A Level Mathematics 9709 Paper 43, Oct/Nov 2025

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9709/43 · Oct/Nov 2025

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1

The diagram shows the velocity-time graph for the motion of an athlete. The athlete runs in a straight line from point A to point B, runs back to point A and finishes at point B. The graph consists of four straight line segments.

Figure 1 of 1 for question 1
a

Find the deceleration of the athlete 7 seconds after leaving A.

[1]
b

Find the total distance travelled by the athlete.

[2]
2

The engine of a motorcycle can generate a maximum power of 105 kW. The mass of the motorcycle and its rider is 400 kg. The total resistance to the motion of the motorcycle and its rider is cv N, where v ms−1 is the motorcyclist's speed and c is a constant. The motorcyclist travels along a straight horizontal road under maximum engine power. When the motorcyclist's speed is 35 ms−1, his acceleration is 0.5 ms−2.

a

Show that c=80.

[3]
b

The motorcyclist now travels up a straight hill under maximum engine power. The hill makes an angle of sin⁡−1(320) with the horizontal. Find the steady speed at which the motorcyclist travels up the hill.

[3]
3

A block of mass 4 kg is pulled along a rough horizontal road by a constant force of magnitude 25 N acting at an angle of 36∘ above the horizontal. The block moves in a straight line passing through two points A and B on the road, where AB=120 m. The coefficient of friction between the block and the road is 0.4.

a

Find the work done against friction in moving the block from A to B.

[3]
b

The speed of the block at A is 7 ms−1. Use an energy method to find the speed of the block at B.

[4]
4

A particle P moves in a straight line. At time t s after passing through a point O on the line, the displacement of P from O is s m, where s=0.01t3−0.3t2−2.07t.

a

Find the value of s when P has its minimum velocity, and find also the speed of P at this instant.

[6]
b

Find the acceleration of P at the instant when the direction of motion of P changes.

[4]
5

Particles A and B, of masses 2 kg and 6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley and the particles hang vertically below the pulley. Both particles are initially held at rest at a height of 3.2 m above horizontal ground (see diagram). Particle A is projected vertically downwards with a speed of 1.2 ms−1.

a

Find the tension in the string and the magnitude of the acceleration of the particles.

[4]
b

In the subsequent motion, A does not hit the ground and neither particle reaches the pulley. When B hits the ground, it does not rebound. Find the time that B is in motion.

[2]
c

Find the distance between the lowest and highest points that A reaches.

[4]
6

Particle P of mass m kg and particle Q of mass 2 kg are free to move on a smooth horizontal plane. P and Q are moving directly towards each other with speeds u ms−1 and 3u ms−1 respectively. P and Q collide and the direction of motion of each particle is reversed by the collision. Immediately after the collision the speed of P is u3 ms−1.

a

Find, in terms of m and u, an expression for the velocity of Q after the collision and hence show that m>4.5.

[3]
b

Q subsequently hits a vertical wall which is perpendicular to the direction of motion of Q. The speed of Q after the impact with the wall is a quarter of its speed before the impact with the wall. There are no further collisions between P and Q. Given that m is an integer, determine the largest possible value of m.

[4]
7

A particle P of mass 6 kg lies on a rough plane inclined at an acute angle θ to the horizontal. A horizontal force of magnitude 12 N acts on P as shown in the diagram. The line of action of this horizontal force lies in a vertical plane which contains the line of greatest slope of the plane that passes through P. The coefficient of friction between P and the plane is μ. P is in equilibrium and on the point of sliding down the plane.

Figure 1 of 1 for question 7
a

Show that μ=5tan⁡θ−15+tan⁡θ.

[5]
b

Hence find the set of possible values of θ.

[2]

Mark scheme

9709/43 · Oct/Nov 2025

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1a · 1 mark

  • B1

    1.6 ms−2

1b · 2 marks

  • M1

    [3×]12×10×8 or [6×]12×5×8

  • A1

    120 m

2a · 3 marks

  • B1

    Driving force F=10500035 [=3000]

  • M1

    F−35×c=400×0.5

  • A1

    10500035−35×c=400×0.5⇒c=80 (2800=35c⇒c=80)

2b · 3 marks

  • B1

    105000v−80×v−400g×320[=0]⇒80v2+600v−105000[=0]

  • M1

    OR 80v2+600v=105000

  • A1

    Speed = 32.7 m s−1 ONLY

3a · 3 marks

  • M1

    R+25sin⁡36=4g [⇒R=25.30536...]

