October/November 2025 Paper 43 Worked Answers (A-Level Maths 9709 AS)

15 questions · 50 marks · 75 minutes

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1. Question 1a

   1 marksVelocity-Time Graph (Deceleration)
   Step 1: At $t = 8$, the runner is on the segment from $(4, 6)$ to $(10, 0)$. Step 2: The gradient of this segment is $\dfrac{0 - 6}{10 - 4} = \dfrac{-6}{6} = -1$ m/s$^2$. Step 3: The deceleration (magnitude of negative acceleration) is $1$ m/s$^2$.
   Method:

   Find the gradient of the line segment from $(4, 6)$ to $(10, 0)$.
   Examiner tips

   • Deceleration is the magnitude of the gradient when the gradient is negative
   • Identify which line segment the given time falls on

2. Question 1b

   2 marksTotal Distance from Velocity-Time Graph
   Step 1: Break the graph into regions and compute absolute areas. Step 2: Triangle from $(0,0)$ to $(4,6)$ to $(10,0)$: base $= 10$, height $= 6$, area $= \dfrac{1}{2}(10)(6) = 30$ m. Step 3: Trapezium from $(10,0)$ to $(14,-6)$ to $(20,-6)$ to $(24,0)$: parallel sides $= 6$ and $14$, height $= 6$, area $= \dfrac{1}{2}(6 + 14)(6) = 60$ m. Step 4: Triangle from $(24,0)$ to $(28,6)$: base $= 4$, height $= 6$, area $= \dfrac{1}{2}(4)(6) = 12$ m. Step 5: Total distance $= 30 + 60 + 12 = 102$ m.
   Method:

   Calculate the absolute area of each region under the v-t graph and sum them.
   Examiner tips

   • Distance requires taking absolute values of all areas, including those below the time axis
   • Break the graph into triangles and trapeziums for easy calculation

3. Question 2a

   3 marksPower and Resistance
   Step 1: Driving force at $30$ m/s: $F = \dfrac{P}{v} = \dfrac{60000}{30} = 2000$ N. Step 2: Apply Newton's second law: $F - cv = ma$, so $2000 - 30c = 500 \times 0.4 = 200$. Step 3: Solve: $30c = 1800$, so $c = 60$.
   Method:

   Find $F$ from $P = Fv$, then use $F - cv = ma$ to solve for $c$.
   Examiner tips

   • Always convert kW to W before using $P = Fv$
   • The resistance $cv$ acts against the driving force

4. Question 2b

   3 marksSteady Speed on an Incline
   Step 1: At steady speed, acceleration $= 0$. Driving force $=$ resistance $+$ weight component: $\dfrac{P}{v} = cv + mg\sin\theta$. Step 2: Substitute: $\dfrac{60000}{v} = 60v + 800 \times 10 \times \dfrac{1}{10} = 60v + 800$. Step 3: Multiply by $v$: $60000 = 60v^2 + 800v$. Step 4: Rearrange: $60v^2 + 800v - 60000 = 0$, i.e. $3v^2 + 40v - 3000 = 0$. Step 5: $v = \dfrac{-40 + \sqrt{1600 + 36000}}{6} = \dfrac{-40 + \sqrt{37600}}{6} = \dfrac{-40 + 193.9}{6} = 25.7$ m/s.
   Method:

   Set $P/v = cv + mg\sin\theta$, multiply by $v$, and solve the quadratic.
   Examiner tips

   • At maximum steady speed, the net force is zero
   • Remember to include the weight component down the slope

5. Question 3a

   3 marksWork Done Against Friction
   Step 1: Resolve vertically to find the normal reaction: $R + 20\sin 30^\circ = 3g$, so $R = 30 - 10 = 20$ N. Step 2: Friction force: $F = \mu R = 0.4 \times 20 = 8$ N. Step 3: Work done against friction $= 8 \times 80 = 640$ J.
   Method:

   Find $R$ by resolving vertically, then $F = \mu R$, then $WD = Fd$.
   Examiner tips

   • The upward component of the pulling force reduces the normal reaction
   • Work done against friction = friction force x distance

