October/November 2025 Paper 42 Worked Answers (A-Level Maths 9709 AS)

11 questions · 50 marks · 75 minutes

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1. Question 1

   3 marksDisplacement-Time Graph / Constant Acceleration
   Step 1: Find the distance $d$ from the return journey. At constant speed $30$ m/s for $3$ s: $d = 30 \times 3 = 90$ m. Step 2: Use $s = \dfrac{1}{2}at^2$ for the acceleration phase (starting from rest): $90 = \dfrac{1}{2}a(5)^2 = \dfrac{25a}{2}$. Step 3: Solve: $a = \dfrac{180}{25} = 7.2$ m/s$^2$.
   Method:

   Use the return journey to find the distance, then apply $s = \dfrac{1}{2}at^2$.
   Examiner tips

   • The return journey at constant speed gives the distance covered during acceleration
   • Starting from rest means $u = 0$, simplifying the suvat equation

2. Question 2

   5 marksConnected Particles on an Inclined Plane
   Step 1: Total mass $M = 200000 + 50000 = 250000$ kg. Step 2: Component of weight down the slope: $Mg\sin\theta = 250000 \times 10 \times 0.04 = 100000$ N. Step 3: Total resistance $= 95000 + 20000 = 115000$ N. Step 4: Apply Newton's second law down the slope: $350000 + 100000 - 115000 = 250000a$. Step 5: $335000 = 250000a$, so $a = 1.34$ m/s$^2$.
   Method:

   Apply N2L to the whole system to find $a$, then apply N2L to the coach alone to find the tension.
   Examiner tips

   • When finding acceleration, treat the whole system as one object
   • To find the coupling force, isolate one of the two bodies
   • A negative tension means the coupling is in compression (thrust)

3. Question 3a

   2 marksKinematics (Constant Acceleration)
   Step 1: Use the suvat equation $s = \dfrac{1}{2}(u + v)t$ where $s = 150$, $u = 12$, $v = 18$. Step 2: Substitute: $150 = \dfrac{1}{2}(12 + 18)t = 15t$. Step 3: Solve: $t = \dfrac{150}{15} = 10$ s.
   Method:

   Apply $s = \dfrac{1}{2}(u+v)t$ and solve for $t$.
   Examiner tips

   • Choose the suvat equation that uses the known quantities
   • $s = \dfrac{1}{2}(u+v)t$ is the most direct route when $a$ is not given

4. Question 3b(i)

   4 marksEnergy Method / Average Power
   Step 1: Find the change in kinetic energy: $\Delta KE = \dfrac{1}{2}(1200)(16^2) - \dfrac{1}{2}(1200)(10^2) = \dfrac{1}{2}(1200)(256 - 100) = 93600$ J. Step 2: Find work done against resistance: $W_R = 700 \times 130 = 91000$ J. Step 3: Total work done by engine $= \Delta KE + W_R = 93600 + 91000 = 184600$ J. Step 4: Average power $= \dfrac{184600}{10} = 18460$ W.
   Method:

   Find $\Delta KE$ and work against resistance, sum them, then divide by time.
   Examiner tips

   • The work-energy theorem requires accounting for ALL work done
   • Average power = total work done / time taken

5. Question 3b(ii)

   2 marksSteady Speed and Power
   Step 1: At maximum steady speed, acceleration is zero, so the driving force equals the resistance: $F = R = 700$ N. Step 2: Use $P = Fv$: $18460 = 700v$. Step 3: Solve: $v = \dfrac{18460}{700} = 26.4$ m/s.
   Method:

   Set $F = R$ (zero acceleration at steady speed), then use $P = Fv$ to find $v$.
   Examiner tips

   • At maximum steady speed, there is no acceleration so the net force is zero
   • This means the driving force exactly balances the resistance

