A Level Mathematics 9709 Paper 42, Oct/Nov 2025

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9709/42 · Oct/Nov 2025

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The diagram shows the displacement-time graph for the motion of a particle. The particle starts from rest at a point $O$ and travels with constant acceleration $a m s^{- 2}$ , taking $4 s$ to move a distance of $x m$ . The particle then returns to $O$ with constant speed of $40 m s^{- 1}$ , over a period of $2 s$ . Find the value of $x$ and the value of $a$ .

[3]

A railway locomotive of mass $240 000 kg$ is towing a coach of mass $36 000 kg$ down a hill inclined at an angle of $\sin^{- 1} 0.04$ to the horizontal. The driving force produced by the locomotive is $450 000 N$ and there are resistances to motion of $120 000 N$ on the locomotive and $15 000 N$ on the coach. The coupling between the locomotive and the coach is light, rigid and parallel to the hill. Find the acceleration of the locomotive and the tension in the coupling.

[5]

A car of mass $1400 kg$ travelling on a straight horizontal road accelerates uniformly from a speed of $15 m s^{- 1}$ to $20 m s^{- 1}$ over a distance of $175 m$ .

Find the time taken to travel the $175 m$ .

[2]

b(i)

There is a constant resistance force of $800 N$ to the motion of the car. Use an energy method throughout to find the average power of the car's engine as the speed increases from $15 m s^{- 1}$ to $20 m s^{- 1}$ .

[4]

b(ii)

Find the steady speed that the car could maintain on the horizontal road if the engine is working at the power found in (b)(i).

[2]

A particle $P$ of mass $0.1 kg$ is projected vertically upwards with speed $30 m s^{- 1}$ from horizontal ground. At the same instant a particle $Q$ of mass $0.4 kg$ is projected vertically upwards with speed $10 m s^{- 1}$ from a height of $15 m$ above the ground. $P$ and $Q$ move in the same vertical line.

Find the height above the ground at which $P$ and $Q$ collide.

[4]

When $P$ and $Q$ collide, they coalesce. Find the speed of the combined particle at the instant that it reaches the ground.

[5]

A block of mass $5 kg$ is being pulled straight down a line of greatest slope of a rough plane by a force of magnitude $20 N$ . The plane is inclined at an angle of $10^{\circ}$ to the horizontal and the $20 N$ force acts at an angle of $35^{\circ}$ above the line of greatest slope of the plane (see diagram). The coefficient of friction between the block and the plane is $0.4$ . The speed of the block when it passes a point $O$ is $2 m s^{- 1}$ . Find the speed of the block when it has moved $3 m$ down the plane from $O$ .

[6]

A particle starts from rest at a point $O$ . The acceleration of the particle at time $t s$ after leaving $O$ is $a m s^{- 2}$ , where $a = 2 (t + 1)^{- \frac{1}{2}} - 1 for t ⩾ 0.$ Find the distance that the particle travels from $O$ until the time at which its acceleration is zero.

[8]

Four coplanar forces of magnitudes $20 N$ , $30 N$ , $15 N$ and $25 N$ act at a point $P$ in the directions shown in Fig. 7.1. The forces act in a vertical plane. The resultant of these forces has magnitude $S N$ and acts at an angle $α^{\circ}$ below the horizontal as shown in Fig. 7.2.

Find the value of $S$ and the value of $α$ .

[6]

A small ring of mass $0.6 kg$ is threaded on a rough straight horizontal wire. The four forces shown in Fig. 7.1 act on the ring and are in the same vertical plane as the wire. The ring starts from rest and takes $3 s$ to travel a distance of $2 m$ along the wire. Find the coefficient of friction between the ring and the wire.

