m is between 14.0 m and 17.0 m, so the water lands in the flower bed
✓ 1
The decision must be consistent with the candidate’s distance. Do not award MP3 for a bare “yes”.
Part (d)
13 m s−1 «same as the launch speed»
✓ 1
horizontal component unchanged AND vertical component has the same magnitude as at launch «by symmetry» OR water returns to the same height so the gravitational potential energy and hence the kinetic energy are the same as at launch
✓ 1
OWTTE.
Part (e)
time to reach 6.0 m: t = 6.0 / 11.0 = 0.545 s
✓ 1
Allow ECF from (a)(ii).
height = 6.89 × 0.545 − ½ × 9.81 × 0.545² = 2.30 m
✓ 1
Accept 2.3–2.4 m «using 7 m s−1». Award [2] for CNA. Award [1 max] for 3.76 m, which omits the ½gt² term.
Answers: (a)(i) 6.89 m s−1 · (a)(ii) 11.0 m s−1 · (b) 2.42 m · (c) 15.5 m; yes · (d) 13 m s−1 · (e) 2.30 m(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the behaviour of projectiles in the absence of fluid resistance, and the application of the equations of motion resolved into vertical and horizontal components Command term: Calculate
9A-2-03
Two-body kinematics·A.1 Kinematics
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksDetermine
In a relay race the incoming runner X runs at a constant speed of 9.0 m s−1. The outgoing runner Y waits at the start of a 20 m exchange zone. Y starts from rest at the instant that X is a distance d behind Y, and accelerates uniformly at 2.5 m s−2 until reaching 9.0 m s−1. Y then runs at this constant speed.
The baton is passed at the instant X draws level with Y. Both runners move along the same straight line.
Figure 1 — axes for (d)
(a)
Calculate the time taken for Y to reach a speed of 9.0 m s−1.
(1)
(b)
Show that, during this time, X runs about 16 m further than Y.
(2)
(c)
The coach sets d = 12 m. Determine the distance from the start of the exchange zone at which the baton is passed.
(3)
(d)
Sketch, on Figure 1, velocity–time graphs for X and for Y from t = 0 to t = 5.0 s when d = 12 m. Ignore any change in motion caused by the exchange. Label each graph.
(3)
(e)
(i)
Explain why, if X has not drawn level with Y by t = 3.6 s, X will never draw level with Y.
(2)
(ii)
The coach considers increasing d to 18 m. Determine whether the baton can then be passed.
(2)
(f)
Suggest one advantage of choosing d = 15 m rather than d = 12 m.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
t = 9.0 / 2.5 = 3.6 s
✓ 1
Part (b)
X: 9.0 × 3.6 = 32.4 m AND Y: ½ × 2.5 × 3.6² = 16.2 m
✓ 1
Must see both displacements.
difference = 32.4 − 16.2 «= 16.2 m»
✓ 1
Accept the area between the two velocity–time graphs, ½ × 3.6 × 9.0. Must see full substitution OR answer to 3 s.f.
Part (c)
draws level when 9.0t = 12 + ½ × 2.5t²
✓ 1
Accept 1.25t² − 9.0t + 12 = 0.
t = 1.77 s «smaller root»
✓ 1
The larger root is later than 3.6 s, after Y stops accelerating, so it is rejected; X has already drawn level.
distance = ½ × 2.5 × 1.77² = 3.9 m
✓ 1
Accept 3.8–4.0 m. Award [3] for CNA.
Part (d)
X: horizontal line at 9.0 m s−1 from 0 to 5.0 s
✓ 1
Y: straight line from the origin reaching 9.0 m s−1 at 3.6 s
✓ 1
Y: horizontal at 9.0 m s−1 from 3.6 s, coinciding with X’s line; both graphs labelled
✓ 1
Allow ECF from (a) for the time.
Part (e)(i)
before this time X is faster than Y, so the gap closes «by the area between the graphs»
✓ 1
after 3.6 s the speeds are equal so the gap stays constant «and is never closed»
✓ 1
OWTTE.
Part (e)(ii)
the greatest distance X can gain is 16.2 m
✓ 1
Allow ECF from (b).
16.2 m < 18 m so X never draws level AND the baton cannot be passed
✓ 1
ALTERNATIVE: 1.25t² − 9.0t + 18 = 0 has no real solution «81 − 90 < 0» so X never draws level.
Part (f)
the exchange happens later when Y is running faster «6.6 m s−1 instead of 4.4 m s−1», so less time is lost passing the baton
✓ 1
Accept “still within the zone «at 8.6 m» but closer to X’s speed”. OWTTE.
Answers: (a) 3.6 s · (b) 16.2 m · (c) 3.9 m (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion; motion with uniform and non-uniform acceleration Command term: Determine
10A-1A-14
Motion graphs·A.1 Kinematics
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
The graph shows how the displacement of a body varies with time.
At which labelled point is the body momentarily at rest?
Graph drawn to scaleShow mark scheme
Marking point
Mark
Notes
Step 1Velocity is the gradient of a displacement–time graph, so the body is at rest wherever that graph is horizontal — at Q.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThe graph rises steadily here, so the displacement is changing and the body is moving at 10 m s⁻¹.
BThis is the steepest rising section, so the body is moving fastest here, not slowest.
CThe graph is falling, so the body is moving back towards the origin at 10 m s⁻¹. A negative velocity is not a zero velocity.
DCorrect: the graph is horizontal, so the displacement is not changing. The gradient — and therefore the velocity — is zero.
Syllabus understandingA.1 — velocity is the rate of change of position, and acceleration is the rate of change of velocity Command term: Identify
11A-1A-15
Distance, displacement & averages·A.1 Kinematics
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDetermine
A hiker walks 3.0 km due north and then 4.0 km due east.
What are the distance travelled and the magnitude of the displacement?
Distance / kmDisplacement / km
Show mark scheme
Marking point
Mark
Notes
Step 1Distance is the total path length: 3.0 + 4.0 = 7.0 km. Displacement is the straight line from start to finish: √(3.0² + 4.0²) = 5.0 km.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: the distance is 3.0 + 4.0 = 7.0 km along the path, and since the legs are at right angles the displacement is √(3.0² + 4.0²) = 5.0 km.