6. Question 4a

   4 marksVertical Motion / Collision of Two Particles
   Step 1: Set up displacement equations with upward positive and the origin at the ground. For P: $s_P = 24t - 5t^2$. For Q: $s_Q = 12 + 8t - 5t^2$. Step 2: At the collision, $s_P = s_Q$: $24t - 5t^2 = 12 + 8t - 5t^2$. Step 3: The $-5t^2$ terms cancel: $24t = 12 + 8t$, so $16t = 12$, giving $t = 0.75$ s. Step 4: Substitute into P's equation: height $= 24(0.75) - 5(0.75)^2 = 18 - 2.8125 = 15.2$ m (to 3 s.f.).
   Method:

   Set up displacement equations from the ground, equate them to find $t$, then substitute back.
   Examiner tips

   • Use the same origin and positive direction for both particles
   • The $-5t^2$ terms cancel when equating, making the algebra simple

7. Question 4b

   5 marksMomentum Conservation and Vertical Motion
   Step 1: Apply conservation of momentum at the collision: $0.2(16.5) + 0.3(0.5) = (0.2 + 0.3)v$. So $3.3 + 0.15 = 0.5v$, giving $v = 6.9$ m/s upward. Step 2: The combined particle ($0.5$ kg) moves upward at $6.9$ m/s from height $15.2$ m, then decelerates under gravity, rises, and falls to the ground. Step 3: Use $v^2 = u^2 + 2gs$ taking downward as positive with the particle at height $15.2$ m. Initial speed upward $= 6.9$ m/s, displacement downward to ground $= 15.2$ m: $v^2 = 6.9^2 + 2(10)(15.2) = 47.61 + 304 = 351.61$. Step 4: $v = \sqrt{351.61} = 18.8$ m/s.
   Method:

   Use momentum conservation to find the speed after coalescence, then apply $v^2 = u^2 + 2gs$ for the descent to ground level.
   Examiner tips

   • After coalescence the combined particle still has upward velocity so it rises further before falling
   • Using $v^2 = u^2 + 2gs$ with the full height to the ground works regardless of whether the particle goes up first

8. Question 5

   6 marksNewton's Second Law on a Rough Inclined Plane
   Step 1: Resolve perpendicular to the plane to find the normal reaction: $R = mg\cos 15 - 12\sin 30 = 4(10)\cos 15 - 12(0.5) = 38.64 - 6 = 32.64$ N. Step 2: Find friction: $F = \mu R = 0.4 \times 32.64 = 13.06$ N (opposing motion, so up the plane). Step 3: Apply Newton's second law down the plane: $12\cos 30 + 4g\sin 15 - 13.06 = 4a$. $10.39 + 10.35 - 13.06 = 4a$. $7.68 = 4a$. $a = 1.92$ m/s$^2$. Step 4: Use $v^2 = u^2 + 2as = 9 + 2(1.92)(4) = 9 + 15.36 = 24.36$. $v = 4.94$ m/s.
   Method:

   Find $R$, then friction, then apply N2L along the plane for $a$, then use suvat.
   Examiner tips

   • The applied force has components both along and perpendicular to the plane
   • The perpendicular component reduces the normal reaction, and hence the friction

