A.1 Kinematics Questions | IB Physics HL 2025

1A-1A-01

Projectile motion·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A projectile is launched with speed u at 60° above the horizontal. Air resistance is negligible.

What is the ratio (kinetic energy at maximum height) / (initial kinetic energy)?

Show mark scheme

Marking point	Mark	Notes
Step 1At the highest point the vertical component of velocity is zero, so the only velocity left is the horizontal component u cos 60°, which never changes.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Kinetic energy is proportional to speed squared, so the ratio is (u cos 60°)²/u² = cos²60°.	—
Step 3cos 60° = 0.5, so cos²60° = 0.25 = 1/4.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: only the horizontal component u cos 60° = 0.5u survives at the top, and (0.5)² = 1/4.
BThis is cos 60°, the ratio of the speeds. Kinetic energy depends on the square of the speed.
CThis is sin²60°, the fraction of the kinetic energy that has been converted to potential energy, not the fraction that remains.
DThis is sin 60°, the fraction of the velocity that is vertical at launch — it has all gone at the top.

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; A.3 — kinetic energy as given by E_k = ½mv² Command term: Determine

2A-1A-02

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A car accelerates uniformly from rest. It travels a distance d during the first t seconds.

What distance does it travel during the next t seconds?

Show mark scheme

Marking point	Mark	Notes
Step 1From rest with constant acceleration, s = ½at², so the distance covered is proportional to the square of the elapsed time.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2After 2t the total distance is (2)² × d = 4d.	—
Step 3The second interval alone covers 4d − d = 3d.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis assumes the speed simply doubles and stays constant in the second interval. The car is still accelerating throughout.
BCorrect: total distance after 2t is 4d, so the second interval covers 4d − d = 3d.
CThis is the total distance after 2t, not the distance in the second interval alone.
DThis would be the total distance after 3t. The question asks only about the second interval.

Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by s = ut + ½at² Command term: Determine

3A-1A-16

Projectile motion·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two identical balls P and Q are projected from the same point on level ground with the same speed. P is projected at 30° above the horizontal and Q at 60° above the horizontal. Air resistance is negligible.

Which statements are correct?
I. P and Q land at the same distance from the point of projection.
II. P and Q spend the same time in the air.
III. The maximum height reached by Q is three times the maximum height reached by P.

Show mark scheme

Marking point	Mark	Notes
Step 1Range = (u cos θ)(2u sin θ/g) ∝ sin θ cos θ. sin 30° cos 30° = sin 60° cos 60°, so the ranges are equal: I is correct.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Time of flight = 2u sin θ/g ∝ sin θ: the ratio is sin 60°/sin 30° = √3, so the times are different: II is wrong.	—
Step 3Maximum height = (u sin θ)²/2g ∝ sin²θ: (√3/2)²/(1/2)² = (3/4)/(1/4) = 3, so III is correct.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis assumes that equal ranges need equal times of flight. Q has a smaller horizontal component and stays in the air √3 times longer.
BCorrect: complementary angles give equal ranges, and the heights are in the ratio sin²60° : sin²30° = 3 : 1.
CThis rejects I, perhaps believing the steeper launch goes further; the range depends on sin 2θ, which is the same for 30° and 60°.
DII is wrong: the times of flight are in the ratio sin 60° : sin 30° = √3 : 1.

4A-1A-18

Motion graphs·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark4 steps to full marksDetermine

A trolley moves along a straight track. The graph shows how its velocity v varies with time t from t = 0 to t = 10 s.

What is the total distance travelled by the trolley in the 10 s?

Velocity v of the trolley against time t (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The velocity is positive from 0 to 8.0 s (where the line crosses the axis) and negative from 8.0 s to 10 s: the trolley reverses.	—	All 4 steps must be completed — there is no mark for a part-answer.
Step 2Distance moved forwards = area of the triangle above the axis = ½ × 8.0 × 4.0 = 16 m.	—
Step 3Distance moved backwards = area of the triangle below the axis = ½ × 2.0 × 2.0 = 2.0 m.	—
Step 4Total distance = 16 + 2.0 = 18 m (the displacement would be 16 − 2.0 = 14 m).	✓ 1	Answer D

Answer: D · 4 stages of work, one mark

Every option, and why

AThis subtracts the area below the axis and also omits the ½ for that triangle: 16 − 2.0 × 2.0 = 12 m.
BThis is the displacement (net signed area, 16 − 2.0 = 14 m). Distance adds the magnitudes of both areas.
CThis counts only the motion before the trolley reverses; it moves another 2.0 m after 8.0 s.
DCorrect: 16 m forwards plus 2.0 m backwards = 18 m.

Syllabus understandingA.1 — the difference between distance and displacement; that the area under a velocity–time graph gives displacement Command term: Determine

5A-1A-23

Relative velocity and vectors·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Car P travels due north at speed v. Car Q travels due east at speed 2v.

What is the velocity of Q relative to P?

Show mark scheme

Marking point	Mark	Notes
Step 1The velocity of Q relative to P is v_Q − v_P = (2v east) + (v south).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The two components are perpendicular: magnitude = √((2v)² + v²) = √5 v.	—
Step 3The east component is larger and the other is southward, so the direction is south of east (tan⁻¹(1/2) = 27° south of east).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: v_Q − v_P has components 2v east and v south.
BThis adds v_P instead of subtracting it: the magnitude happens to be the same, but the direction is reversed in the north–south sense.
CThis subtracts the squares of the speeds, √((2v)² − v²), as if the velocities were not perpendicular.
DThis adds the speeds as scalars and ignores their directions.

Syllabus understandingA.1 — that the motion of bodies through space and time can be described and analysed in terms of position, velocity, and acceleration; Tool 3 — add and subtract vectors in the same plane Command term: Determine

6A-1A-32

Instantaneous and average values·A.1 Kinematics

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A runner moves at constant speed around a circular track of radius r. She completes half a lap in a time t.

What is the magnitude of her average acceleration over this half lap?

Show mark scheme

Marking point	Mark	Notes
Step 1Her speed is v = πr/t (half the circumference in time t).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2After half a lap her velocity is reversed, so the change in velocity has magnitude v − (−v) = 2v = 2πr/t.	—
Step 3Average acceleration = Δv/Δt = 2πr/t². (Her instantaneous acceleration, v²/r = π²r/t², is larger: averaging the direction-changing vector reduces it.)	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the velocity reverses, so |Δv| = 2v = 2πr/t, divided by t.
BThis is the magnitude of the instantaneous centripetal acceleration v²/r, not the average acceleration over the half lap.
CThis takes the change in velocity as v instead of 2v: a reversed velocity changes by twice its magnitude.
DConstant speed does not mean zero acceleration: the direction of the velocity has changed.

Syllabus understandingA.1 — the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them; A.2 — circular motion, centripetal acceleration a = v²/r Command term: Determine

7A-1A-33

Motion graphs·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A motorised camera dolly moves from rest along a straight rail. The graph shows its position x against time t. The tangent to the curve at P, where t = 6.0 s, is drawn.

Which row gives the instantaneous speed of the dolly at t = 6.0 s and its average speed for the first 6.0 s?

Position x of the dolly against time t. The dashed line is the tangent to the curve at P.

Instantaneous speed at 6.0 s / m s⁻¹Average speed 0–6.0 s / m s⁻¹

Show mark scheme

Marking point	Mark	Notes
Step 1Instantaneous speed = gradient of the tangent at P = (10.8 − 0)/(8.0 − 4.0) = 2.7 m s⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Average speed = total distance/total time = 5.4/6.0 = 0.90 m s⁻¹ (the gradient of the chord from the origin to P).	—
Step 3The two differ because the acceleration is not uniform — the curve gets steeper and steeper.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis reads the coordinates of P and divides them (5.4/6.0) for the instantaneous speed — that is the chord gradient, not the tangent gradient.
BThis swaps the two: the tangent gives the instantaneous value (2.7 m s⁻¹), the chord from the origin gives the average (0.90 m s⁻¹).
CCorrect: tangent gradient 2.7 m s⁻¹; chord gradient 0.90 m s⁻¹.
DThis takes the average speed as half the final speed (2.7/2 = 1.35 m s⁻¹), which is only true for uniform acceleration from rest.

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

8A-1A-34

Motion graphs·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

An electric scooter starts from rest and moves in a straight line. The graph shows how its acceleration a varies with time t.

Which row gives the speed of the scooter at t = 4.0 s and the time at which its speed is greatest?

Acceleration a of the scooter against time t.

Speed at 4.0 s / m s⁻¹Speed is greatest

Show mark scheme

Marking point	Mark	Notes
Step 1The change in velocity is the area under the acceleration–time graph: ½ × 3.0 × 4.0 = 6.0 m s⁻¹. Starting from rest, the speed at 4.0 s is 6.0 m s⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration is positive (though decreasing) all the way to 4.0 s, so the speed keeps increasing until then.	—
Step 3After 4.0 s the acceleration is zero, so the speed stays constant at its maximum value of 6.0 m s⁻¹.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

A0.75 is the gradient of the a–t graph (3.0/4.0), which has no useful meaning here; and the speed is zero, not greatest, at t = 0.
BThe area is right, but this confuses the greatest acceleration (at t = 0) with the greatest speed.
CCorrect: area = 6.0 m s⁻¹; the speed rises while a > 0 and is then constant.
DThis uses v = at with the initial acceleration (3.0 × 4.0 = 12 m s⁻¹), treating the non-uniform acceleration as uniform.

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 (area under an acceleration–time graph) Command term: Deduce

9A-1A-35

Projectile motion·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A crew member throws a lifebuoy horizontally from the deck of a ship, 15.0 m above the sea, with a speed of 8.0 m s⁻¹. Air resistance is negligible and g = 9.81 m s⁻².

What is the angle between the velocity of the lifebuoy and the horizontal just before it reaches the sea?

Show mark scheme

Marking point	Mark	Notes
Step 1The horizontal component stays at 8.0 m s⁻¹ because no horizontal force acts.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Vertically from rest: v_y² = 2gs = 2 × 9.81 × 15.0, so v_y = 17.2 m s⁻¹ downwards.	—
Step 3tan θ = 17.2/8.0, so θ = 65° below the horizontal.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis is the angle between the velocity and the vertical: tan⁻¹(8.0/17.2) = 25°.
BThis is the direction of the displacement, not the velocity: the lifebuoy lands 14.0 m out, and tan⁻¹(15.0/14.0) = 47°.
CCorrect: tan⁻¹(v_y/v_x) = tan⁻¹(17.2/8.0) = 65°.
DThis forgets the square root and uses v_y² = 294 in place of v_y: tan⁻¹(294/8.0) = 88°.

10A-1A-36

Fluid resistance and projectiles·A.1 Kinematics

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A badminton shuttlecock is hit upwards at an angle to the horizontal. Air resistance on the shuttlecock is significant.