BThese are the right two numbers the wrong way round. Distance is the length of the path walked, which cannot be less than the displacement.
C4.0 − 3.0 would be the displacement only if the two legs were along the same line in opposite directions.
DThis adds the two legs as though they were in the same direction. Displacement is a vector, so perpendicular legs must be combined with Pythagoras.
Syllabus understandingA.1 — the difference between distance and displacement Command term: Determine
12A-1A-16
Equations of motion·A.1 Kinematics
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDeduce
A body moves in a straight line with uniform acceleration. Its speed increases from u to v while it travels a distance s.
What is the time taken?
Show mark scheme
Marking point
Mark
Notes
Step 1For uniform acceleration the average speed is (u + v)/2, so s = (u + v)t/2 and t = 2s/(u + v).
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis divides by the sum of the speeds rather than by their mean, so the time is half the true value.
BThis uses the change in speed, v − u, in place of the sum. The distance depends on the average speed, not on how much the speed changes.
CCorrect: s = ½(u + v)t, so t = 2s/(u + v).
DThis inverts the relation; it has the units of s−1, not of time.
Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by s = (u + v)t/2 Command term: Deduce
13A-1A-17
Distance, displacement & averages·A.1 Kinematics
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A cyclist travels a distance d along a straight road at a constant speed v, and then a further distance d at a constant speed 2v.
What is the average speed of the cyclist for the whole journey?
Show mark scheme
Marking point
Mark
Notes
Step 1Total time = d/v + d/2v = 3d/2v.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Average speed = total distance ÷ total time = 2d ÷ (3d/2v) = 4v/3.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis divides only one of the two distances, d, by the total time. The total distance is 2d.
BCorrect: the journey takes 3d/2v, so the average speed is 2d ÷ 3d/2v = 4v/3.
CThis is the mean of the two speeds, which would be right for equal times at each speed. The cyclist spends twice as long at the lower speed, so the average is nearer to v.
DThis is the instantaneous speed during the second half, not the average speed over the journey.
Syllabus understandingA.1 — the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them Command term: Deduce
14A-1A-18
Projectile motion·A.1 Kinematics
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A ball is projected horizontally from the top of a vertical cliff and lands on the level ground below. Air resistance is negligible.
The ball is projected horizontally again from the same point with twice the speed. What happens to the time of flight and to the horizontal distance travelled?
Time of flightHorizontal distance
Show mark scheme
Marking point
Mark
Notes
Step 1The vertical motion starts from zero vertical velocity in both cases and covers the same height, so the time of flight is unchanged: × 1.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The horizontal velocity is constant, so the horizontal distance = uxt doubles: × 2.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis assumes a faster ball reaches the ground sooner, so that the two effects cancel. The horizontal velocity has no effect on the vertical motion.
BThis assumes the faster ball also stays in the air longer. The time of flight depends only on the height and on g.
CThe time is right, but the distance is taken to depend on the square of the launch speed. With the time fixed, the horizontal distance is simply proportional to ux.
DCorrect: the time of flight is set by the vertical motion alone, √(2h/g), so it is unchanged, and the horizontal distance uxt doubles.
Syllabus understandingA.1 — the behaviour of projectiles in the absence of fluid resistance, and the application of the equations of motion resolved into vertical and horizontal components Command term: Determine
15A-1A-19
Projectile motion·A.1 Kinematics
Paper 1AHard1 mark
Multiple choice · 1 mark4 steps to full marksDetermine
A ball is projected from level ground at 25 m s−1 at 40° above the horizontal. Air resistance is negligible. Take g = 9.81 m s−2.
How far from the launch point does it land?
Show mark scheme
Marking point
Mark
Notes
Step 1Resolve the launch velocity: ux = 25 cos 40° = 19.15 m s⁻¹ and uy = 25 sin 40° = 16.07 m s⁻¹.
—
All 4 steps must be completed — there is no mark for a part-answer.
Step 2The ball lands at the same height, so its vertical displacement is zero: 0 = uyt − ½gt², giving t = 2uy / g.
—
Step 3t = 2 × 16.07 / 9.81 = 3.276 s.
—
Step 4Range = uxt = 19.15 × 3.276 = 62.7 ≈ 63 m.
✓ 1
Answer A
Answer: A · 4 stages of work, one mark
Every option, and why
ACorrect: u_x = 19.15 m s⁻¹, the time of flight is 2 × 25 sin 40° / 9.81 = 3.276 s, and 19.15 × 3.276 = 62.7 ≈ 63 m.
BThis uses sin 40° instead of sin 80°. The range formula contains sin 2θ, not sin θ.
CThis is the horizontal distance at the top of the flight — half the range. The ball is still in the air for the same time again on the way down.
DThis is the maximum height, (25 sin 40°)² / 2g, not the horizontal range.
Syllabus understandingA.1 — the behaviour of projectiles in the absence of fluid resistance, and the application of the equations of motion resolved into vertical and horizontal components Command term: Determine
16A-1A-20
Fluid resistance & terminal speed·A.1 Kinematics
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A skydiver falls vertically. The air resistance on her is proportional to the square of her speed, and her terminal speed is vT.
What is the magnitude of her acceleration, as a fraction of g, at the instant her speed is ½vT?
Show mark scheme
Marking point
Mark
Notes
Step 1At terminal speed the air resistance equals the weight mg. Air resistance ∝ v², so at ½vT it is (½)² × mg = ¼mg.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Resultant force = mg − ¼mg = ¾mg, so a = ¾g.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
A¼ is the air resistance as a fraction of the weight. The acceleration comes from the resultant force, mg − ¼mg.
BThis takes the air resistance to be proportional to v, so that it halves to ½mg. The stem says it is proportional to v².
CCorrect: the air resistance is (½)² = ¼ of the weight, so the resultant force is ¾mg and a = ¾g.
DThis ignores the air resistance, which is still a quarter of the weight at this speed.
