May/June 2025 Paper 42 Worked Answers (A-Level Maths 9709 AS)

13 questions · 50 marks · 75 minutes

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1. Question 1a

   1 marksSpeed from Distance and Time
   Step 1: The crate moves at constant speed, so speed $= \dfrac{\text{distance}}{\text{time}}$. Step 2: Speed $= \dfrac{15}{6} = 2.5$ m/s.
   Method:

   Divide the distance travelled by the time taken.
   Examiner tips

   • Constant speed means distance divided by time gives the speed directly

2. Question 1b

   2 marksWork Done by an Angled Force
   Step 1: The work done by a force at an angle to the direction of motion is $W = Fd\cos\theta$. Step 2: $W = 30 \times 15 \times \cos 25^\circ = 450 \times 0.9063 = 408$ J (to 3 s.f.).
   Method:

   Use $W = Fd\cos\theta$ with $F = 30$, $d = 15$, $\theta = 25^\circ$.
   Examiner tips

   • Always resolve the force in the direction of displacement when calculating work done

3. Question 1c

   1 marksPower from Force and Velocity
   Step 1: The constant speed is $v = \dfrac{15}{6} = 2.5$ m/s. Step 2: Power $= Fv\cos\theta = 30 \times 2.5 \times \cos 25^\circ = 75 \times 0.9063 = 68.0$ W (to 3 s.f.).
   Method:

   Calculate $v = 15/6 = 2.5$ m/s, then $P = 30 \times 2.5 \times \cos 25^\circ$.
   Examiner tips

   • Power equals the rate of doing work, or equivalently the component of force in the direction of motion times the speed

4. Question 2

   4 marksMomentum Conservation and Kinetic Energy Loss
   Step 1: Both particles move in the same direction (P towards Q, Q away from P). Apply conservation of momentum: $0.3 \times 6 + 0.15 \times 3 = (0.3 + 0.15)v$. Step 2: $1.8 + 0.45 = 0.45v$, so $v = 5$ m/s. Step 3: KE before $= \dfrac{1}{2}(0.3)(36) + \dfrac{1}{2}(0.15)(9) = 5.4 + 0.675 = 6.075$ J. Step 4: KE after $= \dfrac{1}{2}(0.45)(25) = 5.625$ J. Step 5: Loss in KE $= 6.075 - 5.625 = 0.45$ J.
   Method:

   Use conservation of momentum to find the combined velocity, then calculate the difference in total KE.
   Examiner tips

   • When particles coalesce, kinetic energy is always lost
   • Both particles move in the same direction here, so momenta are added with the same sign

5. Question 3

   4 marksEquilibrium of a Suspended Particle
   Step 1: The string has length $1.3$ m and the vertical distance to the ceiling is $1.2$ m. The horizontal displacement is $\sqrt{1.3^2 - 1.2^2} = \sqrt{0.25} = 0.5$ m. So $\cos\theta = \dfrac{1.2}{1.3} = \dfrac{12}{13}$ and $\sin\theta = \dfrac{0.5}{1.3} = \dfrac{5}{13}$. Step 2: Resolve horizontally: $T\sin\theta = 40$, so $T \times \dfrac{5}{13} = 40$, giving $T = 104$ N. Step 3: Resolve vertically: $T\cos\theta = mg$, so $104 \times \dfrac{12}{13} = 10m$, giving $96 = 10m$, so $m = 9.6$ kg.
   Method:

   Find the angle from string geometry, then resolve horizontally and vertically to find $T$ and $m$.
   Examiner tips

   • Always find the triangle dimensions first to get exact trig ratios
   • Resolve in two perpendicular directions to find two unknowns

6. Question 4b

   6 marksVelocity-Time Graph Area and Kinematics
   Step 1: Speed after the acceleration phase: $v = 18 + 0.3 \times 40 = 30$ m/s. Step 2: Distance during acceleration (trapezium): $s_1 = \dfrac{1}{2}(18 + 30)(40) = 960$ m. Step 3: Distance during constant speed: $s_2 = 30 \times 3T = 90T$ m. Step 4: Speed at $B$: $v_B = 30 - 0.15T$ m/s. Distance during deceleration (trapezium): $s_3 = \dfrac{1}{2}(30 + 30 - 0.15T)(T) = 30T - 0.075T^2$. Step 5: Total distance: $960 + 90T + 30T - 0.075T^2 = 3330$, so $0.075T^2 - 120T + 2370 = 0$. Step 6: Multiply by $\dfrac{40}{3}$: $T^2 - 1600T + 31600 = 0$. Factorising: $(T - 20)(T - 1580) = 0$. Since $v_B > 0$ requires $T < 200$, $T = 20$. Step 7: Check: $s_1 = 960$, $s_2 = 90(20) = 1800$, $s_3 = 30(20) - 0.075(400) = 600 - 30 = 570$. Total $= 960 + 1800 + 570 = 3330$ m.
   Method:

   Find speed after acceleration, express each phase distance in terms of $T$, sum to total distance, and solve the resulting equation.
   Examiner tips

   • Use the areas under the velocity-time graph for each phase
   • Check that the speed at B is positive with the value of T found

7. Question 4c

   3 marksDeceleration to Rest
   Step 1: From $B$ to $C$, use $v^2 = u^2 + 2as$ with $v = 0$, $u = 20$, $a = -0.4$. Step 2: $0 = 400 - 0.8s$, so $s = 500$ m. Step 3: Total distance from $A$ to $C$ $= 2400 + 500 = 2900$ m.
   Method:

   Use $v^2 = u^2 + 2as$ with $v = 0$ to find $BC$, then add to $AB$.
   Examiner tips

   • When a particle decelerates to rest, $v = 0$ in the SUVAT equations

8. Question 5a

   3 marksConnected Particles with Pulleys
   Step 1: For particle $C$ (mass $6$ kg, accelerating downward at $1.5$ m/s$^2$): $6g - T_{BC} = 6 \times 1.5$, so $60 - T_{BC} = 9$, giving $T_{BC} = 51$ N. Step 2: For particle $B$ (mass $5$ kg, on smooth horizontal surface, accelerating at $1.5$ m/s$^2$): $T_{BC} - T_{AB} = 5 \times 1.5$, so $51 - T_{AB} = 7.5$, giving $T_{AB} = 43.5$ N.
   Method:

   Apply Newton's second law to $C$ to find $T_{BC}$, then to $B$ to find $T_{AB}$.
   Examiner tips

   • Start with the particle that has the fewest unknowns
   • The tensions on either side of a pulley are equal for light strings over smooth pulleys

9. Question 5b

   4 marksCoefficient of Friction on an Inclined Plane
   Step 1: For particle $A$ (mass $4$ kg) moving up the $30^\circ$ incline, apply Newton's second law along the plane. The forces along the plane are: tension $T_{AB} = 43.5$ N (up the plane), weight component $4g\sin 30^\circ = 20$ N (down), and friction $F$ (down, opposing upward motion). Step 2: $43.5 - F - 20 = 4 \times 1.5 = 6$, so $F = 43.5 - 20 - 6 = 17.5$ N. Step 3: Resolve perpendicular to the plane: $R = 4g\cos 30^\circ = 40 \times \dfrac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.64$ N. Step 4: $\mu = \dfrac{F}{R} = \dfrac{17.5}{20\sqrt{3}} = \dfrac{7}{8\sqrt{3}} = \dfrac{7\sqrt{3}}{24} \approx 0.505$.
   Method:

   Use Newton's second law along the plane to find $F$, resolve perpendicular to find $R$, then $\mu = F/R$.
   Examiner tips

   • Friction acts down the plane when A moves up the plane
   • The normal reaction depends on the component of weight perpendicular to the plane

10. Question 5c

    5 marksTwo-Phase Motion After String Breaks
    Step 1: Phase 1 (first $2$ s): Starting from rest with $a = 1.5$ m/s$^2$. Distance $s_1 = \dfrac{1}{2}(1.5)(2^2) = 3$ m. Speed at end: $v = 1.5 \times 2 = 3$ m/s. Step 2: Phase 2 (after force removed, still moving up): Friction $= \mu R = \dfrac{7\sqrt{3}}{24} \times 4g\cos 30^\circ = \dfrac{7\sqrt{3}}{24} \times 20\sqrt{3} = \dfrac{420}{24} = 17.5$ N (down the plane). Weight component $= 4g\sin 30^\circ = 20$ N (down the plane). Step 3: Deceleration $= \dfrac{17.5 + 20}{4} = \dfrac{37.5}{4} = 9.375$ m/s$^2$. Step 4: Using $v^2 = u^2 + 2as$ with $v = 0$, $u = 3$, $a = -9.375$: $0 = 9 - 18.75s_2$, so $s_2 = 0.48$ m. Step 5: Total distance $= 3 + 0.48 = 3.48$ m.
    Method:

    Find distance and speed after phase 1, calculate the new deceleration (friction + weight component), then find the extra distance to rest.
    Examiner tips