What is the acceleration of the shuttlecock at the highest point of its path?

Show mark scheme

Marking point	Mark	Notes
Step 1At the highest point the velocity is horizontal (vertical component zero, horizontal component not zero).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Two forces act: the weight mg vertically downwards and the air resistance horizontally backwards, opposite to the velocity.	—
Step 3The resultant is the vector sum, so the acceleration has a vertical component g and a horizontal component backwards: its magnitude is √(g² + (F_d/m)²) > g and it points downwards and backwards.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AOnly the vertical component of the velocity is zero at the top; the weight still acts, so the acceleration cannot be zero.
BAt the top the air resistance is horizontal, so it cannot reduce the vertical component of the acceleration below g.
CThis would be true without air resistance; here the horizontal drag adds a backwards component to the acceleration.
DCorrect: weight (down) plus drag (backwards) give a resultant larger than mg, directed downwards and backwards.

Syllabus understandingA.1 — the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed; A.2 — free-body diagrams analysed to find the resultant force Command term: Deduce

11A-1A-62

Distance and displacement·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A survey drone starts at O and flies 400 m due north in 20 s. It hovers for 10 s and then flies 300 m due east in 20 s, finishing at F, as shown in the plan view.

What is the magnitude of the average velocity of the drone for the whole 50 s?

Plan view of the path of the drone (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1Average velocity = displacement/total time. The displacement is the straight line from O to F.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Displacement = √(400² + 300²) = 500 m.	—
Step 3The total time includes the 10 s hover: average velocity = 500/50 = 10 m s⁻¹.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: displacement 500 m divided by the total time of 50 s.
BThis leaves out the 10 s of hovering: 500/40 = 12.5 m s⁻¹. The average is taken over the whole time interval, including time at rest.
CThis is the average speed, distance/time = 700/50 = 14 m s⁻¹. Average velocity uses the displacement, not the distance travelled.
DThis uses the distance instead of the displacement and also leaves out the hover: 700/40 = 17.5 m s⁻¹.

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 Command term: Determine

12A-1A-63

Projectile motion·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

A ball is thrown from a balcony with a speed of 10.0 m s⁻¹ at 30.0° below the horizontal. It lands on the level ground below 1.20 s later. Air resistance is negligible and g = 9.81 m s⁻².

What is the height of the balcony above the ground?

Show mark scheme

Marking point	Mark	Notes
Step 1Initial vertical component = 10.0 sin 30.0° = 5.00 m s⁻¹, directed downwards.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Taking downwards as positive: s = ut + ½at² = 5.00 × 1.20 + ½ × 9.81 × 1.20².	—
Step 3s = 6.00 + 7.06 = 13.1 m.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis treats the initial vertical component as upwards: −6.00 + 7.06 = 1.06 m. The ball is thrown below the horizontal, so it starts with a downward vertical velocity.
BThis ignores the initial vertical component, as if the ball were thrown horizontally: ½ × 9.81 × 1.20² = 7.06 m.
CCorrect: 5.00 × 1.20 + ½ × 9.81 × 1.20² = 13.1 m.
DThis uses the horizontal component 10.0 cos 30.0° = 8.66 m s⁻¹ as the vertical one: 8.66 × 1.20 + 7.06 = 17.5 m.

13A-1A-64

Motion graphs·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

In a brake test, an electric scooter moving in a straight line decelerates uniformly to rest. The graph shows how the square of its speed, v², varies with its displacement s from the point where the brakes are applied.

Which row gives the magnitude of the deceleration and the time taken to stop?

Square of the speed against displacement for the scooter (drawn to scale).

deceleration / m s⁻²time to stop / s

Show mark scheme

Marking point	Mark	Notes
Step 1From v² = u² + 2as, a graph of v² against s is a straight line of gradient 2a.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradient = −100/20 = −5.0 m s⁻², so a = −2.5 m s⁻²: the deceleration is 2.5 m s⁻².	—
Step 3The intercept gives u² = 100 m² s⁻², so u = 10 m s⁻¹, and t = u/a = 10/2.5 = 4.0 s. (Check: s = ½(u + v)t = ½ × 10 × 4.0 = 20 m.)	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the gradient is 2a, so a = 2.5 m s⁻², and t = 10/2.5 = 4.0 s.
BThe deceleration is right, but the time is found as distance/initial speed, 20/10 = 2.0 s, as if the scooter kept its initial speed. Its average speed is only 5.0 m s⁻¹.
CThe time is right, but the gradient of the v²–s graph has been taken as the acceleration itself; it is 2a.
DThis takes the gradient as the acceleration (5.0 m s⁻²) and then uses t = u/a = 10/5.0 = 2.0 s consistently with that error.

Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by v² = u² + 2as and s = (u + v)t/2; interpreting the gradient and intercept of a graph Command term: Determine

14A-1A-74

Motion graphs·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

An object moves along a straight line. The graph shows how its position x varies with time t. Four points P, Q, R and S are marked on the graph.

At which point is the speed of the object decreasing?

Position x of the object against time t (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The velocity is the gradient of the position–time graph and the speed is the magnitude of the gradient.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At R the gradient is positive and becoming less steep, so the speed is decreasing.	—
Step 3At S the gradient is negative and becoming steeper, so the speed is increasing even though the position is decreasing (velocity and acceleration are both negative).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AAt P the gradient is small but increasing: the object is moving slowly, but it is speeding up. A low speed is not the same as a decreasing speed.
BAt Q the graph is a straight line, so the gradient, and hence the speed, is constant.
CCorrect: at R the gradient is positive and decreasing in magnitude, so the object is slowing down.
DAt S the position is decreasing, but the gradient is getting steeper: the object is moving back towards the origin with increasing speed. A decreasing position is not a decreasing speed.

Syllabus understandingA.1 — velocity is the rate of change of position, and acceleration is the rate of change of velocity; motion of bodies described in terms of position, velocity and acceleration (gradient of a position–time graph) Command term: Deduce

15A-1A-75

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A trolley pulls a paper tape through a ticker-timer that makes 50 dots per second. The diagram shows every fifth dot, so successive dots shown were made 0.10 s apart. The trolley accelerates uniformly.

What is the acceleration of the trolley?

Every fifth dot on the tape; the first dot made is on the left (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The mean speed in each 0.10 s interval is 0.12, 0.22, 0.32 and 0.42 m s⁻¹. For uniform acceleration each mean speed is the instantaneous speed at the middle of its interval.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The middles of successive intervals are 0.10 s apart, so the speed increases by 0.10 m s⁻¹ every 0.10 s.	—
Step 3a = 0.10/0.10 = 1.0 m s⁻². (Equivalently, each gap is longer than the one before by Δs = aT² = 1.0 × 0.10² = 0.010 m.)	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis divides the increase in gap length, 0.010 m, by 0.10 s only once. That gives the change in mean speed in m s⁻¹, not the acceleration.
BThis takes the speed change of 0.10 m s⁻¹ to happen over 0.20 s, the whole duration of two intervals, instead of the 0.10 s between their middles.
CCorrect: the mean speeds rise by 0.10 m s⁻¹ every 0.10 s.
DThis treats the extra 0.010 m as if it were covered from rest, using 0.010 = ½a(0.10)². The extra length per interval is aT², not ½aT².

Syllabus understandingA.1 — the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them; the equations of motion for uniformly accelerated motion Command term: Determine

16A-1A-76

Projectile motion·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A delivery drone flies horizontally at constant velocity. It releases a package and continues at the same velocity. Air resistance on the package is negligible.

Which row describes where the package is relative to the drone while it falls, and the shape of its path as seen by an observer on the ground?

Position relative to the dronePath seen from the ground

Show mark scheme

Marking point	Mark	Notes
Step 1At release the package has the same horizontal velocity as the drone.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2No horizontal force acts on the package, so its horizontal velocity stays equal to that of the drone: it stays vertically below the drone.	—
Step 3Relative to the ground the package has a constant horizontal velocity and a uniform vertical acceleration g, so its path is a parabola.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the horizontal velocities stay equal, and constant horizontal velocity with uniform vertical acceleration gives a parabola.
BThis assumes the package gradually loses its horizontal velocity. With no air resistance there is no horizontal force, so the horizontal velocity cannot change.
CThe position is right, but a vertical straight line is the path seen from the drone. The ground observer also sees the horizontal motion.
DThis assumes the package loses its horizontal velocity at release and falls straight down. It keeps the velocity it had when it was released.

17A-1A-77

Projectile motion·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A ball is projected horizontally with speed u from a height h above level ground. It lands a horizontal distance d from the point directly below the point of projection. Air resistance is negligible.

The ball is now projected horizontally with speed 2u from a height h/2. What is the new horizontal distance?

Show mark scheme

Marking point	Mark	Notes
Step 1The time of fall comes from the vertical motion alone: h = ½gt², so t = √(2h/g) ∝ √h.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The horizontal distance is ut, so d ∝ u√h.	—
Step 3New distance = 2 × √(1/2) × d = √2 d.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis takes the time of fall to be proportional to h, so halving the height halves the time and cancels the doubled speed. The time is proportional to √h.
BCorrect: d ∝ u√h, so the factor is 2 × 1/√2 = √2.
CThis doubles the distance for the doubled speed but ignores the shorter time of fall from the lower height.
DThis inverts the height dependence, multiplying by √2 instead of dividing by √2: it treats the time of fall as ∝ 1/√h.

18A-1A-78

Distance and displacement·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

A ball is dropped from rest from a height of 5.00 m above a floor. It rebounds vertically and is caught at the top of its bounce, 3.20 m above the floor. The time in contact with the floor is negligible, air resistance is negligible and g = 9.81 m s⁻².

What is the magnitude of the average velocity of the ball between its release and the catch?

Show mark scheme

Marking point	Mark	Notes
Step 1Time to fall 5.00 m = √(2 × 5.00/9.81) = 1.010 s; time to rise 3.20 m = √(2 × 3.20/9.81) = 0.808 s; total = 1.82 s.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Displacement = 5.00 − 3.20 = 1.80 m downwards.	—
Step 3Average velocity = 1.80/1.82 = 0.99 m s⁻¹ (downwards).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: displacement 1.80 m divided by the total time of 1.82 s.
BThis divides the displacement by the time of the fall only, 1.01 s. The average is taken over the whole time from release to catch.
CThis uses the 5.00 m drop as the displacement. The ball finishes 3.20 m above the floor, so its displacement is only 1.80 m.
DThis is the average speed: total distance 8.20 m divided by 1.82 s. Average velocity uses the displacement.

19A-1A-79

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

A train slows down uniformly on a straight track. In 40 s its speed falls from 24 m s⁻¹ to 8.0 m s⁻¹.

What distance does the train travel during the last 10 s of this 40 s interval?