Syllabus understandingA.1 — the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed; A.2 — Newton's three laws of motion Command term: Determine
17A-1A-21
Non-uniform acceleration·A.1 Kinematics
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A body starts from rest and moves in a straight line. Its acceleration increases uniformly with time from zero, as shown on the graph. The speed of the body at time T is v.
What is the speed of the body at time 2T?
Graph NOT drawn to scaleShow mark scheme
Marking point
Mark
Notes
Step 1The change in speed is the area under the acceleration–time graph. Up to time T the area is a triangle: v = ½ × T × kT = ½kT², where a = kt.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Up to time 2T the triangle has twice the base and twice the height: area = ½ × 2T × 2kT = 2kT² = 4v.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis assumes the speed is proportional to time, which is true only for uniform acceleration. Here the acceleration itself grows.
BThis is the area between T and 2T only, ½(kT + 2kT)T = 3v: the increase in speed. The speed v already gained at T must be added.
CCorrect: the area under the graph grows as t², so doubling the time multiplies the speed by 4.
DThis cubes the time ratio. A quantity proportional to t³ here is the displacement, not the speed.
Syllabus understandingA.1 — motion with uniform and non-uniform acceleration; velocity is the rate of change of position, and acceleration is the rate of change of velocity Command term: Determine
18A-1B-07
Non-uniform acceleration·A.1 Kinematics
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
A student places a smartphone flat on the floor of a lift and uses its accelerometer app to record the vertical acceleration a while the lift travels upwards from rest to rest. Upwards is positive. The graph shows the smoothed data-logger trace.
Before the lift starts to move, the phone gives a steady reading of +0.06 m s−2.
Graph drawn to scale
(a)
(i)
State the type of error that the reading before the lift starts reveals.
(1)
(ii)
Outline how the student should correct the readings.
(1)
(b)
Outline how the graph shows that the lift moves with constant velocity between t = 3.4 s and t = 12.0 s.
(1)
(c)
Determine the maximum speed of the lift.
(2)
(d)
The value of the offset is known to ±0.02 m s−2. Determine the absolute uncertainty in the maximum speed.
(1)
(e)
The manufacturer states that the lift travels at 1.25 m s−1. Comment on whether the result of (c) agrees with this.
(1)
(f)
The floors of the building are 3.50 m apart. Estimate the distance travelled by the lift and hence deduce the number of floors it rises.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
systematic error «zero / offset error»
✓ 1
Do not accept 'random error'.
Part (a)(ii)
subtract 0.06 m s−2 from every reading
✓ 1
OWTTE, eg re-zero (calibrate) the sensor before the journey and repeat.
Part (b)
the corrected acceleration is zero «reading equals the offset» in that interval, so the velocity does not change
✓ 1
Part (c)
area under the corrected graph from 1.0 s to 3.4 s = ½ × (2.4 + 0.8) × 0.80
✓ 1
Award [1] max for 1.42 m s−1 from the uncorrected trace.
vmax = 1.28 m s−1
✓ 1
Accept 1.24–1.32 m s−1.
Part (d)
Δv = 0.02 × 2.4 = 0.048 ≈ 0.05 m s−1
✓ 1
Accept 0.04–0.05 m s−1.
Part (e)
range 1.23–1.33 m s−1 includes 1.25 m s−1, so the result agrees «within uncertainty»
✓ 1
Allow ECF from (c) and (d).
Part (f)
while speeding up and slowing down the average speed is ½vmax «symmetrical trace» OR uses the area under a trapezium-shaped v–t graph
✓ 1
distance = 1.28 × (8.6 + 1.2 + 1.2) = 14.1 m
✓ 1
Accept 13.6–14.5 m. Allow ECF from (c).
14.1 / 3.50 = 4.0, so the lift rises 4 floors
✓ 1
Must be a whole number.
Answers: (c) 1.28 m s−1 · (d) ±0.05 m s−1 · (f) 14 m, 4 floors (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — motion with uniform and non-uniform acceleration; velocity is the rate of change of position, and acceleration is the rate of change of velocity Command term: Determine
19A-1B-08
Projectile motion·A.1 Kinematics
Paper 1BHard12 marks
Data-based question12 steps to full marksDetermine
A student films a small ball thrown upwards at an angle and analyses the video with tracking software. A metre rule placed in the picture is used to set the scale. The software gives the horizontal displacement x and the vertical displacement y of the ball from its release point at time t after release. Air resistance is negligible.
The uncertainty in each value of y is ±0.010 m. The graph shows y/t plotted against t; the error bars show the uncertainty in y/t.
t / s
x / m
y / m
(y/t) / m s−1
0.10
0.305
0.390
3.90
0.20
0.599
0.674
3.37
0.30
0.904
0.887
0.40
1.199
0.989
2.47
0.50
1.507
1.015
2.03
0.60
1.806
0.930
1.55
0.70
2.104
0.772
0.80
2.409
0.504
0.63
Graph drawn to scale
(a)
(i)
Outline how the data show that the horizontal component of velocity is constant.
(1)
(ii)
Show that the horizontal component of velocity is about 3.0 m s−1.
(1)
(b)
Complete the table.
(2)
(c)
The vertical launch velocity is uy. Explain why a graph of y/t against t is a straight line, and state what its gradient represents.
(2)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine the acceleration of free fall g from the graph.
(2)
(f)
The lines of maximum and minimum gradient give an uncertainty in g of ±0.2 m s−2. Comment on the result of (e).
(1)
(g)
The metre rule was held closer to the camera than the plane in which the ball moved. Explain how this accounts for the result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
x increases by equal amounts «≈ 0.30 m» in equal time intervals
✓ 1
OWTTE.
Part (a)(ii)
ux = 2.409 / 0.80 = 3.01 m s−1
✓ 1
Must see a full substitution OR answer to 3 s.f. Any row or a gradient of x against t is acceptable.
Part (b)
0.887/0.30 = 2.96
✓ 1
0.772/0.70 = 1.10
✓ 1
Award [1] for one correct value. Accept 3 s.f.
Part (c)
y = uyt − ½gt² so y/t = uy − ½gt, which is linear in t
✓ 1
gradient = −g/2 «intercept = uy»
✓ 1
Accept 'gradient = −½g'.