[5]

Mark scheme

9709/42 · Oct/Nov 2025

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1 · 3 marks

M1
Any method to find $x$ , or $a$ or getting an equation relating $x$ and $a$ For example: $\frac{x}{2} = 40$ or $x = 2 \times 40$ or $x = 40 \times 2 + \frac{1}{2} (0) 2^{2}$ or $\frac{x - 0}{4 - 6} = - 40$ or $x = \frac{1}{2} \times a \times 4^{2}$
A1
$x = 80$
A1
$a = 10$

2 · 5 marks

M1
Attempt at Newton's second law at least once
A2
$450000 + 240000 g \times 0.04 - 120000 - T = 240000 a [\begin{array}{c} 450000 + 96000 - 120000 - T = 240000 a \\ 426000 - T = 240000 a \end{array}]$ $T + 36000 g \times 0.04 - 15000 = 36000 a [\begin{array}{c} T + 14400 - 15000 = 36000 a \\ T - 600 = 36000 a \end{array}]$ $450000 + (240000 + 36000) g \times 0.04 - 120000 - 15000 = (240000 + 36000) a [\begin{array}{c} 450000 + 110400 - 120000 - 15000 = 276000 a \\ 425400 = 276000 a \end{array}]$
1 mark
Solving for $a$ or $T$ , from equation(s) with the correct number of relevant terms
A1
Acceleration = $1.54 {ms}^{- 2}$ and Tension = $56100 N$

3a · 2 marks

M1
$175 = \frac{1}{2} (15 + 20) t$
A1
Time = 10 s

3b(i) · 4 marks

B1
Change in KE $= \pm (\frac{1}{2} \times 1400 \times 20^{2} - \frac{1}{2} \times 1400 \times 15^{2}) [= \pm (280000 - 157500) = \pm 122500]$ OR Work done against resistance $= \pm 800 \times 175 [= \pm 140000]$
M1
Work done by engine $= \frac{1}{2} \times 1400 \times 20^{2} - \frac{1}{2} \times 1400 \times 15^{2} + 800 \times 175$
1 mark
Av power $\times (t h e i r 10) = \frac{1}{2} \times 1400 \times 20^{2} - \frac{1}{2} \times 1400 \times 15^{2} + 800 \times 175$
A1
Average power = $26 250 W$

3b(ii) · 2 marks

M1
$t h e i r 26250 = 800 v$
A1
Speed = 32.8125 or $\frac{525}{16} {m s}^{- 1}$

4a · 4 marks

M1
[Height gained by $P$ $(s_{P}) =] 30 t - \frac{1}{2} g t^{2}$ [Height lost by $Q$ $(s_{Q}) =] 10 t - \frac{1}{2} g t^{2}$
1 mark
Meet when $30 t - \frac{1}{2} g t^{2} = 10 t - \frac{1}{2} g t^{2} + 15$
A1
$t = \frac{3}{4} or 0.75$
A1
Height = $19.6875 m or \frac{315}{16} m$

4b · 5 marks

B1
$[v_{P} =] \pm (30 - \frac{3}{4} g) [= \pm 22.5] [v_{Q} =] \pm (10 - \frac{3}{4} g) [= \pm 2.5]$ OR $[v_{P} =] \pm \sqrt{30^{2} - 2 g \times 19.6875} [= \pm 22.5]$ $[v_{Q} =] \pm \sqrt{10^{2} - 2 g \times (19.6875 - 15)} [= \pm 2.5]$
1 mark
$0.1 \times 22.5 + 0.4 \times 2.5 = (0.1 + 0.4) \times v_{P Q}$
A1
$v_{P Q} = 6.5$
1 mark
$v^{2} = t h e i r {6.5}^{2} + 2 g \times t h e i r 19.6875$ OR $0 = t h e i r {6.5}^{2} + 2 (- g) s ⟹ s = 2.1125$ , height above ground $= 2.1125 + 19.6875 = 21.8$ hence $v^{2} = 0^{2} + 2 g \times 21.8$
A1
Speed = $20.9 {ms}^{- 1}$ or $2 \sqrt{109} {ms}^{- 1}$

5 · 6 marks

M1
For resolving in any direction to form an equation
A1
$R = 5 g \cos 10 - 20 \sin 35 [= 37.76885892]$
A1
$20 \cos 35 + 5 g \sin 10 - F = 5 a [25.06544977 - F = 5 a]$
1 mark
For use of $F = 0.4 R$ to get an equation in $a$ only
1 mark
$v^{2} = 2^{2} + 2 \times (t h e i r 1.99) \times 3$
A1
Speed = $3.99 {ms}^{- 1}$