Show mark scheme

Marking point	Mark	Notes
Step 1Deceleration = (24 − 8.0)/40 = 0.40 m s⁻².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Speed at the start of the last 10 s (t = 30 s) = 24 − 0.40 × 30 = 12 m s⁻¹.	—
Step 3s = (u + v)t/2 = (12 + 8.0) × 10/2 = 100 m.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses s = ½at² = ½ × 0.40 × 10², a formula for motion starting from rest. The train is moving throughout.
BThis uses the final speed, 8.0 m s⁻¹, as if it were constant for the whole 10 s.
CCorrect: the average speed over the last 10 s is (12 + 8.0)/2 = 10 m s⁻¹.
DThis uses the speed at t = 30 s, 12 m s⁻¹, as if it were constant for the whole 10 s.

Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by s = (u + v)t/2 and v = u + at Command term: Calculate

20A-1A-80

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A car brakes with a constant deceleration. Its speed falls from u to u/3 while it travels a distance d.

What further distance does the car travel before it stops?

Show mark scheme

Marking point	Mark	Notes
Step 1Using v² = u² + 2as for the first stage: u² − u²/9 = 8u²/9 = 2ad.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2For the last stage: (u/3)² = u²/9 = 2ax.	—
Step 3Dividing: x/d = (1/9)/(8/9) = 1/8, so x = d/8.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis compares (u/3)² with u², i.e. with the distance to stop from u, instead of with the distance for the speed to fall from u to u/3, which is proportional to u² − u²/9.
BCorrect: the distances are in the ratio of the changes in v², (1/9) : (8/9).
CThis compares the mean speeds of the two stages, u/6 and 2u/3, as if both stages took the same time. The last stage takes only half as long as the first.
DThis makes the distance proportional to the change in speed (u/3 compared with 2u/3). That is the ratio of the times; the car moves more slowly in the last stage.

Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by v² = u² + 2as Command term: Determine

21A-1A-81

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two stones X and Y are released at the same instant from the edge of a vertical cliff with the same speed u. X is thrown vertically upwards and Y is thrown vertically downwards. Both land on level ground at the foot of the cliff. Air resistance is negligible.

Which statements are correct?
I. X and Y reach the ground with the same speed.
II. X reaches the ground a time 2u/g after Y.
III. X rises a height u²/g above the edge of the cliff.

Show mark scheme

Marking point	Mark	Notes
Step 1I: X returns to the edge of the cliff after 2u/g with speed u downwards (symmetry of the motion with uniform acceleration). From there its motion is identical to that of Y, so both land with the same speed √(u² + 2gH), where H is the height of the cliff.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: because the motion of X after it returns to the edge copies the motion of Y, X lands exactly 2u/g later.	—
Step 3III: at the top v = 0, so 0 = u² − 2gh and h = u²/2g, not u²/g. III is false.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: I and II are true; III is wrong by a factor of 2.
BThis accepts III, which omits the factor 2 in v² = u² + 2as, and rejects II, missing that X passes the edge of the cliff moving down at speed u after a time 2u/g.
CThis rejects I, assuming the stone thrown downwards must land faster. Both start from the same height with the same speed, so they land with the same speed. It also accepts the wrong height in III.
DThis accepts III. The height risen is u²/2g: the factor 2 from v² = u² + 2as has been dropped.

Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by v = u + at and v² = u² + 2as; constant g near the surface of the Earth Command term: Deduce

22A-1A-82

Projectile motion·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

A ball is kicked from level ground with a speed of 16.0 m s⁻¹ at 50.0° above the horizontal. It lands on a flat roof 4.00 m above the ground while it is moving downwards. Air resistance is negligible and g = 9.81 m s⁻².

What is the horizontal distance travelled by the ball before it lands on the roof?

Show mark scheme

Marking point	Mark	Notes
Step 1Components: u_x = 16.0 cos 50.0° = 10.28 m s⁻¹; u_y = 16.0 sin 50.0° = 12.26 m s⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Vertical motion: 4.00 = 12.26t − 4.905t². The roots are t = 0.386 s (rising) and t = 2.113 s (falling).	—
Step 3The ball is moving downwards, so t = 2.113 s and the horizontal distance = 10.28 × 2.113 = 21.7 m.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses the smaller root, 0.386 s, when the ball passes 4.00 m on the way up. The stem says it lands while moving downwards.
BCorrect: the larger root of the quadratic, multiplied by the constant horizontal velocity.
CThis is the range on level ground, u² sin 2θ/g. The ball is stopped earlier by the raised roof.
DThis multiplies the correct time by the launch speed 16.0 m s⁻¹ instead of its horizontal component.

23A-1A-83

Instantaneous and average values·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A projectile is launched from level ground with speed u at an angle θ above the horizontal and lands on the same level ground. Air resistance is negligible.

Which row gives the magnitude of the average velocity and the magnitude of the average acceleration of the projectile over its whole flight?

Average velocityAverage acceleration

Show mark scheme

Marking point	Mark	Notes
Step 1Average velocity = displacement/time. The displacement is the horizontal range u cos θ × T, so the average velocity is u cos θ.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Average acceleration = change in velocity/time. The velocity changes from (u cos θ, u sin θ) to (u cos θ, −u sin θ): a change of 2u sin θ downwards.	—
Step 3With T = 2u sin θ/g, the average acceleration is g (as expected for a constant acceleration).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis makes two errors: it takes the displacement as zero because the projectile returns to the same level, and the change of velocity as zero because the final speed equals the initial speed.
BThis takes the displacement as zero because the projectile returns to the same level. Only the vertical displacement is zero; the horizontal displacement is the range.
CThis takes the change of velocity as zero because the landing speed equals the launch speed. Velocity is a vector: its vertical component has reversed.
DCorrect: the displacement is the range, and the average of a constant acceleration is that acceleration.

Syllabus understandingA.1 — the difference between instantaneous and average values of velocity, speed and acceleration, and how to determine them; the behaviour of projectiles in the absence of fluid resistance Command term: Deduce

24A-1A-84

Fluid resistance and projectiles·A.1 Kinematics

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A ball is thrown vertically upwards. Air resistance on the ball is significant. Upwards is taken as positive.

Which graph shows how the velocity v of the ball varies with time t from the moment it is thrown until it returns to the point of release? The dashed lines mark ± the launch speed.

Velocity–time graphs W, X, Y and Z, drawn on identical scales (upwards positive).

Show mark scheme

Marking point	Mark	Notes
Step 1Rising: air resistance and weight both act downwards, so the deceleration is greater than g and is largest at the start, where the speed is largest. The gradient is steepest at t = 0 and gets less steep.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the top the speed is zero, so the acceleration is g. Falling: air resistance acts upwards, so the acceleration is less than g and decreases as the speed grows; the graph keeps getting less steep.	—
Step 3Energy is transferred to the air, so the ball returns at a speed less than the launch speed. Only X shows all of these features.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AW has a constant gradient −g and returns at the launch speed: this is the motion with no air resistance.
BCorrect: the gradient magnitude is greatest at the start, equal to g at the top, and decreasing on the way down, with a return speed below the launch speed.
CY treats air resistance as acting downwards throughout, so the downward acceleration keeps growing as the ball falls. Air resistance always opposes the velocity, so on the way down it acts upwards.
DZ treats air resistance as acting upwards throughout, so the ball decelerates by less than g while rising. On the way up the air resistance acts downwards, adding to the weight.

Syllabus understandingA.1 — the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed; motion with uniform and non-uniform acceleration Command term: Identify

25A-1A-85

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Ball P is thrown vertically upwards from the ground with speed u. At the same instant ball Q is released from rest at a height H vertically above P. The balls collide while both are in the air. Air resistance is negligible.

How long after release do the balls collide?

Show mark scheme

Marking point	Mark	Notes
Step 1Both balls have the same acceleration g downwards, so the acceleration of P relative to Q is zero.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2P therefore approaches Q at a constant relative speed u and must close the gap H.	—
Step 3t = H/u. Check: ut − ½gt² = H − ½gt² gives ut = H.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis assumes the balls meet when P is at its highest point. The meeting time does not depend on g at all.
BThis is the time for Q to fall all the way to the ground. The balls meet before that.
CThis takes the closing speed to be 2u, as if Q also moved at speed u. Q starts from rest; both balls gain the same downward velocity, so the closing speed stays u.
DCorrect: the relative acceleration is zero, so the gap H closes at the constant rate u.

Syllabus understandingA.1 — the equations of motion for solving problems with uniformly accelerated motion as given by s = ut + ½at²; motion of bodies described in terms of position, velocity and acceleration Command term: Determine

26A-1A-86

Instantaneous and average values·A.1 Kinematics

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A trolley accelerates uniformly along a straight track. A card of length L fixed to the trolley interrupts the beam of a light gate for a time Δt.

The quantity L/Δt is equal to the instantaneous speed of the trolley at which instant?

Show mark scheme

Marking point	Mark	Notes
Step 1L/Δt is the average velocity of the trolley while the card crosses the beam.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2For uniform acceleration the velocity increases linearly with time, so the average velocity equals the instantaneous velocity at the middle of the time interval.	—
Step 3The trolley covers the first half of the card more slowly than the second half, so the midpoint of the card passes the beam later than halfway through Δt, when the speed is already greater than L/Δt.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThe speed when the leading edge arrives is the lowest speed during the interval, so it is less than the average.
BThis assumes the speed at the halfway position equals the average speed. With acceleration the halfway position is reached after the halfway time, at a higher speed.
CThe speed when the trailing edge leaves is the highest speed during the interval, so it is greater than the average.
DCorrect: for uniform acceleration the average velocity over an interval equals the velocity at the middle of the time interval.

27A-1B-10

Stroboscopic analysis of a projectile·A.1 Kinematics

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

A small steel ball rolls off the edge of a horizontal bench and is photographed against a square grid using a stroboscope that flashes every 0.10 s. The first image (gold) is at the instant the ball leaves the bench. The horizontal position x and the vertical distance fallen y of each image are read from the photograph, each with an uncertainty of ±0.01 m. Air resistance is negligible.

The figure shows the photograph and the table gives the measurements.

t / s	0	0.10	0.20	0.30	0.40	0.50
x / m	0	0.20	0.40	0.60	0.80	1.00
y / m	0	0.05	0.20	0.44	0.78	1.23

Successive images of the ball at intervals of 0.10 s; the first image (gold) is at the moment of launch.

(a)

(i)

Determine the initial speed of the ball, using the full range of the data.

(1)

(b)

(i)

Test, using at least three images, the hypothesis that the vertical motion is uniformly accelerated from rest.

(2)

(ii)

Hence determine the acceleration of free fall.

(1)

(iii)

Determine the speed of the ball at t = 0.50 s and the angle that its velocity makes with the horizontal.

