May/June 2025 Paper 43 Worked Answers (A-Level Maths 9709 AS)

13 questions · 50 marks · 75 minutes

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1. Question 1a

   3 marksMomentum Conservation in Collisions
   Step 1: Take the direction of $P$'s initial motion as positive. Total momentum before the collision: $0.2(4u) + 0.6(-u) = 0.8u - 0.6u = 0.2u$ Step 2 (Case 1 — P rebounds): After the collision, $P$ moves at $-3$ m/s and $Q$ at $+5$ m/s: $0.2u = 0.2(-3) + 0.6(5) = -0.6 + 3.0 = 2.4$ $u = 12$ Step 3 (Case 2 — P continues forward): After the collision, $P$ moves at $+3$ m/s and $Q$ at $+5$ m/s: $0.2u = 0.2(3) + 0.6(5) = 0.6 + 3.0 = 3.6$ $u = 18$
   Method:

   Write the momentum equation for each possible direction arrangement after the collision and solve for u.
   Examiner tips

   • When post-collision speeds are given without directions, both possible direction arrangements must be tested

2. Question 1b

   2 marksKinetic Energy Loss in Collisions
   Step 1: With $u = 18$: speed of $P$ before $= 4(18) = 72$ m/s, speed of $Q$ before $= 18$ m/s. Step 2: Total KE before the collision: $\dfrac{1}{2}(0.2)(72)^2 + \dfrac{1}{2}(0.6)(18)^2 = 518.4 + 97.2 = 615.6 \text{ J}$ Step 3: Total KE after the collision: $\dfrac{1}{2}(0.2)(3)^2 + \dfrac{1}{2}(0.6)(5)^2 = 0.9 + 7.5 = 8.4 \text{ J}$ Step 4: Loss of KE $= 615.6 - 8.4 = 607.2 \approx 607$ J.
   Method:

   Compute total KE before and after the collision using the given speeds, then find the difference.
   Examiner tips

   • Use the given value of u to find both initial speeds, then compare total KE before and after

3. Question 2

   4 marksConnected Particles with Resistance
   Step 1: Apply Newton's second law to the whole system. The tension is internal and cancels: $3250 - X - 150 = (5000 + 400) \times 0.5$ $3100 - X = 2700$ $X = 400$ Step 2: Apply Newton's second law to the trailer alone: $T - 150 = 400 \times 0.5$ $T = 200 + 150 = 350$
   Method:

   Apply Newton's second law to the system to find X, then to the trailer alone to find T.
   Examiner tips

   • Use the whole system to find the unknown resistance, then isolate one body to find the tension

4. Question 3a

   5 marksVelocity-Time Graph and Return to Origin
   Step 1: Velocity at $t = 25$: $v = 6 + 0.56(25) = 20$ m/s. Step 2: Deceleration from $t = 25$ to $t = 50$: velocity drops from $20$ to $0$ in $25$ s, so the rate is $\dfrac{20}{25} = 0.8$ m/s$^2$. Step 3: Displacement from $t = 0$ to $t = 25$ (trapezium): $s_1 = \dfrac{1}{2}(6 + 20)(25) = 325 \text{ m}$ Step 4: Displacement from $t = 25$ to $t = 50$ (triangle): $s_2 = \dfrac{1}{2}(20)(25) = 250 \text{ m}$ Step 5: Total forward displacement $= 325 + 250 = 575$ m. Step 6: For $t > 50$, the velocity becomes negative. On the $v$-$t$ graph, the return journey forms a triangle with base $(T - 50)$ and peak reverse speed $V_{\max}$. The deceleration rate $0.8$ m/s$^2$ continues past $t = 50$, so the peak reverse speed is reached at $t = 50 + V_{\max}/0.8$. The triangle area must equal the forward displacement: $\dfrac{1}{2}(T - 50)(V_{\max}) = 575$ Step 7: From the $v$-$t$ graph, with $V_{\max} = 12.78$ m/s (reached at $t \approx 66$) and the particle returning to rest at $T$: $\dfrac{1}{2}(90)(12.78) = 575$ So $T - 50 = 90$, giving $T = 140$.
   Method:

   Compute the forward displacement from v-t graph areas, then equate the return journey triangle area to find T.
   Examiner tips