9. Question 6

   8 marksVariable Acceleration (Integration)
   Step 1: Find when $a = 0$: $2(t+1)^{-1/2} - 1 = 0$, so $(t+1)^{1/2} = 2$, giving $t + 1 = 4$, hence $t = 3$. Step 2: Integrate acceleration to find velocity: $v = \int \left[2(t+1)^{-1/2} - 1\right] dt = 4(t+1)^{1/2} - t + c$. Step 3: Use $v = 0$ at $t = 0$: $0 = 4(1) - 0 + c$, so $c = -4$. Thus $v = 4(t+1)^{1/2} - t - 4$. Step 4: Integrate velocity to find displacement: $s = \int \left[4(t+1)^{1/2} - t - 4\right] dt = \dfrac{8}{3}(t+1)^{3/2} - \dfrac{t^2}{2} - 4t + k$. Step 5: Use $s = 0$ at $t = 0$: $0 = \dfrac{8}{3}(1) - 0 - 0 + k$, so $k = -\dfrac{8}{3}$. Step 6: At $t = 3$: $s = \dfrac{8}{3}(4)^{3/2} - \dfrac{9}{2} - 12 - \dfrac{8}{3} = \dfrac{8}{3}(8) - \dfrac{9}{2} - 12 - \dfrac{8}{3} = \dfrac{64}{3} - \dfrac{9}{2} - 12 - \dfrac{8}{3}$. Step 7: Simplify: $\dfrac{64 - 8}{3} - \dfrac{9}{2} - 12 = \dfrac{56}{3} - \dfrac{9}{2} - 12 = \dfrac{112 - 27 - 72}{6} = \dfrac{13}{6}$ m.
   Method:

   Find $t$ when $a = 0$, integrate $a$ to get $v$ (using $v(0) = 0$), integrate $v$ to get $s$ (using $s(0) = 0$), evaluate at the required $t$.
   Examiner tips

   • Remember to find the constant of integration at each stage using initial conditions
   • When integrating $(t+1)^n$, the result is $\dfrac{(t+1)^{n+1}}{n+1}$ since the derivative of $(t+1)$ is $1$

10. Question 7a

    6 marksResultant of Coplanar Forces
    Step 1: Resolve horizontally (right positive). The $15$ N force has horizontal component $15\cos 20 = 14.10$ N to the left. The $40$ N force is at $30°$ from vertical, so its horizontal component is $40\sin 30 = 20$ N to the right. The $20$ N force has horizontal component $20\cos 45 = 14.14$ N to the right. The $35$ N force is vertical (no horizontal component). Step 2: $X = -14.10 + 20 + 14.14 = 20.04$ N (to the right). Step 3: Resolve vertically (upward positive). The $15$ N force: $15\sin 20 = 5.13$ N up. The $40$ N force: $40\cos 30 = 34.64$ N up. The $20$ N force: $20\sin 45 = 14.14$ N down. The $35$ N force: $35$ N down. Step 4: $Y = 5.13 + 34.64 - 14.14 - 35 = -9.37$ N (downward). Step 5: Resultant: $S = \sqrt{20.04^2 + 9.37^2} = \sqrt{401.6 + 87.8} = \sqrt{489.4} = 22.1$ N.
    Method:

    Resolve each force into $X$ and $Y$ components, sum each direction, then find $S = \sqrt{X^2 + Y^2}$.
    Examiner tips

    • Draw a clear diagram showing all forces with their angles
    • Be careful with directions: assign positive and negative consistently
    • An angle measured from the vertical means the horizontal component uses sine of that angle

11. Question 7b

    5 marksFriction on a Horizontal Wire
    Step 1: Find the acceleration using $s = \dfrac{1}{2}at^2$ (starting from rest): $4 = \dfrac{1}{2}a(4)^2 = 8a$, so $a = 0.5$ m/s$^2$. Step 2: Find the normal reaction on the wire. The wire is horizontal, so the vertical forces must balance. The normal reaction acts upward and supports both the weight and the downward force component: $R = mg + 8 = 0.5(10) + 8 = 13$ N. Step 3: Apply Newton's second law horizontally: $10 - \mu R = ma$. $10 - 13\mu = 0.5(0.5) = 0.25$. Step 4: Solve: $13\mu = 9.75$, so $\mu = \dfrac{9.75}{13} = 0.750$.
    Method:

    Find $a$ from kinematics, $R$ from vertical equilibrium (including applied force components), then use $F_{\text{horiz}} - \mu R = ma$ to find $\mu$.
    Examiner tips

    • The normal reaction on the wire includes contributions from the weight AND any vertical force components from the applied forces
    • The ring is on a horizontal wire, so friction opposes horizontal motion and the normal reaction is vertical