(2)

(iv)

Explain why the image at 0.50 s gives a more reliable value of the acceleration than the image at 0.10 s.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
u = 1.00/0.50 = 2.0 m s⁻¹ (constant horizontal velocity: equal horizontal spacing)	✓ 1	Using a single 0.10 s interval (0.20/0.10) also gives 2.0 m s⁻¹ but must be justified by the equal spacing.
Part (b)(i)
For uniform acceleration from rest y = ½at², so y/t² should be constant; calculates at least three values, e.g. 5.00, 4.89, 4.88, 4.92 m s⁻²	✓ 1	Equivalent: successive differences in y increase by a constant 0.10 m. Two points only scores [1 max].
The values are the same (≈ 4.9 m s⁻²) within the reading uncertainty, so the hypothesis is supported	✓ 1	Conclusion must be consistent with the values.
Part (b)(ii)
g = 2 × y/t², e.g. 2 × 1.23/0.50² = 9.8 m s⁻²	✓ 1	Allow ECF from (b)(i). Accept 9.6–10.0 m s⁻².
Part (b)(iii)
v_y = gt = 9.8 × 0.50 = 4.9 m s⁻¹; speed = √(2.0² + 4.9²) = 5.3 m s⁻¹	✓ 1	Allow ECF from (a)(i) and (b)(ii).
Angle = tan⁻¹(4.9/2.0) = 68° below the horizontal	✓ 1	Allow ECF. Accept 67–69°.
Part (b)(iv)
The reading uncertainty ±0.01 m is the same for every image: it is 20 % of 0.05 m but under 1 % of 1.23 m (and the timing is equally precise), so the fractional uncertainty in g is much smaller for the larger distance	✓ 1	Allow ECF from (b)(ii). Must compare fractional (percentage) uncertainties.

Answers: (a)(i) 2.0 m s⁻¹ · (b)(i) y/t² ≈ 4.9 m s⁻² · (b)(ii) 9.8 m s⁻² · (b)(iii) 5.3 m s⁻¹ at 68° below the horizontal (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — the motion of projectiles in the absence of fluid resistance as independent horizontal and vertical motions; the equations of motion for uniform acceleration; Tools 3 — testing a relationship with several data points, choosing the data range to reduce the percentage uncertainty Command term: Determine

28A-1B-15

Non-uniform acceleration·A.1 Kinematics

Paper 1BHard7 marks

Data-based question7 steps to full marksDetermine

An electric racing kart and its driver, of total mass 180 kg, start from rest on a straight level track. A radar gun records the speed v at times t; each speed is ±0.2 m s⁻¹. The motor delivers a constant mechanical power P and resistive forces may be neglected, so PΔt = Δ(½mv²) and a graph of v² against t has gradient 2P/m.

During the run the battery supplies a current of 160 A at a terminal potential difference of 72.0 V.

t / s	1.0	1.5	2.0	2.5	3.0	3.5
v / m s⁻¹	7.7	10.5	12.7	14.5	16.2	17.5

Graph drawn to scale. The point for t = 3.5 s has not been plotted.

(a)

(i)

Calculate v² at t = 3.5 s together with its absolute uncertainty.

(2)

(ii)

Determine P, using the full range of the data.

(2)

(iii)

Determine the efficiency with which the motor transfers energy from the battery to the kinetic energy of the kart.

(1)

(iv)

The line of best fit crosses the time axis at about 0.4 s instead of passing through the origin. The maximum forward force that friction between the tyres and the track can provide is 1.4 × 10³ N. Explain the intercept, using your answer to (a)(ii).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v² = 17.5² = 306 m² s⁻²	✓ 1
Δ(v²) = 2 × (0.2/17.5) × 306 = ±7 m² s⁻²	✓ 1	Absolute uncertainty to 1 s.f. ±3.5 (not doubled) scores 0 for this mark.
Part (a)(ii)
Gradient of the line from (1.0, 59) to (3.5, 306): (306 − 59)/2.5 = 99 m² s⁻³	✓ 1	Allow ECF from (a)(i). Accept 95–104 m² s⁻³.
P = gradient × m/2 = 99 × 180/2 = 8.9 × 10³ W	✓ 1	Accept 8.6–9.4 kW. Award [0] for gradient × m.
Part (a)(iii)
Efficiency = P/IV = 8.9 × 10³/(160 × 72.0) = 8.9 × 10³/1.15 × 10⁴ = 0.77	✓ 1	Allow ECF from (a)(ii). Accept 0.75–0.82 (or 75–82 %).
Part (a)(iv)
Delivering power P needs a driving force F = P/v, which exceeds 1.4 × 10³ N below v = 8.9 × 10³/1.4 × 10³ = 6.4 m s⁻¹	✓ 1	Allow ECF from (a)(ii). Accept 6.1–6.7 m s⁻¹.
So at first the force is limited (wheels would spin) and the power delivered (Fv) is less than P: less kinetic energy is gained than the model predicts, so the early v² values are low and the line is displaced to later times	✓ 1	Do not accept "reaction time" or "random error".

Answers: (a)(i) 306 ± 7 m² s⁻² · (a)(ii) 8.9 × 10³ W · (a)(iii) 0.77 · (a)(iv) 6.4 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — motion with non-uniform acceleration; A.3 — power P = Fv, kinetic energy, efficiency; B.5 — electric power P = IV; Tools 3 — propagation of uncertainty for a power, error bars, linearising a relationship, gradient using the full data range, interpreting an intercept Command term: Determine

29A-1B-24

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1BEasy7 marks

Data-based question6 steps to full marksDetermine

A steel ball is held by an electromagnet above a trapdoor switch. Switching off the electromagnet starts an electronic timer; the timer stops when the ball hits the trapdoor. The height h fallen is measured with a metre rule to ±0.002 m, and each time t is the mean of five drops with an uncertainty of ±0.003 s. Air resistance is negligible.

The magnetism of the electromagnet takes a short, constant time t₀ to decay, so the ball is released slightly after the timer starts. The time recorded is then t = √(2/g) √h + t₀. The graph shows t against √h.

h / m	√h / m^½	t / s
0.200	0.447	0.242
0.400	0.632	0.326
0.600	0.775	0.389
0.800		0.443
1.000	1.000	0.491
1.200	1.095	0.533

Graph drawn to scale. The line of best fit has been extended to √h = 0.

(a)

(i)

Calculate √h for h = 0.800 m, together with its absolute uncertainty.

(2)

(ii)

Determine the gradient of the line of best fit. Give its unit.

(2)

(b)

(i)

Hence determine g.

(2)

(ii)

Determine t₀, and state why it does not affect your answer to (b)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
√0.800 = 0.894 m^½	✓ 1
Percentage uncertainty halves for a square root: ½ × (0.002/0.800) × 100 % = 0.125 %, so Δ√h = ±0.001 m^½	✓ 1	Accept ±0.0011. ±0.002 m^½ (uncertainty of h copied unchanged) scores 0 for this mark.
Part (a)(ii)
Gradient using the full range of the line, e.g. (0.533 − 0.242)/(1.095 − 0.447) = 0.449	✓ 1	Accept 0.44–0.46. A triangle spanning less than half the line scores [1 max].
Unit: s m^−½	✓ 1	Accept s/m^½ or s m^−0.5.
Part (b)(i)
Gradient = √(2/g), so g = 2/gradient²	✓ 1	Allow ECF from (a)(ii).
g = 2/0.449² = 9.9 m s⁻²	✓ 1	Accept 9.4–10.4 m s⁻² (from a gradient in the range 0.44–0.46). g = 1/gradient² or 2 × gradient scores [1 max].
Part (b)(ii)
Intercept on the t axis: t₀ = 0.04 s (accept 0.035–0.045 s); it adds the same time to every reading, so it shifts the line without changing its gradient	✓ 1	Allow ECF from (a)(ii) for an intercept calculated from the candidate's gradient and one data point. Both the value and the reason are needed.

Answers: (a)(i) 0.894 ± 0.001 m^½ · (a)(ii) 0.449 s m^−½ · (b)(i) 9.9 m s⁻² · (b)(ii) t₀ ≈ 0.04 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; Tools 3 — linearising a relationship; gradient and intercept with units; propagating uncertainty for a power (square root); systematic error shown by an intercept Command term: Determine

30A-1B-27

Projectile launched horizontally·A.1 Kinematics

Paper 1BEasy7 marks

Data-based question6 steps to full marksDetermine

A small steel ball is released from the same mark on a curved track fixed to the top of a laboratory jack, so that it always leaves the horizontal end of the track with the same speed u. The height h of the end of the track above the floor is changed by adjusting the jack and is measured to ±0.002 m. For each height, the horizontal distance R from the end of the track to the landing point is found from the marks on carbon paper; each R is the mean of five landings and has an uncertainty of ±0.005 m. Air resistance is negligible.

Theory predicts that R = u√(2h/g). The graph shows R² against h with the line of best fit for all six heights.

h / m	R / m	R² / m²
0.400	0.514	0.264
0.550	0.604	0.365
0.700	0.679	0.461
0.850	0.747
1.000	0.811	0.658
1.150	0.869	0.755

Graph drawn to scale. The point for h = 0.850 m has not been plotted.

(a)

(i)

Calculate R² for h = 0.850 m, together with its absolute uncertainty.

(2)

(ii)

Determine the gradient of the line of best fit. State its unit.

(2)

(b)

(i)

Hence determine u.

(2)

(ii)

Explain why R² is plotted against h, rather than R against h, to test the prediction.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
R² = 0.747² = 0.558 m²	✓ 1
Fractional uncertainty doubles: 2 × (0.005/0.747) = 1.3 %, so Δ(R²) = ±0.007 m²	✓ 1	Accept ±0.007 or ±0.008 m². Copying ±0.005 m², or using 0.005/0.747 without doubling (±0.004), scores 0 for this mark.
Part (a)(ii)
Gradient from two well-separated points on the line, e.g. (0.755 − 0.265)/(1.150 − 0.400) = 0.653	✓ 1	Accept 0.63–0.68. A triangle using less than half the line scores [1 max].
Unit: m (m²/m)	✓ 1	Accept "metre".
Part (b)(i)
R² = (2u²/g)h, so the gradient = 2u²/g	✓ 1	Allow ECF from (a)(ii).
u = √(g × gradient/2) = √(9.81 × 0.653/2) = 1.79 m s⁻¹	✓ 1	Accept 1.76–1.83 m s⁻¹ (from a gradient of 0.63–0.68). Using gradient = u²/g gives 2.53 m s⁻¹ and scores [1 max].
Part (b)(ii)
The prediction gives R² ∝ h, so a straight line through the origin tests it directly (a graph of R against h would be a curve, which is hard to judge), and the gradient gives u	✓ 1	Both the straight line through the origin and the reason it is useful are needed.