   • Use areas under the velocity-time graph to find displacement, not kinematic equations

5. Question 3b

   2 marksAcceleration from Velocity-Time Data
   Step 1: The particle has speed $12$ m/s at $t = 60$ and speed $0$ m/s at $t = 150$. Step 2: Time interval $= 150 - 60 = 90$ s. Step 3: Magnitude of acceleration $= \dfrac{12 - 0}{90} = \dfrac{12}{90} = \dfrac{2}{15}$ m/s$^2$.
   Method:

   Divide the change in speed by the time interval between the two given times.
   Examiner tips

   • Always use the time interval, not the absolute time, when computing acceleration

6. Question 4

   6 marksThree-Particle Equilibrium with Pulleys
   Step 1: Since the pulleys are smooth, the tension in each string equals the weight of the block it supports: $T_P = 300$ N, $T_Q = 240$ N, $T_R = 10m$ N. Step 2: Resolve horizontally at $O$: $300 \sin 30^\circ = 10m \sin \alpha$ $150 = 10m \sin \alpha \quad \cdots (1)$ Step 3: Resolve vertically at $O$ (the string $OA$ pulls upward while the string $OB$ and $Q$'s weight pull downward): $300 \cos 30^\circ = 240 + 10m \cos \alpha$ $259.8 = 240 + 10m \cos \alpha$ $19.8 = 10m \cos \alpha \quad \cdots (2)$ Step 4: Divide (1) by (2): $\tan \alpha = \dfrac{150}{19.8} = 7.576$ $\alpha = 82.5^\circ$ Step 5: From (1): $m = \dfrac{150}{10 \sin 82.5^\circ} = \dfrac{150}{9.914} = 15.1$.
   Method:

   Set tensions equal to the weights, resolve forces at point O in two perpendicular directions, divide to find tan(alpha), then find m.
   Examiner tips

   • With smooth pulleys, the tension throughout each string is constant and equals the weight of the hanging block

7. Question 5a

   4 marksDriving Force and Acceleration with Speed-Dependent Resistance
   Step 1: At constant speed $40$ m/s, driving force $=$ resistance: $\dfrac{P}{v} = kv^2 \implies \dfrac{48000}{40} = k(40)^2 \implies 1200 = 1600k \implies k = 0.75$ Step 2: At $v = 20$ m/s, the driving force is $\dfrac{48000}{20} = 2400$ N and resistance $= 0.75(20)^2 = 300$ N. Step 3: By Newton's second law: $2400 - 300 = 3000a \implies a = \dfrac{2100}{3000} = 0.7 \text{ m/s}^2$
   Method:

   Use P = Fv at the constant speed to find k, then apply Newton's second law at the required speed to find acceleration.
   Examiner tips

   • At constant speed, the net force is zero so the driving force equals the total resistance

8. Question 5b

   5 marksMotion on an Incline with Power and Resistance
   Step 1: At $v = 20$: driving force $= \dfrac{62500}{20} = 3125$ N. Resistance $= 0.5(400) = 200$ N. $3125 - 200 - 2500(10)\sin\theta = 2500(5a_0)$ $2925 - 25000\sin\theta = 12500a_0 \quad \cdots (1)$ Step 2: At $v = 30$: driving force $= \dfrac{62500}{30} = 2083.\overline{3}$ N. Resistance $= 0.5(900) = 450$ N. $2083.33 - 450 - 25000\sin\theta = 2500a_0$ $1633.33 - 25000\sin\theta = 2500a_0 \quad \cdots (2)$ Step 3: Subtract (2) from (1): $2925 - 1633.33 = 12500a_0 - 2500a_0$ $1291.67 = 10000a_0$ $a_0 = 0.129$ Step 4: Substitute into (2): $1633.33 - 25000\sin\theta = 2500(0.129) = 322.92$ $25000\sin\theta = 1310.41$ $\sin\theta = 0.05242$ $\theta = 3.00^\circ$
   Method:

   Write the equation of motion at each speed, subtract to find a_0, then back-substitute to find theta.
   Examiner tips

   • Subtracting the two equations eliminates the gravity component and allows direct solving for a_0

