A Level Mathematics 9709 Paper 41, May/June 2025

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Question paper

9709/41 · May/June 2025

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1

A block of mass 12 kg is being pulled by a rope up a rough plane. The plane is inclined at an angle of 20∘ above the horizontal. The rope pulling the block is parallel to a line of greatest slope of the plane. The coefficient of friction between the block and the plane is 0.4. The acceleration of the block is 2 m s−2. Find the tension in the rope.

[4]
Figure 1 of 1 for question 1
2

Three coplanar forces of magnitudes P N, 5 N and 10 N act at a point O, as shown in the diagram. The resultant of the three forces has magnitude Q N and acts in a direction perpendicular to the force of magnitude P N. Find the value of P and the value of Q.

[4]
3

The diagram shows the velocity-time graph of the motion of a cyclist. The graph consists of three straight line segments. The cyclist passes a point O with speed 3 m s−1 and then accelerates for 10 s with constant acceleration 0.5 m s−2. He then travels at constant speed for 30 s before decelerating, coming to rest at point P, covering a distance of 80 m whilst decelerating.

a

Find the total time taken for the journey from O to P.

[3]
b

On the given axes, sketch a displacement-time graph for the cyclist’s journey from O to P, showing on your graph the distances travelled after 10 s and 40 s.

[4]
4

A lorry of mass 18 000 kg is travelling along a straight road.

a(i)

On a horizontal section of the road, the power of the lorry’s engine is constant. There is a constant resistance to motion of 1600 N. The steady speed which the lorry can maintain with the engine working at power P W is 30 m s−1. Find the value of P.

[1]
a(ii)

At an instant when the speed of the lorry is 16 m s−1, its engine is working at a power of 40 kW. Find the acceleration of the lorry at this instant.

[2]
b

When the lorry has reached a speed of 20 m s−1, it begins to ascend a section of road inclined at an angle α∘ to the horizontal. The engine now works at a power of 120 kW. There is no change in the lorry’s speed as it ascends the hill. The constant resistance to motion remains 1600 N. Find the value of α.

[3]
5
a

When a particle P of mass m kg has speed u m s−1, its momentum is 4 N s and its kinetic energy is 16 J. Find the value of m and the value of u.

[3]
b

P is now projected on a smooth horizontal surface with speed v m s−1 directly towards a particle Q of mass 1.25 kg which is stationary. After P and Q collide, the velocity of P is w m s−1 and the velocity of Q is 2w m s−1. The loss of kinetic energy in the collision is 25 J. Find the value of v and the value of w.

[4]
6

Two particles, P and Q, of masses 0.3 kg and 0.6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth pulley fixed at a point B where the inclined planes AB and BC meet. P lies on the smooth plane AB which is inclined at an angle θ∘ to the horizontal where sin⁡θ∘=0.4. Q lies on the plane BC which is inclined at 30∘ to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram). The particles are released from rest.

Figure 1 of 1 for question 6
a

It is given that the plane BC is smooth. Find the tension in the string and the acceleration of Q.

[5]
b

It is given instead that the plane BC is rough. The work done against the frictional force when Q moves 2 m down the plane is 1.8 J. You should assume that P does not reach the pulley and that Q does not reach C. Use an energy method to find the speed of Q when it has moved 2 m down the plane.

[4]
7

A particle X moves along a straight track, starting from a point O at time t=0. The displacement of X from O at time t s is s m, where s=3t32−6t.

a

Find the time at which X is instantaneously at rest, and hence find the total distance travelled by X between t=0 and t=16.

[6]
b

A second particle Y moves along another straight track, starting from a point P at time t=0. The acceleration of Y at time t s is a m s−2, where a=0.8−0.6t. The velocity of Y when it leaves P is 7.5 m s−1. When the velocity of Y is −9.6 m s−1, show that the displacement of X from O is equal to the displacement of Y from P.

[7]

Mark scheme

9709/41 · May/June 2025

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1 · 4 marks

  • B1

    R=12gcos⁡20

  • 1 mark

    T−F−12gsin⁡20=12×2

  • 1 mark

    T−0.4×12gcos⁡20−12gsin⁡20=12×2 [T−45.1052⋯−41.0424⋯=24]

  • A1

    Tension = 110 N [110.147…]

2 · 4 marks

  • M1

    P+5sin⁡60−10sin⁡30=0

  • A1

    P=5−532 or 0.670 [0.66987…]

  • M1

    Q=±(10cos⁡30−5cos⁡60)

  • A1

    Q=1032−52 or 6.16 [6.16025…]

3a · 3 marks

  • B1

    Velocity at t=10 is 8 [m s−1]

  • M1

    12×8(T−40)=80 OR 12×8×t=80 [⇒t=20] and T=their t+40

  • A1

    Total time = 60s

3b · 4 marks

  • 1 mark

    Distance after 10s = 55m

  • 1 mark

    Distance after 40s = 295m

  • B1

    For at least one quadratic the right way up or a straight line with positive gradient between t=10 and t=40.

  • B1

    For fully correct, smooth and continuous at t=10 and t=40.

4a(i) · 1 mark

  • B1

    P30−1600=0⇒P=48000 only

4a(ii) · 2 marks

  • M1

    4000016−1600=18000a

  • A1

    Acceleration = 0.05 ms−2

4b · 3 marks

  • M1

    For attempt to resolve up the hill to form an equation

  • A1

    12000020−1600−18000gsin⁡α=0

  • A1

    α=1.40 [1.4007… ]

5a · 3 marks

  • B1

    12mu2=16

  • B1

    mu=4

  • B1

    u=8,m=0.5

5b · 4 marks

  • 1 mark

    0.5×v=0.5w+1.25×2w [v=6w]

  • 1 mark

    12×0.5×v2−(12×0.5×w2+12×1.25×(2w)2)=±25

  • 1 mark

    9w2−14w2−2.5w2−25=0 [254w2=25]

  • A1

    w=2,v=12

6a · 5 marks

  • 1 mark

    Use of Newton’s second law for P or Q or system

  • A2

    0.6gsin⁡30−T=0.6a T−0.3g×0.4=0.3a 0.6gsin⁡30−0.3g×0.4=(0.3+0.6)a

  • 1 mark

    For attempt to solve for T or a

  • A1

    a=2 ms−2 T=1.8N

6b · 4 marks

  • B1

    PE change for P=±0.3g×2×0.4[=±2.4] PE change for Q=±0.6g×2sin⁡30[=±6]

  • B1

    KE change = 12×(0.3+0.6)v2

  • M1

    12×0.9v2=0.6g×2sin⁡30−0.3g×2×0.4−1.8

  • A1

    Speed = 2ms−1

7a · 6 marks

  • 1 mark

    For an attempt at differentiation

  • A1

    v=92t12−6

  • M1

    v=0⇒t=169

  • 1 mark

    (3×1632−6×16)−(3×(169)32−6×169)=[8969] Or (3×(169)32−6×169)−0=[−329]

  • 1 mark

    For both i.e. F(16)−F(their 169)−[F(their 169)−[F(0)]] OR F(16)−F(their 169)+∣F(their 169)−[F(0)]∣ OR F(16)−2×F(their 169)

  • A1

    9289 m

7b · 7 marks

  • 1 mark

    For integration

  • A1

    v=0.8t−0.3t2+7.5

  • 1 mark

    v=−9.6⇒0.3t2−0.8t−17.1=0

  • A1

    t=9 [or −193]

  • 1 mark

    s=0.4t2−0.1t3+7.5t(+c)

  • B1

    At t=9 sX=3×932−6×9=27

  • B1

    At t=9 sY=0.4×92−0.1×93+7.5×9=27

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