Answers: (a)(i) 0.558 ± 0.007 m² · (a)(ii) 0.65 m · (b)(i) 1.79 m s⁻¹ (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 (projectiles launched horizontally); Tools 3 — linearising a relationship; gradient with units; propagating uncertainty for a power Command term: Determine

31A-1B-28

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

A spring launcher fires a small ball vertically upwards. A light gate fixed just above the launcher measures the speed v of the ball as it passes through the gate; v is changed by changing the compression of the spring and has an uncertainty of ±0.02 m s⁻¹. The maximum height H of the ball above the top of the launcher is read from a video recorded against a vertical scale, to ±0.005 m. Air resistance is negligible.

A student expects H = v²/2g. The graph shows H against v² with the line of best fit for all six results.

v / m s⁻¹	v² / m² s⁻²	H / m
2.00	4.00	0.254
2.40	5.76	0.349
2.80	7.84	0.454
3.20		0.570
3.60	12.96	0.709
4.00	16.00	0.863

Graph drawn to scale. The point for v = 3.20 m s⁻¹ has not been plotted. The line of best fit has been extended to v² = 0.

(a)

(i)

Calculate v² for v = 3.20 m s⁻¹, together with its absolute uncertainty.

(2)

(b)

(i)

Determine the gradient of the line of best fit. State its unit.

(2)

(ii)

Hence determine g.

(1)

(c)

(i)

The line of best fit does not pass through the origin. Determine its intercept on the H axis and suggest the cause.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v² = 3.20² = 10.24 m² s⁻²	✓ 1
Δ(v²) = 2 × (0.02/3.20) × 10.24 = ±0.1 m² s⁻²	✓ 1	Accept ±0.13. ±0.04 (doubling the absolute uncertainty) or ±0.02 scores 0 for this mark.
Part (b)(i)
Gradient from two well-separated points on the line, e.g. (0.863 − 0.257)/(16.0 − 4.0) = 0.0505	✓ 1	Accept 0.0485–0.0525. A triangle using less than half the line scores [1 max].
Unit: s² m⁻¹	✓ 1	Accept m⁻¹ s².
Part (b)(ii)
g = 1/(2 × gradient) = 1/(2 × 0.0505) = 9.9 m s⁻²	✓ 1	Allow ECF from (b)(i). Accept 9.5–10.3 m s⁻². 1/gradient (20 m s⁻²) scores 0.
Part (c)(i)
Intercept ≈ 0.055 m	✓ 1	Accept 0.04–0.07 m. Allow ECF from (b)(i) for an intercept calculated from the candidate's gradient and a point on the line.
The speed is measured at the light gate, which is about 0.05 m above the top of the launcher: the ball is already that high when it has speed v, so H = v²/2g + (height of the gate). This systematic error shifts the line but does not change the gradient (or g)	✓ 1	Answers such as "random error", "air resistance" or "reaction time" score 0. The link to the position of the gate is needed.

Answers: (a)(i) 10.2 ± 0.1 m² s⁻² · (b)(i) 0.0505 s² m⁻¹ · (b)(ii) 9.9 m s⁻² · (c)(i) ≈ 0.055 m (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; Tools 3 — linearising a relationship; gradient and intercept with units; propagating uncertainty for a power; identifying a systematic error from an intercept Command term: Determine

32A-1B-29

Equations of uniformly accelerated motion·A.1 Kinematics

Paper 1BHard7 marks

Data-based question6 steps to full marksDetermine

A cyclist stops pedalling on a straight road of uniform slope and freewheels down it. Marker posts are placed every 20.0 m along the road. A video records the time t at which the bicycle passes each post, measured from the instant it passes the first post, where it is already moving with speed u. Each time has an uncertainty of ±0.10 s; the distances s from the first post may be taken as exact. The acceleration a is assumed to be uniform.

The graph shows s/t against t with error bars, the line of best fit and the lines of maximum and minimum gradient (dashed).

s / m	t / s	s/t / m s⁻¹
20.0	4.31	4.64
40.0	7.71	5.19
60.0	10.70	5.61
80.0	13.36	5.99
100.0	15.67	6.38
120.0	17.92

Graph drawn to scale. The dashed lines are the steepest and the least steep straight lines that pass through all the error bars.

(a)

(i)

Show that s/t = u + ½at, and state what the intercept on the vertical axis represents.

(1)

(b)

(i)

Calculate s/t for s = 120.0 m, together with its absolute uncertainty.

(2)

(ii)

Use the graph to determine a, together with its absolute uncertainty.

(2)

(c)

(i)

The road is inclined at 3.0° to the horizontal and the cyclist and bicycle have a total mass of 90 kg. Determine the average resistive force on them.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
s = ut + ½at², and dividing by t gives s/t = u + ½at; the intercept is u, the speed at the first post	✓ 1	Both the division and the identification of the intercept are needed.
Part (b)(i)
s/t = 120.0/17.92 = 6.70 m s⁻¹	✓ 1
Fractional uncertainty = 0.10/17.92 = 0.56 %, so Δ(s/t) = ±0.04 m s⁻¹	✓ 1	Accept ±0.037. Quoting ±0.10 m s⁻¹ scores 0 for this mark.
Part (b)(ii)
Gradient of the best-fit line = 0.151 m s⁻², so a = 2 × gradient = 0.30 m s⁻²	✓ 1	Accept a gradient of 0.145–0.156 m s⁻² (a = 0.29–0.31 m s⁻²). a = gradient (0.15 m s⁻²) scores 0 for this mark.
Gradients of the dashed lines ≈ 0.158 and 0.140 m s⁻²; Δa = 2 × ½(0.158 − 0.140) = ±0.02 m s⁻²	✓ 1	Accept ±0.02 (or ±0.018). Δa = ±0.01 (the gradient uncertainty not doubled) scores 0 for this mark.
Part (c)(i)
Component of the weight down the slope = 90 × 9.81 × sin 3.0° = 46 N; resultant force = 90 × 0.30 = 27 N	✓ 1	Allow ECF from (b)(ii).
Resistive force = 46 − 27 = 19 N	✓ 1	Accept 17–21 N (from a = 0.28–0.32 m s⁻²). Using cos 3.0° instead of sin 3.0° scores 0 for this mark.

Answers: (b)(i) 6.70 ± 0.04 m s⁻¹ · (b)(ii) a = 0.30 ± 0.02 m s⁻² · (c)(i) 19 N (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; A.2 — free-body diagrams analysed to find the resultant force (components of the weight on a slope); Tools 3 — linearising a relationship; error bars; lines of maximum and minimum gradient; uncertainty in the gradient Command term: Determine

33A-2-12

Projectile motion·A.1 Kinematics

Paper 2Medium12 marks

Short answer & extended response7 steps to full marksDetermine

An astronaut strikes a golf ball on the Moon. The ball leaves the level ground at 28.0 m s⁻¹ at 35.0° to the horizontal. The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m and no atmosphere.

Height against horizontal distance for the same strike on the Earth (drawn to scale).

(a)

(i)

Show that the gravitational field strength at the surface of the Moon is about 1.6 N kg⁻¹.

(2)

(ii)

Calculate the time of flight of the ball.

(2)

(iii)

Calculate the horizontal distance travelled by the ball.

(1)

(b)

(i)

Show that, for the same launch velocity and no air resistance, the ratio of the range on the Moon to the range on the Earth is g_Earth/g_Moon.

(2)

(ii)

The graph shows the path of the same strike on the Earth with and without air resistance. Use the graph to determine the percentage reduction in range caused by air resistance.

(2)

(iii)

Outline two ways, other than the range, in which the path with air resistance differs from the path without it, and give a reason for each.

(2)

(c)

(i)

Explain why the ball cannot escape from the Moon, however hard a golf ball could be struck by an astronaut.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g = GM/R² = 6.67 × 10⁻¹¹ × 7.35 × 10²²/(1.74 × 10⁶)²	✓ 1
g = 1.62 N kg⁻¹	✓ 1	Must see 1.62 or better.
Part (a)(ii)
Vertical component = 28.0 sin 35.0° = 16.1 m s⁻¹	✓ 1
t = 2 × 16.06/1.62 = 19.8 s	✓ 1	Allow ECF from (a)(i). Accept 19.6–20.1 s.
Part (a)(iii)
28.0 cos 35.0° × 19.8 = 455 m	✓ 1	Allow ECF from (a)(ii). Accept 450–460 m.
Part (b)(i)
Time of flight 2u sin θ/g, so range = u cos θ × 2u sin θ/g = u² sin 2θ/g	✓ 1	Symbolic working required.
Only g differs, so the range ∝ 1/g and the ratio is g_Earth/g_Moon	✓ 1
Part (b)(ii)
Ranges read: about 75 m without and about 60 m with air resistance	✓ 1	Accept 58–62 m for the range with air resistance.
Reduction = (75 − 60)/75 × 100 ≈ 20 %	✓ 1	Accept 16–24 %.
Part (b)(iii)
Lower maximum height: the drag force has a downward component while the ball rises, so it decelerates faster than g vertically	✓ 1
Asymmetric path, steeper on the way down: the horizontal velocity decreases throughout the flight because the drag has a backward horizontal component	✓ 1	Accept: the peak occurs at more than half the range.
Part (c)(i)
Escape speed = √(2GM/R) = 2374 m s⁻¹, about 85 times the launch speed of 28 m s⁻¹ and far beyond any speed an astronaut could give a golf ball	✓ 1	A calculated escape speed is required.

Answers: (a)(ii) 19.8 s · (a)(iii) 455 m · (b)(ii) ≈ 20 % (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 horizontal and vertical components; the qualitative effect of fluid resistance on projectiles; D.1 — gravitational field strength g = GM/r²; escape speed Command term: Determine

34A-2-18

Projectile motion·A.1 Kinematics

Paper 2Hard12 marks

Short answer & extended response8 steps to full marksDetermine

In a compressed-air launcher, a projectile of mass 0.25 kg fits closely inside a barrel. Air behind the projectile, initially at a pressure of 6.0 × 10⁵ Pa and volume 0.50 × 10⁻³ m³, pushes it along the barrel. The projectile leaves the barrel when the volume of air behind it is 2.0 × 10⁻³ m³. The air in front of the projectile stays at atmospheric pressure, 1.0 × 10⁵ Pa. The graph shows how the pressure of the air behind the projectile varies with its volume.

Pressure p of the air behind the projectile against its volume V (drawn to scale).

(a)

(i)

Estimate the work done by the air behind the projectile.

(2)

(ii)

Show that the net work done on the projectile by the air on both sides of it is about 270 J.

(2)

(b)

(i)

Friction in the barrel dissipates 30 % of the net work. Calculate the speed of the projectile as it leaves the barrel.

(2)

(ii)

The launcher fires the projectile at 30° above the horizontal towards a vertical wall that is 75 m away horizontally. Determine the height above the launch point at which the projectile hits the wall. Air resistance is negligible.

