Cambridge AS & A Level Mathematics 9709/42 is the May/June 2026 Paper 4 (Mechanics), variant 2. It lasts 1 hour 15 minutes and is worth 50 marks, and it counts towards AS Level (with Paper 1) or the full A Level (with Papers 1, 3 and 5). It had seven compulsory questions, answered on the question paper. The biggest was Question 3 (10 marks), on particles connected over a pulley and the work done by friction after the string breaks; Questions 4 and 7 were worth 9 marks each. This guide goes through the paper question by question with the final answers and what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all six 9709 papers from this series, see A Level Maths 9709 May/June 2026 past papers: what came up and how it was marked.

9709/42 at a glance

Question	Topic	Marks
1	Power at constant speed, then the acceleration after a power increase	4
2	Two particles coalesce, then rebound from a wall	4
3	Connected particles over a pulley: tension, coefficient of friction, work done by friction	10
4	Rough slope: energy method for the distance up, then the total time up and down	9
5	One particle released, another projected downwards a second later	6
6	A crate pulled by a rope at an angle, then sliding to rest	8
7	Variable acceleration: find a constant, then a distance by integration	9
	Total	50

Question-by-question walkthrough

The mark scheme uses three main types of mark. M marks are for a valid method applied to this problem (quoting a formula is not enough; you must substitute into it). A marks are for accurate answers and cannot be given without the M mark before them. B marks are for a correct result or statement and stand alone. A mark with a star (M1*, B1*) has later marks depending on it; those later marks are written DM or DB. FT (follow through) means an A or B mark can be earned from your own earlier wrong value. The cover asks for non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees, and for g = 10 m s⁻² throughout.

Question 1: power of a car (4 marks)

A 960 kg car on a level road against a constant 620 N resistance.

• (a) Power at 25 m s⁻¹ (2 marks): at constant speed the driving force equals the resistance, so P/25 − 620 = 0 and P = 15 500 W (15.5 kW). M1 for Newton’s second law using P = Fv, then A1. The bare number 15 500 was accepted without units, but 15.5 needed “kW”.
• (b) Acceleration after a 12 kW increase (2 marks): the new driving force is 27 500 ÷ 25 = 1100 N, so 1100 − 620 = 960a and a = 0.5 m s⁻². M1 for Newton’s second law with the new power, then A1FT from your power in part (a).

Question 2: coalescing particles and a wall (4 marks)

A (2.5 kg) moving at 3 m s⁻¹ hits B (3.5 kg), which is at rest 9 m from a wall. They coalesce to form C, which hits the wall and rebounds with its speed reduced by 25%.

• (a) Speed after the rebound (2 marks): conservation of momentum gives 2.5 × 3 = 6v, so C moves at 1.25 m s⁻¹; after the wall it moves at 0.75 × 1.25 = 0.9375 m s⁻¹ (15/16). M1 for the momentum equation, then A1 (0.938 accepted). Putting weights (mg) into the momentum equation capped the part at M1 A0.
• (b) Time until C is 9 m from the wall again (2 marks): 9 ÷ 1.25 = 7.2 s to the wall and 9 ÷ 0.9375 = 9.6 s back, so 16.8 s. M1 for both times or the total, then A1. The track is smooth, so the speeds are constant; working out an acceleration was M0.

Question 3: connected particles over a pulley (10 marks)

A (0.5 kg) is on a rough horizontal plane and B (0.8 kg) on a smooth plane at 55°, joined by a string over a pulley. From rest, B moves 0.3 m down the slope in 0.4 s.

• (a) Tension (4 marks): 0.3 = ½a(0.4)² gives a = 3.75 m s⁻². For B, 0.8g sin 55° − T = 0.8 × 3.75, so T = 3.55 N (3.553…). B1* for a = 3.75, M1* for Newton’s second law for B (correct number of terms; sign errors and a sin/cos mix allowed), DM1 for the correct equation, A1. An energy equation for B, using its speed of 1.5 m s⁻¹ after 0.3 m, earned the same marks.
• (b) Coefficient of friction (3 marks): for A, T − F = 0.5 × 3.75, so F = 1.678 N; the normal reaction is 0.5g = 5 N, so μ = 0.336 (0.335 if you used T = 3.55). M1 for Newton’s second law for A or the whole system, A1FT for the equation with R = 0.5g, A1 for the answer.
• (c) Work done by friction after the string breaks (3 marks): A’s speed when the string breaks is 3.75 × 0.4 = 1.5 m s⁻¹, and friction removes all of A’s kinetic energy, so the work done is ½ × 0.5 × 1.5² = 0.5625 J (9/16 J). B1FT* for the speed, DM1 for “work done by friction = change in kinetic energy” for the motion after the break only, then A1 (−0.5625 also accepted). The scheme said do not ISW (ignore subsequent working) here, so going on to change a correct value cost the mark. Finding the deceleration and the stopping distance (0.335 m) first, then multiplying by the friction force, was an accepted alternative.