    • After the string breaks, friction still opposes the motion (acts down the plane) along with the weight component
    • Check if the particle continues moving in the same direction after the force is removed

11. Question 6a

    5 marksMaximum Velocity with Variable Acceleration
    Step 1: Differentiate: $\dfrac{dv}{dt} = \dfrac{3}{2} \times 4 \times (4t+1)^{1/2} - 16t = 6(4t+1)^{1/2} - 16t$. Step 2: Set $\dfrac{dv}{dt} = 0$: $6(4t+1)^{1/2} = 16t$, so $3(4t+1)^{1/2} = 8t$. Step 3: Square both sides: $9(4t+1) = 64t^2$, giving $64t^2 - 36t - 9 = 0$. Step 4: Using the quadratic formula: $t = \dfrac{36 \pm \sqrt{1296 + 2304}}{128} = \dfrac{36 \pm \sqrt{3600}}{128} = \dfrac{36 \pm 60}{128}$. Step 5: $t = \dfrac{96}{128} = 0.75$ (taking the positive root; the negative root is rejected). Step 6: $v(0.75) = (4 \times 0.75 + 1)^{3/2} - 8(0.75)^2 = 4^{3/2} - 8(0.5625) = 8 - 4.5 = 3.5$ m/s.
    Method:

    Differentiate $v$, set $dv/dt = 0$, square to get a quadratic, solve for $t$, substitute back into $v$.
    Examiner tips

    • Use the chain rule carefully when differentiating $(at+b)^n$
    • After squaring, check the solution satisfies the original unsquared equation

12. Question 6b

    4 marksDisplacement by Integration (Variable Velocity)
    Step 1: Integrate $v = (4t+1)^{3/2} - 8t^2$ to find position $s$: $s = \dfrac{(4t+1)^{5/2}}{4 \times \frac{5}{2}} - \dfrac{8t^3}{3} + C = \dfrac{(4t+1)^{5/2}}{10} - \dfrac{8t^3}{3} + C$. Step 2: At $t = 0$, $P$ is at $A$ so $s = 0$: $0 = \dfrac{1}{10} + C$, giving $C = -\dfrac{1}{10}$. Step 3: At $t = 0.75$: $4(0.75) + 1 = 4$, so $(4)^{5/2} = 32$. And $(0.75)^3 = 0.421875$. $s = \dfrac{32}{10} - \dfrac{8 \times 0.421875}{3} - \dfrac{1}{10} = 3.2 - 1.125 - 0.1 = 1.975$ m. Step 4: The displacement is $1.975 \approx 1.98$ m (to 3 s.f.).
    Method:

    Integrate $v$ to find $s(t)$, apply $s(0) = 0$ to find $C$, then evaluate at $t = 0.75$.
    Examiner tips

    • When integrating $(at+b)^n$, divide by $a \times (n+1)$
    • Always find the constant of integration using initial conditions

13. Question 7

    6 marksEnergy Method on a Rough Incline (Round Trip)
    Step 1: Find the friction force: $R = 4g\cos 30^\circ = 40 \times \dfrac{\sqrt{3}}{2} = 20\sqrt{3}$ N. Friction $= \mu R = \dfrac{\sqrt{3}}{6} \times 20\sqrt{3} = \dfrac{60}{6} = 10$ N. Step 2: Find the distance $d$ travelled up the plane. Using energy going up: $\dfrac{1}{2}(4)(100) = 4g\sin 30^\circ \times d + 10d$, so $200 = 20d + 10d = 30d$, giving $d = \dfrac{20}{3}$ m. Step 3: Total work done against friction for the round trip (up and back down): $W_f = 10 \times 2 \times \dfrac{20}{3} = \dfrac{400}{3}$ J. Step 4: On the round trip, the particle returns to the same height, so the net change in GPE is zero. By the work-energy theorem: $\dfrac{1}{2}(4)(10^2) - \dfrac{1}{2}(4)v^2 = \dfrac{400}{3}$. Step 5: $200 - 2v^2 = \dfrac{400}{3}$, so $2v^2 = 200 - \dfrac{400}{3} = \dfrac{200}{3}$, giving $v^2 = \dfrac{100}{3}$. Step 6: $v = \dfrac{10}{\sqrt{3}} = \dfrac{10\sqrt{3}}{3} \approx 5.77$ m/s.
    Method:

    Find $d$ from energy conservation going up, then apply energy conservation for the round trip with total friction work $= 2Fd$.
    Examiner tips

    • On a round trip the GPE changes cancel, leaving friction as the only source of energy loss
    • Friction acts over the total distance $2d$, not just $d$