Part (d)
single straight line through all error bars «drawn with a ruler»
✓ 1
Part (e)
gradient read from their line ≈ −4.63 m s−2
✓ 1
Sign of the gradient not required for MP1.
g = −2 × gradient = 9.3 m s−2
✓ 1
Accept 9.0–9.45 m s−2.
Part (f)
range 9.1–9.5 m s−2 does not include 9.81 m s−2, so there is a systematic error «not explained by random error»
✓ 1
Allow ECF from (e).
Part (g)
the rule appears larger than it would in the plane of the ball, so the scale gives too few metres per pixel and every measured distance of the ball is too small by the same factor
✓ 1
g «and ux, uy» are proportional to the distance scale, so g is too small «by about 6 %»
✓ 1
OWTTE.
Answers: (a)(ii) 3.01 m s−1 · (b) 2.96 and 1.10 m s−1 · (e) 9.3 m s−2(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the behaviour of projectiles in the absence of fluid resistance, and the application of the equations of motion resolved into vertical and horizontal components; Tool 3 — linearize graphs Command term: Determine
20A-1B-09
Fluid resistance & terminal speed·A.1 Kinematics
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
A paper cone is released from rest directly below a motion sensor. The sensor records the distance s fallen by the cone every 0.10 s. The data are shown in the table and plotted on the graph.
t / s
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
s / m
0.000
0.049
0.176
0.363
0.572
0.805
1.035
1.278
1.512
1.755
1.996
2.231
2.474
Graph drawn to scale
(a)
Draw the curve of best fit on the graph.
(1)
(b)
Assuming that the acceleration of the cone is uniform during the first 0.10 s, show that it is about 10 m s−2.
(1)
(c)
Explain, with reference to the forces acting on the cone, why the gradient of the graph becomes constant.
(2)
(d)
Determine the terminal speed of the cone.
(2)
(e)
Calculate the time the cone would take to fall 2.00 m from rest if there were no air resistance, and compare it with the time taken in the experiment.
(2)
(f)
The student suspects that the cone was released a short time after the sensor started recording. State and explain whether this affects the value found in (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
smooth curve through «or close to» all points, becoming a straight line after about 0.5 s
✓ 1
Do not accept point-to-point joining.
Part (b)
a = 2s/t² = 2 × 0.049 / 0.10² = 9.8 m s−2
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
air resistance increases as the speed increases while the weight stays constant, so the resultant force «and acceleration» decreases
✓ 1
when air resistance equals the weight the resultant force is zero, so the velocity «gradient of s–t» is constant
✓ 1
Part (d)
gradient of the straight part of the graph, using a large triangle «eg from 0.6 s to 1.2 s»
✓ 1
Award [0] for a single point s/t.
terminal speed = 2.40 m s−1
✓ 1
Accept 2.3–2.5 m s−1.
Part (e)
t = √(2 × 2.00 / 9.81) = 0.64 s
✓ 1
the cone takes about 1.00 s, so air resistance increases the time of fall «by about 0.36 s / a factor of 1.6»
✓ 1
Accept 0.98–1.02 s read from the table or graph.
Part (f)
no effect
✓ 1
Award MP1 only if a valid reason is given.
a delay shifts every reading by the same time, moving the graph sideways without changing the gradient of the straight section
✓ 1
OWTTE.
Answers: (b) 9.8 m s−2 · (d) 2.4 m s−1 · (e) 0.64 s (no air) vs 1.00 s (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed Command term: Determine
21A-2-11
Distance, displacement & averages·A.1 Kinematics
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksCalculate
A survey drone flies at a constant height above a field. Starting from point B it flies 400 m due east at a constant speed of 10 m s−1 (leg 1), then 300 m due north at a constant speed of 6.0 m s−1 (leg 2). It then hovers for 20 s before flying in a straight line directly back to B at a constant speed of 12.5 m s−1.
Figure 1 shows legs 1 and 2 drawn to scale.
Figure 1 — drawn to scale on a square grid
(a)
(i)
Draw, on Figure 1, an arrow to represent the displacement of the drone from B at the end of leg 2.
(1)
(ii)
Calculate the magnitude and direction of this displacement.
(2)
(b)
(i)
Show that the whole flight, including the hover, lasts 150 s.
(1)
(ii)
Calculate the average speed of the drone for the whole flight.
(2)
(iii)
State the average velocity of the drone for the whole flight.
(1)
(c)
(i)
Calculate the magnitude of the average velocity of the drone for legs 1 and 2 together.
(1)
(ii)
Explain why your answer to (c)(i) is less than the average speed for legs 1 and 2.
(2)
(iii)
State the instantaneous speed of the drone at t = 60 s after it leaves B.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
single straight arrow from B to the end of leg 2 «(400 m, 300 m)», arrowhead at the end of leg 2
✓ 1
Part (a)(ii)
√(400² + 300²) = 500 m
✓ 1
tan−1(300/400) = 37° north of east
✓ 1
Accept a bearing of 053° or 53° east of north. A direction is needed for MP2.
Must see the return time calculated from the 500 m displacement.
Part (b)(ii)
total distance = 400 + 300 + 500 = 1200 m
✓ 1
Allow ECF from (a)(ii).
average speed = 1200 / 150 = 8.0 m s−1
✓ 1
Award [2] for CNA.
Part (b)(iii)
zero «the displacement for the whole flight is zero»
✓ 1
Part (c)(i)
500 / 90 = 5.6 m s−1
✓ 1
Allow ECF from (a)(ii).
Part (c)(ii)
average speed uses the distance travelled «700 m» but average velocity uses the displacement «500 m» over the same time
✓ 1
the displacement is the straight line from B, which is shorter than the path flown «because the drone changed direction»
✓ 1
OWTTE.
Part (c)(iii)
6.0 m s−1
✓ 1
The drone is on leg 2 from 40 s to 90 s.