Question 4: energy on a rough slope (9 marks)

P is projected at 4 m s⁻¹ up a rough slope with tan θ = 4/3 (so sin θ = 0.8 and cos θ = 0.6) and μ = ⅓, and stops at B.

• (a) Show AB = 0.8 m by an energy method (4 marks): R = mg cos θ = 6m and F = ⅓R = 2m. Kinetic energy lost equals potential energy gained plus work against friction: ½m × 4² = mg × 0.8d + 2md, so 8m = 10md and d = 0.8. B1 for R, M1* for F = μR, DM1 for the three-term work-energy equation, A1. This is AG (answer given on the paper), so A0 if any error is seen. Working with a rounded angle and getting 0.8006 ”≈ 0.8” was not accepted, and setting m to a number without justification, or leaving m out, scored 0.
• (b) Total time from A back to A (5 marks): up the slope, 0.8 = ½(4 + 0)t gives 0.4 s. Down the slope, friction now acts up the plane: a = g sin θ − ⅓g cos θ = 8 − 2 = 6 m s⁻², and 0.8 = ½ × 6 × t² gives t = 0.516 s. Total 0.916 s (exactly (6 + 2√15)/15). B1 for the time up, M1* for Newton’s second law down the plane (no sign errors allowed here), A1 for a = 6, DM1 for the equation in t, A1. An energy route (v² = 9.6 at the bottom) earned the same marks. Using a = 10 down the plane scored M0.

Question 5: two particles reaching the ground together (6 marks)

P is released from rest at height h; one second later Q is projected downwards at 18 m s⁻¹ from the same height. They land at the same time.

• Answer: h = 13.2 (exactly 845/64 = 13.203125).
• Working: h = 5tP² and h = 18tQ + 5tQ², with tP = tQ + 1. Then 5(tQ + 1)² = 18tQ + 5tQ², and the squared terms cancel to leave 10tQ + 5 = 18tQ, so tQ = 5/8 s and tP = 13/8 s. Then h = 5 × (13/8)².
• Marking: B1* for each displacement equation, DM1 for linking the times to get one equation in one time, DM1 for expanding and solving the linear equation, A1 for either time, A1 for h. Each DM depended on all the marks before it. Working with P’s speed (10 m s⁻¹) and drop (5 m) at the moment Q is projected was an accepted alternative. A misread of the one-second gap was not treated as a minor misread.

Question 6: a crate pulled by a rope (8 marks)

The rope pulls with 26 N at θ = sin⁻¹(5/13) above the horizontal (so cos θ = 12/13); the crate accelerates at 1.5 m s⁻² and μ = 0.25.

• (a) Weight W (5 marks): horizontally, 26 × 12/13 − F = (W/10) × 1.5, so 24 − F = 0.15W. Vertically, R + 26 × 5/13 = W, so R = W − 10. With F = 0.25R: 24 − 0.25(W − 10) = 0.15W, so 26.5 = 0.4W and W = 66.25 N (66.3 accepted). M1* for Newton’s second law in either direction, A1 for each of the two resolved equations, DM1 for using F = 0.25R to get one equation, A1. For checking: R = 56.25 N and F = 14.0625 N.
• (b) Distance AB after the rope is removed (3 marks): now R = W, so the deceleration is 0.25g = 2.5 m s⁻², whatever the weight. Then 0 = 8² − 2 × 2.5 × AB, so AB = 12.8 m. B1* for the deceleration, DM1 for v² = u² + 2as with a negative acceleration, A1. A work-energy route was also accepted. Treating W as the mass in F = ma scored nothing.

Question 7: variable acceleration (9 marks)

v = k₁(4t + 11)^(1/2) − ¼(2t + 1)² + k₂, and at t = 1.25 the deceleration is 0.5 m s⁻².

• (a) k₁ (4 marks): a = dv/dt = 2k₁(4t + 11)^(−1/2) − (2t + 1). At t = 1.25, 4t + 11 = 16, so 2k₁ ÷ 4 − 3.5 = −0.5 and k₁ = 6. M1* for differentiating to the right form, A1 for the correct derivative, DM1 for substituting t = 1.25 and the acceleration, A1. A deceleration means a = −0.5; the scheme condoned k₁ = 8 from using +0.5 and followed it through part (b).
• (b) Distance for 0 ≤ t ≤ 1 (5 marks): v = 0 at t = 3.5 gives 6 × 5 − 16 + k₂ = 0, so k₂ = −14. Integrating, s = (4t + 11)^(3/2) − (2t + 1)³/24 − 14t. Since P is at rest only at t = 3.5, it does not change direction between t = 0 and t = 1, so the distance is s(1) − s(0) = 42.970 − 36.441 = 6.53 m. M1* for finding k₂, M1* for integrating to the right form, A1FT for the integral, DM1 for using the limits correctly, A1. A calculator answer with no integration shown earned at most 2 of the 5 marks.