  • 1 mark

    Work done against friction =120×25×(40−25sin⁡36)

  • A1

    1210 or 1215 J

3b · 4 marks

  • B1

    KE change = ±(12×4×vB2−12×4×72)

  • B1

    Work Done by pulling force = ±25cos⁡36×120 [=±2427.050...]

  • M1

    12×4×vB2−12×4×72=25cos⁡36×120−’1214.65...’ [2vB2−98=2427.050...−’1214.65...’]

  • A1

    vB=25.6 m s−1

4a · 6 marks

  • M1

    Attempt to differentiate s=0.01t3−0.3t2−2.07t

  • A1

    [v=]0.03t2−0.6t−2.07

  • A1

    [a=]0.06t−0.6 OR [v=]0.03(t−10)2−5.07

  • 1 mark

    0.06t−0.6=0 (⇒t=10)

  • A1

    s=−40.7

  • A1

    v=−5.07⇒speed is 5.07 m s−1

4b · 4 marks

  • M1

    0.03t2−0.6t−2.07=0

  • A1

    t=23

  • 1 mark

    a=0.06×23−0.6

  • A1

    a=0.78 m s−2 ONLY

5a · 4 marks

  • M1

    Attempt at N2L for either A or B or the system

  • A1

    A: T−2g=2a B: 6g−T=6a System: 6g−2g=(2+6)a

  • 1 mark

    Solving for T

  • A1

    a=0.5g or 5 m s−2 and T=3g or 30 N

5b · 2 marks

  • M1

    −3.2=1.2t+12×(−5)×t2 OR 3.2=−1.2t+12×5×t2

  • A1

    1.4(0) s

5c · 4 marks

  • M1

    Distance A travels from initial position to first time at rest: 0=1.22+2×(−5)×s1 (⇒s1=0.144) OR 0=(−1.2)2+2×(5)×s1 (⇒s1=−0.144) OR s1=1.2×0.24+12×(−5)×0.242 (⇒s1=0.144)

  • M1

    w2=1.22+2×5×3.2 or w2=02+2×5×(3.2+0.144) (w2=33.44,w=5.7827...) OR w=1.2+5(1.1565...−0.24)

  • 1 mark

    Distance A travels from a point 6.4 metres above the ground to rest: 0=33.44+2×(−10)×s2 (⇒s2=1.672)

  • A1

    Difference in distances is s1+3.2+s2=5.016 m

6a · 3 marks

  • M1

    m×u+2×(−3u)=m×(−13u)+2×v

  • A1

    v=23mu−3u

  • A1

    v>0⇒23mu−3u>0∴m>4.5

6b · 4 marks

  • B1

    [Q’s speed is reduced by 75% by its impact with the wall ⇒] speed after impact is 14(23mu−3u)[=16mu−34u]

  • 1 mark

    No further collisions therefore 13u⩾16mu−34u

  • 1 mark

    Solve inequality to obtain m⩽6.5

  • A1

    (4.5<)m⩽6.5⇒m=6

7a · 5 marks

  • M1

    For resolving in any direction to form an equation

  • A1

    Perpendicular to plane: R=6gcos⁡θ+12sin⁡θ

  • A1

    Parallel to plane: F+12cos⁡θ=6gsin⁡θ or 6gsin⁡θ−12cos⁡θ=μ(...)

  • 1 mark

    For use of F=μR to get an equation in μ and θ only

  • A1

    6gsin⁡θ−12cos⁡θ=μ(6gcos⁡θ+12sin⁡θ) leading to μ=5tan⁡θ−15+tan⁡θ With some evidence of intermediate step e.g. 12cos⁡θ(5sin⁡θcos⁡θ−1)12cos⁡θ(5+sin⁡θcos⁡θ) OR 12cos⁡θ(5tan⁡θ−1)12cos⁡θ(5+tan⁡θ) OR ‘divide by cosine’ OR use/mention of tan⁡θ=sin⁡θcos⁡θ OR sin⁡θ=tan⁡θcos⁡θ, etc.

7b · 2 marks

  • M1

    μ>0⇒tan⁡θ>15

  • A1

    11.3<θ(<90)

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