Answers: (a)(ii) 500 m, 37° north of east · (b)(ii) 8.0 m s−1 · (c)(i) 5.6 m s−1 · (c)(iii) 6.0 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the change in position is the displacement; the difference between distance and displacement; the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them Command term: Calculate
22A-2-12
Equations of motion·A.1 Kinematics
Paper 2Medium11 marks
Short answer & extended response11 steps to full marksDetermine
An aircraft starts from rest at one end of a straight, level runway of length 1800 m. It accelerates uniformly at 2.0 m s−2 and leaves the ground when its speed reaches 70 m s−1.
If the pilot abandons the take-off, the brakes give the aircraft a constant deceleration of 3.5 m s−2. The pilot’s reaction time may be neglected.
Figure 1 — axes for (b)(ii)
(a)
(i)
Calculate the time taken for the aircraft to reach its take-off speed.
(1)
(ii)
Show that the distance needed for take-off is about 1.2 km.
(1)
(b)
On one flight the pilot abandons the take-off when the speed is 50 m s−1.
(i)
Calculate the total distance travelled by the aircraft from the start until it stops.
(2)
(ii)
Sketch, on Figure 1, the velocity–time graph for this flight from the start until the aircraft stops. Add values to your graph where appropriate.
(2)
(c)
Determine the greatest speed at which the take-off can be abandoned so that the aircraft stops before the end of the runway.
(3)
(d)
On a wet runway the braking deceleration is smaller, but the acceleration is unchanged. State and explain the effect on your answer to (c).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
t = 70/2.0 = 35 s
✓ 1
Part (a)(ii)
s = 70²/(2 × 2.0) = 1225 «m»
✓ 1
ALTERNATIVE: s = ½ × 2.0 × 35². Must see full substitution OR answer to 3 s.f.
Part (b)(i)
accelerating: 50²/(2 × 2.0) = 625 m AND braking: 50²/(2 × 3.5) = 357 m
✓ 1
total = 982 m
✓ 1
Accept 980 m. Award [2] for CNA. Award [1 max] for 625 m or 357 m alone.
Part (b)(ii)
straight line from the origin to 50 m s−1 at 25 s
✓ 1
straight, steeper line from that point down to zero at 39 s
✓ 1
Accept 39–40 s. Allow ECF from (b)(i).
Part (c)
total distance = v²/(2 × 2.0) + v²/(2 × 3.5) = 1800 m
✓ 1
v² = 4580 «m² s−2»
✓ 1
v = 68 m s−1
✓ 1
Award [3] for CNA. Award [0] for 112 m s−1 «whole runway used for braking».
Part (d)
at any speed the braking distance v²/2a is greater, so less of the runway is left for accelerating
✓ 1
so the greatest speed at which the take-off can be abandoned is lower
✓ 1
MP2 only scores if MP1 scores.
Answers: (a)(i) 35 s · (a)(ii) 1225 m · (b)(i) 982 m · (c) 68 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by v = u + at and v² = u² + 2as; velocity is the rate of change of position, and acceleration is the rate of change of velocity Command term: Determine
23A-2-13
Projectile motion·A.1 Kinematics
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
A delivery drone flies horizontally at a constant velocity of 12 m s−1 at a height of 30 m above level ground. It releases a package, which falls to the ground. Air resistance on the package is negligible unless stated otherwise.
Figure 1 — drawn to scale
(a)
(i)
State the horizontal and the vertical components of the velocity of the package at the instant of release.
(1)
(ii)
Show that the package takes about 2.5 s to reach the ground.
(1)
(iii)
Calculate the horizontal distance travelled by the package before it lands.
(1)
(b)
Determine the velocity of the package just before it lands.
(3)
(c)
Sketch, on Figure 1, the path of the package as seen by an observer at rest on the ground. Mark with a cross the position of the drone at the instant the package lands.
(2)
(d)
The package can survive an impact at a speed of up to 25 m s−1. Determine the greatest height from which the drone, flying at 12 m s−1, can release it safely.
(3)
(e)
In practice air resistance on the package is not negligible. State and explain where the package lands relative to the point directly below the drone at that instant.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
horizontal: 12 m s−1 AND vertical: zero
✓ 1
Both needed.
Part (a)(ii)
t = √(2 × 30 / 9.81) «= 2.47 s»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(iii)
12 × 2.47 = 29.7 m
✓ 1
Accept 30 m from 2.5 s.
Part (b)
vy = 9.81 × 2.47 = 24.3 m s−1
✓ 1
Accept √(2 × 9.81 × 30).
v = √(12² + 24.3²) = 27.1 m s−1
✓ 1
Allow ECF from (a)(ii).
tan−1(24.3/12) = 64° below the horizontal
✓ 1
Direction must be stated as below the horizontal «or 26° to the vertical». Accept 63°–64°.
Part (c)
curve from the release point, horizontal at the start and becoming steeper, reaching the ground at ≈ 30 m
✓ 1
Allow ECF from (a)(iii). Do not accept a straight line.
cross on the drone’s path vertically above the landing point
✓ 1
The drone and the package have the same horizontal velocity throughout.
Award [1 max] for 31.9 m, which ignores the horizontal component of the velocity.
Part (e)
behind the drone «short of the point below it»
✓ 1
air resistance reduces the horizontal velocity of the package while the drone keeps moving at 12 m s−1
✓ 1
MP2 only scores if MP1 scores.
Answers: (a)(ii) 2.47 s · (a)(iii) 29.7 m · (b) 27.1 m s−1 at 64° below the horizontal · (d) 24.5 m (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the behaviour of projectiles in the absence of fluid resistance, and the application of the equations of motion resolved into vertical and horizontal components; the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed Command term: Determine
24A-2-14
Fluid resistance & terminal speed·A.1 Kinematics
Paper 2Medium14 marks
Short answer & extended response14 steps to full marksExplain
A badminton shuttlecock of mass 5.0 g is hit from point O with a speed of 14 m s−1 at 40° above the horizontal. The dashed curve in Figure 1 shows, to scale, the path the shuttlecock would follow if there were no air resistance.
In reality the air resistance on a shuttlecock is large and increases with its speed. A shuttlecock dropped from a great height reaches a terminal speed of 6.8 m s−1.
Figure 1 — graph drawn to scale
(a)
(i)
Calculate the weight of the shuttlecock.