6 · 8 marks

B1
Acceleration = 0 when $[2 (t + 1)^{- \frac{1}{2}} - 1 = 0 ⟹ (t + 1)^{\frac{1}{2}} = 2] t = 3$
M1
For attempt at integration of $(a =) 2 (t + 1)^{- \frac{1}{2}} - 1$
A1
$[v =] \frac{2}{- \frac{1}{2} + 1} (t + 1)^{- \frac{1}{2} + 1} - t [+ c] = 4 (t + 1)^{\frac{1}{2}} - t [+ c]$
1 mark
$v = 0 at t = 0 ⟹ c = - 4 [v = 4 (t + 1)^{\frac{1}{2}} - t - 4]$
1 mark
Attempt to integrate $t h e i r v$ , which has come from integration
A1
$[s =] \frac{4}{\frac{1}{2} + 1} (t + 1)^{\frac{1}{2} + 1} - \frac{1}{1 + 1} t^{1 + 1} - 4 t [+ k] = \frac{8}{3} (t + 1)^{\frac{3}{2}} - \frac{1}{2} t^{2} - 4 t [+ k]$
1 mark
$(\frac{8}{3} \times 4^{\frac{3}{2}} - \frac{1}{2} \times 3^{2} - 4 \times 3) - (\frac{8}{3} - 0 - 0)$ OR evaluate $k$ from using $s = 0$ when $t = 0$ AND substitute $t = t h e i r 3$
A1
Distance = $\frac{13}{6} m or 2.17 m$

7a · 6 marks

M1
For attempt to resolve in any direction to form an expression or equation
A1
$[X =] \pm (30 \cos 30 + 15 \cos 45 - 20 \cos 20)$ OR $[S \cos α =] \pm (30 \cos 30 + 15 \cos 45 - 20 \cos 20)$
A1
$[Y =] \pm (25 + 15 \sin 45 - 20 \sin 20 - 30 \sin 30)$ OR $[S \sin α =] \pm (25 + 15 \sin 45 - 20 \sin 20 - 30 \sin 30)$
1 mark
$S = \sqrt{(t h e i r 17.7935 \dots)^{2} + (t h e i r 13.76661 \dots)^{2}}$ OR $S = \frac{t h e i r 17.7935 \dots}{\cos (t h e i r 37.7278 \dots)}$ OR $S = \frac{t h e i r 13.76661 \dots}{\sin (t h e i r 37.7278 \dots)}$
1 mark
$α = \tan^{- 1} (\frac{t h e i r 13.7661 \dots}{t h e i r 17.7935 \dots})$ OR $α = \cos^{- 1} (\frac{t h e i r 17.7935 \dots}{t h e i r 22.4970 \dots})$ OR $α = \sin^{- 1} (\frac{t h e i r 13.7661 \dots}{t h e i r 22.4970 \dots})$
A1
$[S =] 22.5 [α =] 37.7$

7b · 5 marks

M1
$(t h e i r 17.7935 \dots) - F = 0.6 a$ OR $30 \cos 30 + 15 \cos 45 - 20 \cos 20 - F = 0.6 a$ OR $(t h e i r S) \cos (t h e i r α) - F = 0.6 a [17.8025 - F = 0.6 a]$
M1
$2 = 0 + \frac{1}{2} a \times 3^{2} [a = \frac{4}{9}]$
B1
$R = t h e i r (13.7661 \dots) + 0.6 g [= 19.7661 \dots]$ OR $R = 25 + 15 \sin 45 - 20 \sin 20 - 30 \sin 30 + 0.6 g$ OR $R = (t h e i r S) \sin (t h e i r α) + 0.6 g [R = 13.759 + 0.6 g]$
1 mark
$(t h e i r 17.7935) \dots - μ \times 19.7661 = 0.6 \times (t h e i r \frac{4}{9})$
A1
$μ = 0.887$

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