(2)

(iii)

Deduce whether the projectile is rising or falling when it hits the wall.

(1)

(c)

(i)

The graph was drawn assuming the air behind the projectile stays at constant temperature. In practice the expansion is very rapid. Explain why the projectile leaves the barrel with a lower speed than predicted.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Work = area under the p–V graph between 0.50 and 2.0 × 10⁻³ m³ (counting squares or trapezia)	✓ 1
Work ≈ 4.2 × 10² J	✓ 1	Accept 390–440 J. Powers of ten must be handled: one large square = 50 J.
Part (a)(ii)
The atmosphere in front does negative work p₀ΔV = 1.0 × 10⁵ × 1.5 × 10⁻³ = 150 J	✓ 1	The hidden step: the projectile pushes the outside air out of the barrel.
Net work = 416 − 150 = 266 J	✓ 1	Allow ECF from (a)(i). Must see a value to 3 s.f. or full working.
Part (b)(i)
E_k = 0.70 × 266 = 186 J	✓ 1	Allow ECF from (a)(ii); use 270 J if (a)(ii) was not attempted.
v = √(2 × 186/0.25) = 38.6 m s⁻¹	✓ 1	Accept 38–39 m s⁻¹.
Part (b)(ii)
Horizontal velocity = 38.6 cos 30° = 33.4 m s⁻¹, constant, so the time to reach the wall = 75/33.4 = 2.24 s	✓ 1	Allow ECF from (b)(i).
Height = 19.3 × 2.24 − ½ × 9.81 × 2.24² = 19 m	✓ 1	Accept 18–19 m (17.8–19.1 m for a launch speed of 38–39 m s⁻¹). Using the full launch speed for the vertical motion scores [1 max].
Part (b)(iii)
Vertical velocity at the wall = 19.3 − 9.81 × 2.24 = −2.7 m s⁻¹, so it is falling (it passed its highest point at 19.3/9.81 = 1.97 s)	✓ 1	Allow ECF from (b)(ii). The conclusion must follow from a calculated vertical velocity or a comparison of times.
Part (c)(i)
The expansion is (nearly) adiabatic: there is no time for thermal energy to flow into the air	✓ 1
The air does work on the projectile at the expense of its internal energy, so its temperature falls	✓ 1
So the pressure at each volume is lower than on the isothermal curve, the area under the curve (work done) is smaller, and less kinetic energy is given to the projectile	✓ 1	Allow ECF from (a)(i): reference to a smaller area or less work needed.

Answers: (a)(i) 4.2 × 10² J · (b)(i) 38.6 m s⁻¹ · (b)(ii) 19 m · (b)(iii) falling (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — projectile motion in the absence of fluid resistance, equations of motion resolved into horizontal and vertical components; A.3 — work done and energy transfer; efficiency; B.3 — the pressure and volume of a gas; B.4 — the work done by a gas is the area under a p–V graph; adiabatic processes Command term: Determine

35A-2-19

Motion graphs·A.1 Kinematics

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

The graph shows how the velocity of a sprinter of mass 70 kg varies with time during the first 6.0 s of a race along a straight, level track.

Velocity v of the sprinter against time t (drawn to scale).

(a)

(i)

Determine the instantaneous acceleration of the sprinter at t = 1.0 s.

(2)

(ii)

Estimate the distance run in the first 4.0 s.

(2)

(iii)

Hence calculate the average speed over the first 4.0 s.

(1)

(b)

(i)

Calculate the resultant horizontal force on the sprinter at t = 1.0 s.

(1)

(ii)

Explain, with reference to Newton's third law, how the sprinter is able to accelerate forwards.

(2)

(iii)

Use the graph to determine the minimum coefficient of static friction between the sprinter's shoes and the track at the start. Assume that the vertical force from the track equals the sprinter's weight.

(2)

(c)

(i)

Determine the kinetic energy of the sprinter at t = 6.0 s and the minimum average power developed during the first 6.0 s.

(2)

(ii)

The sprinter's leg muscles transfer chemical energy to the kinetic energy of the sprinter with an efficiency of 25 %. Estimate the rise in temperature of the 30 kg of leg muscle during the 6.0 s. The specific heat capacity of muscle is 3.5 × 10³ J kg⁻¹ K⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Tangent drawn at t = 1.0 s and its gradient found from a large triangle	✓ 1
a = 3.7 m s⁻²	✓ 1	Accept 3.2–4.2 m s⁻².
Part (a)(ii)
Area under the graph from 0 to 4.0 s, by counting squares or trapezia	✓ 1
Distance ≈ 28 m	✓ 1	Accept 26–30 m.
Part (a)(iii)
28/4.0 = 7.0 m s⁻¹	✓ 1	Allow ECF from (a)(ii).
Part (b)(i)
F = 70 × 3.7 = 2.6 × 10² N	✓ 1	Allow ECF from (a)(i). Accept 220–290 N.
Part (b)(ii)
The sprinter's foot pushes backwards on the track (a frictional force)	✓ 1
By Newton's third law the track pushes forwards on the sprinter with an equal force; this external force on the sprinter causes the acceleration	✓ 1	A reference to the forces acting on different bodies is needed.
Part (b)(iii)
Initial acceleration = gradient of the tangent at t = 0 ≈ 7.5 m s⁻²	✓ 1	Accept 6.8–8.0 m s⁻².
Friction provides ma and F_f ≤ μmg, so μ ≥ a/g = 7.5/9.81 = 0.76	✓ 1	Accept 0.69–0.82, consistent with the gradient read.
Part (c)(i)
E_k = ½ × 70 × 10.4² = 3.8 × 10³ J	✓ 1	Accept 10.3–10.5 m s⁻¹ read from the graph (3.7–3.9 × 10³ J).
Average power ≥ 3.79 × 10³/6.0 = 6.3 × 10² W	✓ 1	Allow ECF from the kinetic energy. Minimum because work against air resistance is ignored.
Part (c)(ii)
Thermal energy = 3 × 3.79 × 10³ = 1.1 × 10⁴ J (75 % of the input, three times the useful output)	✓ 1	Allow ECF from (c)(i). Using 75 % of the kinetic energy instead of 3 × scores [1 max].
ΔT = 1.14 × 10⁴/(30 × 3.5 × 10³) = 0.11 K	✓ 1	Accept 0.10–0.12 K.

Answers: (a)(i) 3.7 m s⁻² · (a)(ii) 28 m · (a)(iii) 7.0 m s⁻¹ · (b)(i) 2.6 × 10² N · (b)(iii) 0.76 · (c)(i) 3.8 × 10³ J; 6.3 × 10² W · (c)(ii) 0.11 K (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — the difference between instantaneous and average values of velocity, speed and acceleration; the gradient and area of a velocity–time graph; motion with non-uniform acceleration; A.2 — Newton's three laws of motion; static friction F_f ≤ μ_sF_N; A.3 — kinetic energy, power and efficiency; B.1 — specific heat capacity Command term: Determine

36A-2-38

Projectile motion·A.1 Kinematics

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

In an ink-jet printer, a negatively charged ink droplet of mass 1.2 × 10⁻¹⁰ kg and charge −1.8 × 10⁻¹³ C enters the region between two horizontal parallel plates, midway between them, moving horizontally at 20 m s⁻¹. The plates are 16 mm long and 2.0 mm apart, and the potential difference between them is 1.5 kV, with the upper plate positive. The paper is 12 mm beyond the end of the plates, as shown. The field between the plates is uniform and there is no field outside them. Air resistance is negligible.

Droplet deflected between the charged plates (not to scale).

(a)

(i)

Show that the electric force on the droplet is about 1.4 × 10⁻⁷ N.

(2)

(ii)

Show that the effect of gravity on the motion of the droplet can be ignored.

(1)

(b)

(i)

Calculate the time the droplet spends between the plates and its acceleration there.

(2)

(ii)

Determine the vertical displacement of the droplet as it leaves the plates.

(2)

(iii)

Calculate the vertical component of the velocity of the droplet as it leaves the plates.

(1)

(iv)

Determine the total vertical displacement of the droplet, from its line of entry, when it reaches the paper.

(3)

(c)

(i)

Show that the vertical displacement of a droplet as it leaves the plates is y = qVL²/(2mdu²), where L is the length of the plates and u the speed of entry. Hence state the effect on this displacement of doubling the speed of entry.

(2)

(ii)

Calculate the gain in kinetic energy of the droplet while it is between the plates.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E = V/d = 1.5 × 10³/2.0 × 10⁻³ = 7.5 × 10⁵ V m⁻¹	✓ 1	Field strength must be calculated with d in m.
F = qE = 1.8 × 10⁻¹³ × 7.5 × 10⁵ = 1.35 × 10⁻⁷ N	✓ 1	The answer must be given to at least 3 s.f., or the full substitution shown.
Part (a)(ii)
Weight = 1.2 × 10⁻¹⁰ × 9.81 = 1.2 × 10⁻⁹ N, about 1 % of (or about 115 times smaller than) the electric force	✓ 1	Allow ECF from (a)(i). A comparison is needed.
Part (b)(i)
t = 16 × 10⁻³/20 = 8.0 × 10⁻⁴ s: the horizontal velocity is constant because the force is vertical	✓ 1
a = F/m = 1.35 × 10⁻⁷/1.2 × 10⁻¹⁰ = 1.1 × 10³ m s⁻² (upwards)	✓ 1	Allow ECF from (a)(i). Accept 1125 m s⁻²; 1.17 × 10³ m s⁻² from 1.4 × 10⁻⁷ N.
Part (b)(ii)
The droplet starts with zero vertical velocity: y = ½at²	✓ 1	Allow ECF from (b)(i).
y = ½ × 1125 × (8.0 × 10⁻⁴)² = 3.6 × 10⁻⁴ m (0.36 mm), towards the upper plate	✓ 1	Accept 3.6–3.8 × 10⁻⁴ m.
Part (b)(iii)
v_y = at = 1125 × 8.0 × 10⁻⁴ = 0.90 m s⁻¹	✓ 1	Allow ECF from (b)(i). Accept 0.93–0.94 m s⁻¹ from a = 1.17 × 10³ m s⁻².
Part (b)(iv)
Beyond the plates no force acts, so the droplet moves in a straight line at constant velocity (Newton's first law); time to the paper = 12 × 10⁻³/20 = 6.0 × 10⁻⁴ s	✓ 1	Award this mark for the time with any valid reason for straight-line motion, or for a similar-triangles method.
Extra displacement = 0.90 × 6.0 × 10⁻⁴ = 5.4 × 10⁻⁴ m	✓ 1	Allow ECF from (b)(iii).
Total = 3.6 × 10⁻⁴ + 5.4 × 10⁻⁴ = 9.0 × 10⁻⁴ m (0.90 mm)	✓ 1	Allow ECF from (b)(ii). Accept 9.0–9.4 × 10⁻⁴ m. Continuing with acceleration beyond the plates scores [2 max].
Part (c)(i)
a = qV/md and t = L/u, so y = ½(qV/md)(L/u)² = qVL²/(2mdu²)	✓ 1	Both substitutions must be seen.
y ∝ 1/u², so doubling the speed reduces the displacement to a quarter	✓ 1	Accept "divided by 4".
Part (c)(ii)
Gain = ½mv_y² = ½ × 1.2 × 10⁻¹⁰ × 0.90² = 4.9 × 10⁻¹¹ J	✓ 1	Allow ECF from (b)(iii). Accept the work done by the electric force, Fy = 1.35 × 10⁻⁷ × 3.6 × 10⁻⁴ = 4.9 × 10⁻¹¹ J, Allow ECF from (b)(ii).