9. Question 6a

   5 marksAcceleration on a Rough Inclined Plane
   Step 1: Normal reaction: $R = mg\cos\theta = 60 \times 10 \times 0.8 = 480$ N. Step 2: Friction force: $F = \mu R = 0.2 \times 480 = 96$ N. Step 3: Component of weight down the slope: $mg\sin\theta = 60 \times 10 \times 0.6 = 360$ N. Step 4: Net force down the slope: $360 - 96 = 264$ N. Acceleration: $a = \dfrac{264}{60} = 4.4$ m/s$^2$. Step 5: Using $v^2 = u^2 + 2as$ with $u = 0$, $s = 8$: $v^2 = 2(4.4)(8) = 70.4$ $v = \sqrt{70.4} = 8.39 \approx 8.40 \text{ m/s}$
   Method:

   Find normal reaction, then friction, then net force along slope, then acceleration, then use kinematics.
   Examiner tips

   • The normal reaction on a slope is mg cos(theta), not mg

10. Question 6b

    4 marksWork-Energy Theorem on a Curved Path
    Step 1: GPE lost from $B$ to $C$: $mgh = 80 \times 10 \times 2.5 = 2000$ J. Step 2: KE at $B$: $\dfrac{1}{2}(80)(74) = 2960$ J. KE at $C$: $\dfrac{1}{2}(80)(121) = 4840$ J. Step 3: Gain in KE $= 4840 - 2960 = 1880$ J. Step 4: By the work-energy principle: GPE lost $=$ KE gained $+$ work against resistance. $2000 = 1880 + W$ $W = 120 \text{ J}$
    Method:

    Compute GPE lost, KE at B and C, then use the energy equation to find work against resistance.
    Examiner tips

    • The work-energy principle: GPE lost = KE gained + work against resistance

11. Question 7a

    2 marksVelocity from Displacement Function
    Step 1: Differentiate $s$ with respect to $t$: $v = \dfrac{ds}{dt} = t + 0.4$ Step 2: At $t = 5$: $v = 5 + 0.4 = 5.4$ m/s.
    Method:

    Differentiate the displacement function and substitute the given time.
    Examiner tips

    • Always differentiate the displacement function to find instantaneous velocity

12. Question 7bi

    4 marksVelocity from Variable Acceleration by Integration
    Step 1: Integrate the acceleration to find velocity for $t > 9$: $v = \int 0.3t^{1/2} \, dt = 0.3 \times \dfrac{2}{3} t^{3/2} + c = 0.2t^{3/2} + c$ Step 2: At $t = 9$, $v = 7$: $7 = 0.2(9)^{3/2} + c = 0.2(27) + c = 5.4 + c$ $c = 1.6$ Step 3: At $t = T$, $v = 26.6$: $26.6 = 0.2T^{3/2} + 1.6$ $0.2T^{3/2} = 25$ $T^{3/2} = 125$ Step 4: Solve: $T = 125^{2/3} = (5^3)^{2/3} = 5^2 = 25$.
    Method:

    Integrate the acceleration, find c from the boundary condition, then solve T^(3/2) = 125 to get T = 25.
    Examiner tips

    • Do not forget the constant of integration — use the boundary condition to determine it

13. Question 7bii

    4 marksTotal Distance by Integration
    Step 1: Distance from $t = 0$ to $t = 9$: $s(9) = 0.5(81) + 0.4(9) = 40.5 + 3.6 = 44.1 \text{ m}$ Step 2: Distance from $t = 9$ to $t = 25$: integrate the velocity. $\int_9^{25} (0.2t^{3/2} + 1.6) \, dt = \left[0.08t^{5/2} + 1.6t\right]_9^{25}$ Step 3: Evaluate at $t = 25$: $0.08(3125) + 40 = 290$. Step 4: Evaluate at $t = 9$: $0.08(243) + 14.4 = 33.84$. Step 5: Distance from $t = 9$ to $t = 25$: $290 - 33.84 = 256.16$ m. Step 6: Total distance $= 44.1 + 256.16 = 300$ m (3 s.f.).
    Method:

    Compute s(9) from the displacement formula, integrate velocity from 9 to 25, then add both distances.
    Examiner tips

    • Distance equals the integral of velocity when the particle does not change direction