Where the marks are on 9709/42

• Forces, friction and Newton’s second law carried 20 of the 50 marks: Question 3(a) and (b) (7), Question 4(b) (5) and Question 6 (8).
• Work, energy and power carried 11 marks: power in Question 1 (4), the work done by friction in Question 3(c) (3) and the energy method in Question 4(a) (4).
• Variable acceleration with calculus carried 9 marks (Question 7).
• Constant-acceleration kinematics carried 6 marks (Question 5), although the suvat equations also appeared inside Questions 3, 4 and 6.
• Momentum carried 4 marks (Question 2).

Friction or a resisting force appeared in four of the seven questions (1, 3, 4 and 6). Most of the paper’s method marks were for setting up Newton’s second law or an energy equation with the right number of terms.

How to self-mark 9709/42 with the mark scheme

1. Check every equation for the right number of terms first. The M marks for Newton’s second law (Questions 1, 3, 4 and 6) and the work-energy principle (Questions 3(c) and 4(a)) needed every relevant force and to be dimensionally consistent. Sign errors and a sin/cos mix were usually allowed for the M mark, but sign errors were not allowed for Newton’s second law down the plane in Question 4(b).
2. Follow the stars. In Questions 3, 4, 5, 6 and 7, later DM marks depend on an earlier starred mark. If you miss the starred mark, cross out the dependent marks too.
3. Apply follow through where it is written. Question 1(b) follows your power from (a); Question 3(b) and (c) follow your a and T from (a); Question 7(b) follows your k₁ and k₂.
4. Be strict with the “show that”. In Question 4(a) the answer 0.8 is printed. Any slip, a rounded angle or a missing m costs the A mark or more.
5. Use the method the question asks for. Question 4(a) said “using an energy method”. In Question 7(b), a distance with no integration shown is worth at most 2 marks.
6. Check accuracy. Answers are accepted to 3 significant figures (AWRT, answer which rounds to): 3.55 N, 0.336, 0.916 s, 13.2, 66.3 N, 6.53 m. Keep full calculator values until the end; the scheme accepted 0.335 for μ only when it came from T = 3.55.
7. Log each lost mark as a missing force, a sign or resolving error, a lost dependent mark, or an accuracy slip.

Sit 9709/42 as a timed mock

The paper gives 75 minutes for 50 marks, a minute and a half per mark. A workable plan:

Block	Marks	Time
Questions 1 and 2	8	11 minutes
Question 3 (connected particles)	10	15 minutes
Question 4 (rough slope)	9	13 minutes
Question 5 (two particles)	6	9 minutes
Question 6 (crate and rope)	8	11 minutes
Question 7 (variable acceleration)	9	12 minutes
Check, especially signs and the show-that	–	4 minutes

Draw a force diagram for Questions 3, 4 and 6 before writing any equation; it is the quickest way to make sure no force is missing.

Other May/June 2026 9709 papers

• 9709 Paper 12 (Pure Mathematics 1)
• 9709 Paper 22 (Pure Mathematics 2)
• 9709 Paper 32 (Pure Mathematics 3)
• 9709 Paper 52 (Probability & Statistics 1)
• 9709 Paper 62 (Probability & Statistics 2)

Practise and download

• Download 9709/42 and its mark scheme on the Learning Portal →
• All 9709 May/June 2026 papers: overview and marking patterns
• Try a free A Level Maths quiz

Need help with a 9709 topic? Book a free trial lesson with an A Level Maths tutor →

Frequently asked questions

What topics were on 9709 Paper 42 May/June 2026? +

Seven questions: the power of a car, a collision followed by a rebound from a wall, connected particles over a pulley with friction, an energy method on a rough slope, two particles falling to the ground at the same time, a crate pulled by a rope at an angle, and variable acceleration using calculus.

How many marks is 9709/42 and how long is it? +

It is out of 50 marks and lasts 1 hour 15 minutes. All seven questions are compulsory. Question 3 (connected particles and the work done by friction) is the biggest at 10 marks, followed by Questions 4 and 7 at 9 marks each.

What was the answer to the falling particles question in 9709/42? +

h = 13.2 (exactly 845/64). With P falling for one second longer than Q, the equations 5tP² = h and 18tQ + 5tQ² = h with tP = tQ + 1 give a linear equation, so tQ = 5/8 s and tP = 13/8 s. The question was worth 6 marks.

Why did 9709/42 Question 4(a) need an energy method? +

The question said to use an energy method, so a Newton's second law solution does not answer it as set. The kinetic energy lost (8m J) equals the gain in potential energy plus the work against friction (10md J), giving AB = 0.8 m. Because the answer is printed, any error cost the final mark.

What was the distance travelled in 9709/42 Question 7? +

6.53 m. With k₁ = 6 from the deceleration and k₂ = −14 from v = 0 at t = 3.5, integrating v gives s = (4t + 11)^(3/2) − (2t + 1)³/24 − 14t, and s(1) − s(0) = 42.970 − 36.441. P does not stop between t = 0 and t = 1, so the displacement is the distance.

Where can I download 9709/42 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Mathematics 9709 past papers page.

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