Answers: (a)(i) 1.35 × 10⁻⁷ N · (b)(i) 8.0 × 10⁻⁴ s; 1.1 × 10³ m s⁻² · (b)(ii) 3.6 × 10⁻⁴ m · (b)(iii) 0.90 m s⁻¹ · (b)(iv) 9.0 × 10⁻⁴ m · (c)(ii) 4.9 × 10⁻¹¹ J (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; motion with uniform acceleration; A.2 — Newton's first law; field forces; A.3 — work done by a constant force; D.2 — electric field strength E = V/d between parallel plates and the electric force F = qE Command term: Determine

37A-2-40

Motion graphs·A.1 Kinematics

Paper 2Easy12 marks

Short answer & extended response9 steps to full marksDetermine

A metro train of mass 2.0 × 10⁵ kg travels between two stations along a straight, level track. The graph shows how its velocity v varies with time t from the moment it leaves the first station until it stops at the second.

Velocity v of the train against time t (drawn to scale).

(a)

(i)

Calculate the acceleration of the train during the first 24 s.

(1)

(ii)

Show that the distance between the stations is about 1.3 km.

(2)

(iii)

Calculate the average speed of the train between the stations.

(1)

(iv)

Determine the two times at which the instantaneous speed of the train is equal to its average speed.

(2)

(b)

(i)

Calculate the resultant force on the train during the first 24 s.

(1)

(ii)

A constant resistive force of 8.0 kN acts on the train throughout. Determine the power output of the motors just before t = 24 s.

(2)

(iii)

Between t = 24 s and t = 74 s the motors take their energy from a 750 V supply and have an efficiency of 0.80. Determine the current drawn from the supply during this time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
a = 18/24 = 0.75 m s⁻²	✓ 1
Part (a)(ii)
Distance = area under the graph = ½ × 24 × 18 + 50 × 18 + ½ × 20 × 18	✓ 1	Any correct method of finding the area, e.g. a trapezium ½(50 + 94) × 18.
= 216 + 900 + 180 = 1296 m	✓ 1	Must see 1296 m (or 1.296 km) or the full substitution.
Part (a)(iii)
Average speed = 1296/94 = 13.8 m s⁻¹	✓ 1	Allow ECF from (a)(ii). Use of 1300 m gives 13.8 m s⁻¹.
Part (a)(iv)
While accelerating: t = 13.8/0.75 = 18.4 s	✓ 1	Allow ECF from (a)(i) and (a)(iii). Accept 18–19 s read from the graph.
While braking (deceleration 18/20 = 0.90 m s⁻²): t = 74 + (18 − 13.8)/0.90 = 78.7 s	✓ 1	Allow ECF from (a)(iii). Accept 78–79 s read from the graph.
Part (b)(i)
F = ma = 2.0 × 10⁵ × 0.75 = 1.5 × 10⁵ N	✓ 1	Allow ECF from (a)(i).
Part (b)(ii)
Driving force = 1.5 × 10⁵ + 8.0 × 10³ = 1.58 × 10⁵ N	✓ 1	Allow ECF from (b)(i). The hidden step: the motors must also overcome the resistive force.
P = Fv = 1.58 × 10⁵ × 18 = 2.8 × 10⁶ W	✓ 1	Accept 2.84 × 10⁶ W. Ignoring the resistive force gives 2.7 × 10⁶ W and scores [1 max].
Part (b)(iii)
At constant velocity the driving force equals the resistive force, 8.0 × 10³ N, so the output power = 8.0 × 10³ × 18 = 1.44 × 10⁵ W	✓ 1	Using the driving force from (b)(ii) scores 0 for this mark.
Input power = 1.44 × 10⁵/0.80 = 1.8 × 10⁵ W	✓ 1
I = P/V = 1.8 × 10⁵/750 = 240 A	✓ 1	Allow ECF from an incorrect power. Multiplying by the efficiency gives 154 A and scores [2 max].

Answers: (a)(i) 0.75 m s⁻² · (a)(iii) 13.8 m s⁻¹ · (a)(iv) 18.4 s and 78.7 s · (b)(i) 1.5 × 10⁵ N · (b)(ii) 2.8 × 10⁶ W · (b)(iii) 240 A (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; the gradient and area of a velocity–time graph; A.2 — Newton’s second law, translational equilibrium; A.3 — power P = Fv and efficiency; B.5 — electric power P = IV Command term: Determine

38A-2-41

Motion graphs·A.1 Kinematics

Paper 2Easy11 marks

Short answer & extended response9 steps to full marksDetermine

At t = 0 a cyclist moving at constant velocity passes a bus that is just starting to move away from a bus stop. The bus accelerates uniformly in the same direction along a straight, level road. The graph shows how the displacement s of the cyclist and of the bus from the bus stop varies with time t.

Displacement s from the bus stop against time t for the cyclist and the bus (drawn to scale).

(a)

(i)

Use the graph to determine the speed of the cyclist.

(1)

(ii)

Show that the acceleration of the bus is about 1.2 m s⁻².

(2)

(b)

(i)

Determine the greatest distance by which the cyclist is ahead of the bus.

(3)

(ii)

Show that, whatever the speed of the cyclist and the acceleration of the bus, the bus is moving at twice the speed of the cyclist at the instant it draws level.

(2)

(c)

(i)

The bus has a mass of 1.2 × 10⁴ kg and a constant resistive force of 2.0 kN acts on it. Calculate the work done by the driving force on the bus from the bus stop to the point where it draws level with the cyclist.

(2)

(ii)

Hence determine the fraction of this work that becomes kinetic energy of the bus.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient of the cyclist line, e.g. 84/14 = 6.0 m s⁻¹	✓ 1	Accept 5.8–6.2 m s⁻¹.
Part (a)(ii)
A point read from the bus curve, e.g. (10 s, 60 m), with s = ½at² because the bus starts from rest	✓ 1	Any point on the curve may be used, e.g. (14 s, 118 m).
a = 2s/t² = 2 × 60/10² = 1.20 m s⁻²	✓ 1	Must see 1.20 or better, or the full substitution.
Part (b)(i)
The separation is greatest when the bus and the cyclist have the same velocity (the tangent to the bus curve is parallel to the cyclist line)	✓ 1	Award this mark for the condition stated or used.
Time: t = 6.0/1.2 = 5.0 s	✓ 1	Allow ECF from (a)(i) and (a)(ii).
Separation = 6.0 × 5.0 − ½ × 1.2 × 5.0² = 30 − 15 = 15 m	✓ 1	Allow ECF from the time. Accept 14–16 m read from the graph at 5.0 s.
Part (b)(ii)
When the bus draws level both have the same displacement in the same time, so they have the same average velocity, v_c	✓ 1	Symbolic working is required.
For uniform acceleration from rest the average velocity of the bus is (0 + v_b)/2, so v_b/2 = v_c and v_b = 2v_c	✓ 1	ALT: ½at² = v_ct gives t = 2v_c/a, so v_b = at = 2v_c.
Part (c)(i)
Driving force = 1.2 × 10⁴ × 1.2 + 2.0 × 10³ = 1.64 × 10⁴ N	✓ 1	Allow ECF from (a)(ii).
Work = 1.64 × 10⁴ × 60 = 9.8 × 10⁵ J	✓ 1	The distance 60 m is read from the graph; accept 58–62 m. Omitting the resistive force gives 8.6 × 10⁵ J and scores [1 max].
Part (c)(ii)
Speed of the bus = 2 × 6.0 = 12 m s⁻¹, kinetic energy = ½ × 1.2 × 10⁴ × 12² = 8.64 × 10⁵ J; fraction = 8.64/9.84 = 0.88	✓ 1	Allow ECF from (b)(ii) and (c)(i). Accept 0.87–0.89.

Answers: (a)(i) 6.0 m s⁻¹ · (b)(i) 15 m · (c)(i) 9.8 × 10⁵ J · (c)(ii) 0.88 (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — motion of bodies described in terms of position, velocity and acceleration; the difference between instantaneous and average values of velocity; the equations of motion for uniformly accelerated motion; A.2 — Newton’s second law; A.3 — work done by a force W = Fs cos θ; kinetic energy Command term: Determine

39A-2-42

Projectile motion·A.1 Kinematics

Paper 2Medium13 marks

Short answer & extended response10 steps to full marksDetermine

In a garden fountain, a nozzle at the surface of a pool sends a jet of water into the air at 60° above the horizontal. The water leaves the nozzle at 9.0 m s⁻¹ and falls back into the pool at the level of the nozzle. Water leaves the nozzle at a constant rate of 4.0 × 10⁻⁴ m³ s⁻¹. The density of water is 1.0 × 10³ kg m⁻³. Air resistance is negligible.

(a)

(i)

Show that the maximum height of the jet above the nozzle is about 3.1 m.

(2)

(ii)

Calculate the time for which each element of water is in the air.

(1)

(iii)

Calculate the horizontal distance from the nozzle to the point where the water lands.

(1)

(iv)

Determine the volume of water in the air at any instant.

(2)

(b)

(i)

The electric pump that drives the fountain has an efficiency of 0.30. Determine the minimum electrical power input to the pump.

(2)

(c)

(i)

The owner wants the water to land 9.0 m from the nozzle without changing the speed of the water. Deduce whether this can be done by changing only the angle of the nozzle.