(1)
(ii)
State the magnitude of the air resistance on the shuttlecock when it falls vertically at its terminal speed, and explain your answer.
(2)
(b)
Sketch, on Figure 1, the path of the shuttlecock when air resistance acts. It leaves O with the same velocity as before.
(3)
(c)
At the highest point of its real path the shuttlecock is moving horizontally. State and explain the direction of its acceleration at this point.
(2)
(d)
Explain, in terms of the forces acting, why the shuttlecock takes longer to fall from its highest point than to rise to it.
(3)
(e)
A player argues that, because the range of a projectile is proportional to the square of its launch speed, hitting the shuttlecock twice as fast will make it travel four times as far. Evaluate this argument.
(3)
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Part (a)(i)
W = 5.0 × 10−3 × 9.81 = 4.9 × 10−2 N
✓ 1
Part (a)(ii)
4.9 × 10−2 N «equal to the weight»
✓ 1
Allow ECF from (a)(i).
the velocity is constant so the resultant force is zero «Newton’s first law»
✓ 1
MP2 needs the reason, not just “terminal speed”.
Part (b)
lower maximum height than the dashed curve
✓ 1
shorter range «lands closer to O»
✓ 1
asymmetric path: the highest point is reached before half of the horizontal distance AND the descent is steeper than the ascent «almost vertical near the end»
✓ 1
Both features needed for MP3. Do not award MP3 for a symmetric curve.
Part (c)
weight acts vertically downwards AND air resistance acts horizontally, opposite to the velocity
✓ 1
so the acceleration is directed downwards and backwards «below the horizontal, opposite to the motion», with a magnitude greater than g
✓ 1
Do not accept “vertically downwards”. MP2 only scores if MP1 scores.
Part (d)
on the way up the air resistance has a downward component that adds to the weight, so the vertical deceleration is greater than g
✓ 1
on the way down the air resistance has an upward component that opposes the weight, so the downward acceleration is less than g
✓ 1
the same vertical distance is covered with a smaller average vertical speed on the way down, so the fall takes longer
✓ 1
OWTTE.
Part (e)
the argument holds without air resistance: time of flight and horizontal velocity are each proportional to the launch speed
✓ 1
air resistance increases with speed, so a faster shuttlecock experiences a much larger resistive force and loses speed more rapidly
✓ 1
so the range increases by much less than four times «the argument does not apply to a shuttlecock»
✓ 1
MP3 is for a conclusion supported by the effect of air resistance.
Answers: (a)(i) 4.9 × 10−2 N · (a)(ii) 4.9 × 10−2 N (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed Command term: Explain
25A-2-15
Non-uniform acceleration·A.1 Kinematics
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksSketch
A maglev test vehicle of mass 2.4 × 104 kg starts from rest at t = 0 and moves along a straight, horizontal track. Figure 1 shows how its acceleration a varies with time t for the first 30 s.
Figure 1 — graph drawn to scaleFigure 2 — axes for (b)
(a)
(i)
Calculate the speed of the vehicle at t = 8.0 s.
(1)
(ii)
Deduce the maximum speed of the vehicle.
(2)
(b)
Sketch, on Figure 2, the variation with time of the velocity of the vehicle from t = 0 to t = 30 s.
(3)
(c)
Explain why the speed of the vehicle continues to increase until t = 20 s even though its acceleration decreases from t = 8.0 s onwards.
(2)
(d)
Calculate the resultant force on the vehicle at t = 14 s.
(2)
(e)
Outline why the equation s = ut + ½at² cannot be used to find the distance travelled between t = 8.0 s and t = 20 s, and suggest how this distance could be found.
(2)
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Part (a)(i)
area = 1.5 × 8.0 = 12 m s−1
✓ 1
Part (a)(ii)
area of the triangle from 8.0 s to 20 s = ½ × 12 × 1.5 = 9.0 m s−1
✓ 1
maximum speed = 12 + 9.0 = 21 m s−1
✓ 1
Allow ECF from (a)(i). Award [2] for CNA.
Part (b)
straight line from the origin to 12 m s−1 at 8.0 s
✓ 1
Allow ECF from (a).
curve from 8.0 s to 20 s with decreasing gradient, reaching 21 m s−1 at 20 s
✓ 1
horizontal line at 21 m s−1 from 20 s to 30 s, joined smoothly «zero gradient at 20 s»
✓ 1
Do not award MP3 if the curve has a maximum before 20 s or falls after 20 s.
Part (c)
the acceleration is positive «in the direction of motion» until 20 s, so the speed keeps increasing
✓ 1
a decreasing acceleration means the speed increases at a smaller rate; after 20 s a = 0 so the speed is constant
✓ 1
OWTTE.
Part (d)
a = 0.75 m s−2 «read from the graph»
✓ 1
F = 2.4 × 104 × 0.75 = 1.8 × 104 N
✓ 1
Allow ECF for a.
Part (e)
the equation applies only to uniform acceleration, and a changes during this interval
✓ 1
the distance is the area under the velocity–time graph «e.g. by counting squares»
✓ 1
Answers: (a)(i) 12 m s−1 · (a)(ii) 21 m s−1 · (d) 1.8 × 104 N (the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — motion with uniform and non-uniform acceleration; velocity is the rate of change of position, and acceleration is the rate of change of velocity Command term: Sketch
26A-1A-36
Fluid resistance & terminal speed·A.1 Kinematics
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksIdentify
A ball is projected from level ground. In each graph the dashed line shows how the height h of the ball varies with horizontal distance x when air resistance is negligible.
Which solid line shows the trajectory of the same ball, launched with the same velocity, when air resistance is not negligible?
Sketch graphs, not to scaleShow mark scheme
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Notes
Step 1Air resistance opposes the velocity throughout, so the ball rises less high and travels less far, and the horizontal velocity keeps falling; the path is therefore no longer symmetric: the ball comes down more steeply than it went up, and the highest point lies beyond the middle of the range.
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThe height and range are reduced correctly, but the path is still a symmetric parabola. With air resistance the horizontal velocity falls during the flight, so the descent is steeper than the ascent.
BCorrect: lower maximum height, shorter range, and an asymmetric path whose descent is steeper than its ascent, because the horizontal velocity keeps decreasing.
CThis path is asymmetric the wrong way round: it rises more steeply than it falls. The ball is moving horizontally more slowly on the way down, so the descent is the steeper part (and the launch direction is the same as without air resistance).
DThis assumes air resistance acts only on the vertical motion, so that the range is unchanged. It also opposes the horizontal motion, so the range is reduced.
Syllabus understandingA.1 — the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed Command term: Identify
27A-1A-37
Velocity & acceleration·A.1 Kinematics
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksDeduce
A body moves along a straight line.
Which statement about the motion of the body must be correct?
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Notes
Step 1Acceleration is the rate of change of velocity, so zero acceleration means the velocity, and therefore the speed, does not change; each of the other statements has a counter-example (a ball at the top of its flight; a body speeding up in the negative direction; a body that goes out and comes back).
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AA ball thrown upwards has zero velocity at the top of its flight, yet its acceleration is still g downwards. Zero velocity does not mean zero rate of change of velocity.
BCorrect: zero acceleration means zero rate of change of velocity, so the velocity, and with it the speed, stays constant.
CA negative acceleration slows the body only if the velocity is positive. A body moving in the negative direction with negative acceleration is speeding up.
DThis confuses an average value with instantaneous values. A body that moves out and returns to its start has zero average velocity but was moving.
Syllabus understandingA.1 — velocity is the rate of change of position, and acceleration is the rate of change of velocity; the difference between instantaneous and average values of velocity, speed and acceleration Command term: Deduce
28A-1A-38
Motion graphs·A.1 Kinematics
Paper 1AHard1 mark
Multiple choice · 1 mark4 steps to full marksDetermine
The graph shows how the velocity of a body along a straight line varies with time.
What is the displacement of the body over the 10 s shown?
Graph drawn to scaleShow mark scheme
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Notes
Step 1Displacement is the area under a velocity–time graph, counted as negative below the time axis.
—
All 4 steps must be completed — there is no mark for a part-answer.
Step 2From 0 to 6.0 s the area is ½ × 6.0 × 12 = +36 m.
—
Step 3The line falls from +12 m s⁻¹ to −8.0 m s⁻¹ over 4.0 s, so it crosses zero at 6.0 + 4.0 × 12/20 = 8.4 s. From 6.0 to 8.4 s the area is ½ × 2.4 × 12 = +14.4 m.
—
Step 4From 8.4 to 10 s the area is ½ × 1.6 × 8.0 = 6.4 m below the axis, so the displacement is 36 + 14.4 − 6.4 = 44 m.
✓ 1
Answer B
Answer: B · 4 stages of work, one mark
Every option, and why
AThis adds all three areas as positive, which gives the total distance travelled (56.8 m). Displacement counts area below the axis as negative.
BCorrect: the areas are +36 m (0 to 6.0 s), +14.4 m (6.0 to 8.4 s) and −6.4 m (8.4 to 10 s), giving 36 + 14.4 − 6.4 = 44 m.
CThis is the area of the first triangle only, up to 6.0 s. The motion continues for another 4.0 s.
DThis treats the whole section from 6.0 s to 10 s as a single triangle below the axis. Part of that section is still above the axis.
Syllabus understandingA.1 — the change in position is the displacement; the difference between distance and displacement; motion with uniform and non-uniform acceleration Command term: Determine
29A-1B-17
Motion graphs·A.1 Kinematics
Paper 1BEasy8 marks
Data-based question8 steps to full marksDetermine
A sports-science database publishes the times at which a sprinter passed each 10 m mark in a 100 m race, measured from the starting signal by a laser system. The times are given to the nearest 0.01 s.
The graph shows the average speed in each 10 m interval plotted against the time at the middle of that interval. The error bars show the uncertainty in each average speed.
Distance / m
Split time / s
Time for the 10 m interval / s
Average speed in interval / m s−1
10
2.01
2.01
4.98
20
3.07
1.06
9.43
30
4.04
0.97
40
4.97
0.93
10.8
50
5.90
0.93
10.8
60
6.81
0.91
11.0
70
7.74
0.93
10.8
80
8.69
0.95
10.5
90
9.65
0.96
10.4
100
10.63
0.98
Graph drawn to scale
(a)
Complete the table.
(2)
(b)
Outline why each average speed is plotted at the middle time of its interval.
(1)
(c)
Draw the curve of best fit on the graph.
(1)
(d)
Each split time is uncertain by ±0.01 s. Determine the absolute uncertainty in the average speed for the interval from 50 m to 60 m.
(2)
(e)
The sprinter's coach claims that the sprinter slows down over the last 40 m of the race. Discuss whether the data support this claim.
(2)
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Notes
Part (a)
20 m to 30 m: 10 / 0.97 = 10.3 «m s−1»
✓ 1
90 m to 100 m: 10 / 0.98 = 10.2 «m s−1»
✓ 1
Accept answers to 2 or 3 s.f. Award [1] for one correct value.
Part (b)
the average speed over a short interval is approximately the instantaneous speed at the middle of the interval «exactly so for uniform acceleration»
✓ 1
OWTTE.
Part (c)
smooth curve rising steeply, levelling off near 11 m s−1 and falling slightly at the end, passing through the error bars
✓ 1
Do not accept point-to-point joining.
Part (d)
uncertainty in the interval time = ±0.02 s, so fractional uncertainty = 0.02 / 0.91 = 0.022 «2.2 %»
✓ 1
Award [0] for using ±0.01 s: two readings are subtracted.
Δv = 0.022 × 11.0 = 0.24 ≈ 0.2 m s−1
✓ 1
Accept 0.2–0.25 m s−1.
Part (e)
50–60 m: 11.0 ± 0.2 m s−1; 90–100 m: 10.2 ± 0.2 m s−1, so the ranges do not overlap
✓ 1
Allow ECF from (d). Any fair comparison of an interval near 60 m with the last interval.
the decrease is larger than the uncertainties, so the data support the claim
✓ 1
Do not award MP2 without reference to uncertainty.