(2)

(ii)

Instead, the nozzle is raised to 1.2 m above the surface of the pool, with the same speed and angle. Determine the new horizontal distance from the nozzle to the point where the water lands.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Vertical component = 9.0 sin 60° = 7.79 m s⁻¹	✓ 1
H = 7.79²/(2 × 9.81) = 3.10 m	✓ 1	Must see 3.10 m (or 3.096 m) or the full substitution. Using 9.0 m s⁻¹ (4.13 m) scores 0.
Part (a)(ii)
t = 2 × 7.79/9.81 = 1.59 s	✓ 1	Allow ECF from (a)(i) for the vertical component.
Part (a)(iii)
9.0 cos 60° × 1.59 = 4.5 × 1.59 = 7.2 m	✓ 1	Allow ECF from (a)(ii). Accept 7.1–7.2 m. ALT: u² sin 2θ/g = 7.15 m.
Part (a)(iv)
Each element of water spends 1.59 s in the air, so the volume in the air is the volume that leaves the nozzle in 1.59 s	✓ 1	The hidden step: volume in the air = flow rate × time of flight. Allow ECF from (a)(ii).
Volume = 4.0 × 10⁻⁴ × 1.59 = 6.4 × 10⁻⁴ m³	✓ 1	Accept 6.3–6.4 × 10⁻⁴ m³ (0.63–0.64 litres).
Part (b)(i)
Mass leaving per second = 1.0 × 10³ × 4.0 × 10⁻⁴ = 0.40 kg s⁻¹; kinetic energy given per second = ½ × 0.40 × 9.0² = 16.2 W	✓ 1
Input power = 16.2/0.30 = 54 W	✓ 1	Multiplying by 0.30 (4.9 W) scores [1 max].
Part (c)(i)
The range u² sin 2θ/g is greatest when θ = 45°: maximum range = 9.0²/9.81 = 8.3 m	✓ 1	The limiting case (sin 2θ = 1, θ = 45°) must be used.
8.3 m < 9.0 m, so this is not possible at any angle	✓ 1	The conclusion must be consistent with a calculated maximum range.
Part (c)(ii)
Vertical displacement −1.2 m: −1.2 = 7.79t − 4.905t²	✓ 1	The sign of the displacement (or of u_y) must be consistent. Allow ECF from (a)(i) for the vertical component.
t = [7.79 + √(7.79² + 4 × 4.905 × 1.2)]/(2 × 4.905) = 1.73 s	✓ 1	The positive root must be chosen.
Horizontal distance = 4.5 × 1.73 = 7.8 m	✓ 1	Accept 7.7–7.8 m. Taking the displacement as +1.2 m gives 6.4 m and scores [1 max].

Answers: (a)(ii) 1.59 s · (a)(iii) 7.2 m · (a)(iv) 6.4 × 10⁻⁴ m³ · (b)(i) 54 W · (c)(i) not possible (maximum 8.3 m) · (c)(ii) 7.8 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 (projectiles launched above the horizontal from a height); A.3 — kinetic energy, power and efficiency Command term: Determine

40A-2-43

Non-uniform acceleration·A.1 Kinematics

Paper 2Medium14 marks

Short answer & extended response10 steps to full marksDetermine

A model rocket of mass 0.50 kg is launched vertically from rest. Its engine burns for 1.5 s, during which the rocket has a constant upward acceleration of 24 m s⁻². After the engine stops the rocket moves freely. Assume that the mass of the rocket stays constant and, except in (c), that air resistance is negligible.

(a)

(i)

Calculate the speed of the rocket when the engine stops.

(1)

(ii)

Show that the maximum height reached by the rocket is about 93 m.

(2)

(iii)

Determine the total time from launch until the rocket would hit the ground if no parachute opened.

(3)

(iv)

Calculate the average speed of the rocket for this whole flight.

(1)

(b)

(i)

Determine the upward force (thrust) exerted on the rocket by the exhaust gases while the engine burns.

(2)

(ii)

The exhaust gases leave the rocket at 650 m s⁻¹ relative to the rocket. Estimate the mass of gas ejected while the engine burns, and comment on the assumption that the mass of the rocket is constant.

(2)

(c)

(i)

In practice a parachute opens at the highest point. Explain why the rocket soon falls at a constant speed.

(2)

(ii)

The terminal speed is 4.0 m s⁻¹. Estimate the time the rocket takes to descend from its highest point.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = at = 24 × 1.5 = 36 m s⁻¹	✓ 1
Part (a)(ii)
Height when the engine stops = ½ × 24 × 1.5² = 27.0 m	✓ 1
Further rise = 36²/(2 × 9.81) = 66.1 m, so maximum height = 27.0 + 66.1 = 93.1 m	✓ 1	Allow ECF from (a)(i). Must see 93.1 m or the full substitution. Using g for the powered stage scores 0 for the first mark.
Part (a)(iii)
Time from the engine stopping to the highest point = 36/9.81 = 3.67 s	✓ 1	Allow ECF from (a)(i).
Time to fall from the highest point = √(2 × 93.1/9.81) = 4.36 s	✓ 1	Allow ECF from (a)(ii); use 93 m if (a)(ii) was not attempted.
Total = 1.5 + 3.67 + 4.36 = 9.5 s	✓ 1	Accept 9.5–9.6 s. Omitting the 1.5 s of powered flight gives 8.0 s and scores [2 max].
Part (a)(iv)
Average speed = total distance/time = 2 × 93.1/9.53 = 19.5 m s⁻¹	✓ 1	Allow ECF from (a)(ii) and (a)(iii). Accept 19–20 m s⁻¹. Zero (the average velocity) scores 0.
Part (b)(i)
Resultant force: F − mg = ma	✓ 1	The free-body analysis must include the weight.
F = 0.50 × (24 + 9.81) = 16.9 N	✓ 1	Omitting the weight gives 12 N and scores [1 max].
Part (b)(ii)
Thrust = rate of change of momentum of the gas, so mass ejected per second = 16.9/650 = 0.026 kg s⁻¹, and in 1.5 s about 0.039 kg	✓ 1	Allow ECF from (b)(i). Accept 0.038–0.040 kg.
This is about 8 % of 0.50 kg, so the assumption of constant mass is reasonable (but the true acceleration would rise slightly as the mass falls)	✓ 1	Allow ECF from the mass estimate. The comment must be consistent with the candidate's percentage.
Part (c)(i)
Air resistance on the rocket and parachute increases as its speed increases	✓ 1
When the air resistance equals the weight the resultant force, and so the acceleration, is zero, and the rocket falls at its terminal speed	✓ 1	Vague answers such as "the parachute slows it down" score 0.
Part (c)(ii)
t ≈ 93/4.0 = 23 s, assuming the terminal speed is reached almost at once	✓ 1	Allow ECF from (a)(ii).

Answers: (a)(i) 36 m s⁻¹ · (a)(iii) 9.5 s · (a)(iv) 19.5 m s⁻¹ · (b)(i) 16.9 N · (b)(ii) 0.039 kg (about 8 %) · (c)(ii) 23 s (the remaining parts are explanations — see the table above)

Syllabus understandingA.1 — motion with uniform and non-uniform acceleration; the equations of motion for uniformly accelerated motion; the difference between distance and displacement and between average speed and average velocity; the qualitative effect of fluid resistance, including terminal speed; A.2 — free-body diagrams; Newton’s second law; force as the rate of change of momentum, F = Δp/Δt Command term: Determine

41A-2-44

Projectile motion·A.1 Kinematics

Paper 2Hard12 marks

Short answer & extended response10 steps to full marksDetermine

A ski jumper leaves the end of a take-off ramp horizontally at a speed of 26 m s⁻¹. The landing slope is a straight plane that starts at the take-off point and is inclined at 35° below the horizontal, as shown. The jumper is modelled as a particle and air resistance is negligible.

Take-off ramp, landing slope and path of the jumper (drawn to scale).

(a)

(i)

Show that the time of flight is t = 2u tan α/g, where u is the take-off speed and α the angle of the slope.

(2)

(ii)

Calculate the distance along the slope from the take-off point to the landing point.

(2)

(iii)

Show that, on landing, the velocity of the jumper makes an angle φ with the horizontal where tan φ = 2 tan α.

(2)

(iv)

Hence calculate the angle between the velocity of the jumper and the landing slope at the moment of landing.

(1)

(b)

(i)

The jumper has a mass of 65 kg. On landing, the component of velocity perpendicular to the slope is reduced to zero in 0.40 s. Determine the average normal force exerted by the slope on the jumper during this time.

(3)

(c)

(i)

The take-off speed is increased by 10 %. Deduce the percentage change in the distance along the slope from the take-off point to the landing point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Horizontal distance x = ut and vertical drop y = ½gt² (the jumper leaves horizontally, so the initial vertical velocity is zero)	✓ 1
On landing the jumper is on the slope, so y/x = tan α: ½gt²/(ut) = tan α, giving t = 2u tan α/g	✓ 1	The condition y/x = tan α (the hidden step) must be seen.
Part (a)(ii)
t = 2 × 26 × tan 35°/9.81 = 3.71 s; x = 26 × 3.71 = 96.5 m	✓ 1	Allow ECF from (a)(i).
Distance along the slope = 96.5/cos 35° = 118 m	✓ 1	Accept 117–119 m. ALT: √(x² + y²) with y = 67.6 m. Quoting the horizontal distance 96.5 m scores [1 max].
Part (a)(iii)
On landing the horizontal velocity is u and the vertical velocity is gt	✓ 1
tan φ = gt/u = g(2u tan α/g)/u = 2 tan α	✓ 1	Allow ECF from (a)(i). The substitution of the time of flight must be seen.
Part (a)(iv)
φ = tan⁻¹(2 tan 35°) = 54.5°, so the angle to the slope = 54.5° − 35° = 19.5°	✓ 1	Allow ECF from (a)(iii). Accept 19–20°.
Part (b)(i)
Landing speed = √(26² + (9.81 × 3.71)²) = 44.7 m s⁻¹; perpendicular component = 44.7 sin 19.5° = 14.9 m s⁻¹	✓ 1	Allow ECF from (a)(ii) and (a)(iv). ALT: gt cos 35° − u sin 35° = 14.9 m s⁻¹.
Resultant perpendicular force = mΔv/Δt = 65 × 14.9/0.40 = 2.42 × 10³ N	✓ 1	Allow ECF from the perpendicular component.
The normal force must also balance the component of the weight perpendicular to the slope, 65 × 9.81 × cos 35° = 522 N, so F_N = 2.42 × 10³ + 522 = 2.9 × 10³ N	✓ 1	Accept 2.9–3.0 × 10³ N. The hidden step is adding the weight component; an answer of 2.4 × 10³ N scores [2 max].
Part (c)(i)
t ∝ u and x = ut ∝ u² (the angle of the slope is unchanged), so the distance along the slope ∝ u²	✓ 1	Allow ECF from (a)(i).
1.10² = 1.21: the distance increases by 21 %	✓ 1	An answer of 10 % scores 0.

Answers: (a)(ii) 118 m · (a)(iv) 19.5° · (b)(i) 2.9 × 10³ N · (c)(i) +21 % (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 (projectiles launched horizontally); A.2 — impulse and change of momentum; free-body diagrams analysed to find the resultant force Command term: Determine

A.1 Kinematics: IB Physics HL exam-style questions

15 practice questions on A.1 Kinematics

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