Answers: (a) 10.3 and 10.2 m s−1 · (d) ±0.2 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them; motion with uniform and non-uniform acceleration Command term: Determine
30A-1B-18
Equations of motion·A.1 Kinematics
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
A road-safety agency publishes typical stopping distances for a car on a dry road. The thinking distance is the distance travelled at constant speed during the driver's reaction time; the braking distance dB is the distance travelled after the brakes are applied. The published distances are rounded to the nearest metre.
A student tests the hypothesis that dB = kvn, where v is the speed of the car and k and n are constants.
v / km h−1
v / m s−1
Thinking distance / m
dB / m
lg(v / m s−1)
lg(dB / m)
30
8.33
6
5
0.921
0.699
50
13.9
9
15
1.143
1.176
70
19.4
13
29
90
25.0
17
48
1.398
1.681
110
30.6
21
72
130
36.1
25
100
1.558
2.000
Graph drawn to scale
(a)
Show that the agency assumes a reaction time of about 0.7 s.
(1)
(b)
Complete the table.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine n.
(2)
(e)
Explain how your value of n is consistent with the car having a constant deceleration while braking.
(2)
(f)
Determine the deceleration assumed by the agency.
(2)
(g)
Suggest why the point for 30 km h−1 lies furthest from the line.
(1)
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Notes
Part (a)
eg 17 / 25.0 = 0.68 s
✓ 1
Any row. Must see substitution of a thinking distance and a speed in m s−1.
Part (b)
70 km h−1: 1.289 and 1.462
✓ 1
110 km h−1: 1.485 and 1.857
✓ 1
Accept 2 or 3 decimal places. Award [1] for any three correct values.
Part (c)
single straight line with points evenly scattered on either side
✓ 1
Line need not pass through the 30 km h−1 point.
Part (d)
gradient calculated from a large triangle on the line
✓ 1
n = 2.0
✓ 1
Accept 1.9–2.1.
Part (e)
with final speed zero, v² = u² + 2as gives dB = v²/2a
✓ 1
so for constant a, dB ∝ v², that is n = 2, as found
✓ 1
Allow ECF from (d) if the value is close to 2.
Part (f)
uses a = v²/2dB with a point on the line OR intercept = −lg 2a
✓ 1
Award [0] for use of the thinking distance.
a = 6.5 m s−2
✓ 1
Accept 6.2–6.9 m s−2.
Part (g)
rounding to ±0.5 m is a much larger fraction of 5 m «10 %» than of the larger distances, so lg dB is least precise there
✓ 1
OWTTE.
Answers: (a) 0.68 s · (b) 1.289, 1.462; 1.485, 1.857 · (d) n = 2.0 · (f) 6.5 m s−2(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by v² = u² + 2as; Tool 3 — construct and interpret graphs using logarithmic scales Command term: Determine
31A-2-29
Distance, displacement & averages·A.1 Kinematics
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
A diver jumps from a springboard 3.0 m above the surface of the water in a pool. She leaves the board moving vertically upwards at 4.2 m s−1, rises, falls into the water and slows down until she momentarily stops below the surface. Treat the diver as a point; air resistance is negligible.
The graph shows how her height above the water surface varies with time t from the moment she leaves the board. Upwards is positive.
Graph drawn to scale
(a)
Using the graph, determine the distance travelled by the diver, and her displacement, between leaving the board and reaching her lowest point.
(3)
(b)
(i)
Draw a tangent to the graph at t = 1.00 s and use it to determine the velocity of the diver at this instant.
(2)
(ii)
Calculate the speed of the diver as she reaches the water.
(2)
(c)
Calculate the average velocity of the diver between leaving the board and reaching the water.
(2)
(d)
Describe, with reference to the graph, how the acceleration of the diver changes after she enters the water.
(2)
(e)
Estimate the average acceleration of the diver while she is slowing down in the water.
(2)
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Notes
Part (a)
greatest height 3.9 m, so she rises 0.9 m above the board
✓ 1
Accept 3.8–4.0 m.
distance = 0.9 + 3.9 + 2.4 = 7.2 m
✓ 1
Accept 7.0–7.4 m.
displacement = 5.4 m downwards «−5.4 m»
✓ 1
Direction or sign required. Accept 5.3–5.5 m.
Part (b)(i)
tangent drawn touching the curve at t = 1.00 s
✓ 1
gradient = −5.6 m s−1 «5.6 m s−1 downwards»
✓ 1
Accept 5.2–6.0 m s−1 downwards. The equations of motion give v = 4.2 − 9.81 × 1.00 = −5.61 m s−1.
Part (b)(ii)
v² = 4.2² + 2 × 9.81 × 3.0
✓ 1
Accept energy conservation. Accept v = u − gt with t read from the graph.
v = 8.7 m s−1
✓ 1
Accept 8.75 m s−1.
Part (c)
displacement = −3.0 m in 1.32 s
✓ 1
Accept 1.3 s from the graph.
average velocity = −2.3 m s−1 «2.3 m s−1 downwards»
✓ 1
Direction or sign required. Accept (u + v)/2 = (4.2 − 8.7)/2, valid because the acceleration is uniform in the air.
Part (d)
the acceleration is upwards «opposite to the velocity»: the gradient of the graph decreases in magnitude to zero
✓ 1
the magnitude of the acceleration decreases «as her speed falls», so the acceleration is non-uniform «it is greatest just after she enters the water»
✓ 1
Part (e)
time to stop ≈ 2.1 − 1.3 = 0.8 s
✓ 1
Accept 0.7–0.9 s.
a ≈ 8.7/0.8 = 11 m s−2 «upwards»
✓ 1
Accept 9.5–13 m s−2. Allow ECF from (b)(ii).
Answers: (a) 7.2 m; 5.4 m downwards · (b)(i) −5.6 m s−1 · (b)(ii) 8.7 m s−1 · (c) −2.3 m s−1 · (e) 11 m s−2(the remaining parts are explanations — see the table above)
Syllabus understandingA.1 — the difference between distance and displacement; the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them; motion with uniform and non-uniform acceleration; the equations of motion for solving problems with uniformly accelerated motion Command term: Determine
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