6. Question 3b

   4 marksEnergy Method (Work-Energy Theorem)
   Step 1: Work done by the pulling force: $W_{\text{pull}} = 25 \times 80 = 2000$ J. Step 2: Net work done $= 2000 - 800 = 1200$ J. Step 3: Apply the work-energy theorem: $\dfrac{1}{2}(4)v^2 - \dfrac{1}{2}(4)(25) = 1200$. Step 4: $2v^2 = 1200 + 50 = 1250$, so $v^2 = 625$, giving $v = 25$ m/s.
   Method:

   Apply the work-energy theorem: $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = W_{\text{pull}} - W_{\text{friction}}$.
   Examiner tips

   • Only the horizontal component of the pulling force does work along the surface
   • Include the initial kinetic energy in the work-energy equation

7. Question 4a

   6 marksVariable Acceleration (Differentiation)
   Step 1: Find velocity: $v = \dfrac{ds}{dt} = 0.06t^2 - 0.72t - 1.5$. Step 2: Find acceleration: $a = \dfrac{dv}{dt} = 0.12t - 0.72$. Step 3: Minimum velocity occurs when $a = 0$: $0.12t = 0.72$, so $t = 6$. Step 4: Find velocity at $t = 6$: $v(6) = 0.06(36) - 0.72(6) - 1.5 = 2.16 - 4.32 - 1.5 = -3.66$. Step 5: Speed $= |{-3.66}| = 3.66$ m/s.
   Method:

   Differentiate $s$ to get $v$, differentiate again to get $a$. Set $a = 0$ to find $t$, then evaluate $v$ and $s$ at that $t$.
   Examiner tips

   • Minimum velocity means $dv/dt = 0$, not $v = 0$
   • Speed is the magnitude of velocity

8. Question 4b

   4 marksVariable Acceleration (Direction Change)
   Step 1: Expand: $v = 0.04(t^2 - 16t - 36) = 0.04t^2 - 0.64t - 1.44$. Step 2: Direction changes when $v = 0$: $0.04(t - 18)(t + 2) = 0$, so $t = 18$ (reject $t = -2$). Step 3: Acceleration: $a = \dfrac{dv}{dt} = 0.08t - 0.64$. Step 4: At $t = 18$: $a = 0.08(18) - 0.64 = 1.44 - 0.64 = 0.80$ m/s$^2$.
   Method:

   Set $v = 0$ and solve for positive $t$. Differentiate $v$ to find $a$, then substitute.
   Examiner tips

   • Direction of motion changes when $v = 0$, not when $a = 0$
   • Reject any negative values of $t$

9. Question 5a

   4 marksConnected Particles Over a Pulley
   Step 1: B is heavier, so B moves down and A moves up. Let the acceleration be $a$ m/s$^2$. Step 2: For A (moving up): $T - 3g = 3a$, so $T - 30 = 3a$. Step 3: For B (moving down): $5g - T = 5a$, so $50 - T = 5a$. Step 4: Add the two equations: $20 = 8a$, so $a = 2.5$ m/s$^2$. Step 5: Substitute back: $T = 30 + 3(2.5) = 37.5$ N.
   Method:

   Write N2L equations for each particle and solve simultaneously for $T$ and $a$.
   Examiner tips

   • The heavier particle accelerates downward
   • The tension is the same throughout a light inextensible string

10. Question 5b

    2 marksConnected Particles (Time of Motion)
    Step 1: Taking downward as positive for B: initial velocity $u = -1$ m/s (upward), acceleration $a = 4$ m/s$^2$ (downward), displacement $s = 3$ m (downward). Step 2: Use $s = ut + \dfrac{1}{2}at^2$: $3 = -t + 2t^2$. Step 3: Rearrange: $2t^2 - t - 3 = 0$, i.e. $(2t - 3)(t + 1) = 0$. Step 4: $t = 1.5$ (reject $t = -1$). So B reaches the ground after $1.50$ s.
    Method:

    Use $s = ut + \frac{1}{2}at^2$ with downward positive, $u = -1$, $a = 4$, $s = 3$, and solve the quadratic.
    Examiner tips

    • Be careful with signs: B initially moves upward but the net acceleration is downward
    • B must travel the full height downward to reach the ground

11. Question 5c

    4 marksConnected Particles (Maximum Height)
    Step 1: A is projected downward at $1$ m/s but the system accelerates A upward at $4$ m/s$^2$. A first moves down. Distance down: $0 = 1^2 - 2(4)s_1$, so $s_1 = 0.125$ m. Step 2: B descends $3$ m to reach the ground. A's net displacement is $3$ m upward. Find A's speed when B hits ground: $v^2 = 1^2 + 2(4)(3) = 25$, so $v = 5$ m/s upward. Step 3: After string goes slack, A decelerates under gravity alone: $0 = 25 - 2(10)s_2$, so $s_2 = 1.25$ m. Step 4: Lowest point is $0.125$ m below start. Highest point is $3 + 1.25 = 4.25$ m above start. Distance between lowest and highest $= 0.125 + 4.25 = 4.375 \approx 4.38$ m.
    Method:

    Find the distance A descends, then find A's speed when B hits the ground, then find the extra height A rises under gravity alone. Sum all parts.
    Examiner tips

    • A initially moves downward before the system reverses direction
    • After the string goes slack, A decelerates under gravity alone

12. Question 6a

    3 marksConservation of Momentum (Collision)
    Step 1: Take P's initial direction as positive. Before: P has momentum $mu$, Q has momentum $-6u$. After: P has momentum $-mu/4$, Q has momentum $3v$. Step 2: Conservation of momentum: $mu - 6u = -\dfrac{mu}{4} + 3v$. Step 3: Solve for $v$: $3v = mu + \dfrac{mu}{4} - 6u = \dfrac{5mu}{4} - 6u$. So $v = \dfrac{u(5m - 24)}{12}$. Step 4: For Q to reverse: $v > 0$, so $5m - 24 > 0$, giving $m > 4.8$. Step 5: The minimum integer value of $m$ is $5$.
    Method:

    Apply conservation of momentum, solve for $v_Q$, and require $v_Q > 0$ for reversal.
    Examiner tips

    • Choose a consistent positive direction and apply it to all velocities
    • Reversing direction means the velocity changes sign

13. Question 6b

    4 marksMomentum (No Further Collisions)
    Step 1: After Q rebounds from the wall, Q moves in the negative direction at speed $\dfrac{1}{4} \times \dfrac{(5m-24)u}{12} = \dfrac{(5m-24)u}{48}$. Step 2: P also moves in the negative direction at speed $\dfrac{u}{4}$. Step 3: For no further collision, P must move at least as fast as Q: $\dfrac{u}{4} \geq \dfrac{(5m-24)u}{48}$. Step 4: Simplify: $12 \geq 5m - 24$, so $5m \leq 36$, giving $m \leq 7.2$. Step 5: The largest integer value is $m = 7$.
    Method:

    Find Q's speed after the wall. Set up inequality: speed of P $\geq$ rebounded speed of Q. Solve for $m$ and find the largest integer.
    Examiner tips

    • After the wall, both particles move in the same direction
    • No further collision means P moves at least as fast as Q in that direction

14. Question 7a

    5 marksFriction on an Inclined Plane (Equilibrium)
    Step 1: Resolve perpendicular to the plane: $R = 8g\cos\theta + 32\sin\theta = 80\cos\theta + 32\sin\theta$. Step 2: Resolve along the plane (friction acts up since P is about to slip down): $\mu R + 32\cos\theta = 8g\sin\theta$. Step 3: So $\mu R = 80\sin\theta - 32\cos\theta$. Step 4: $\mu = \dfrac{80\sin\theta - 32\cos\theta}{80\cos\theta + 32\sin\theta}$. Step 5: Divide numerator and denominator by $16\cos\theta$: $\mu = \dfrac{5\tan\theta - 2}{5 + 2\tan\theta}$.
    Method:

    Resolve perpendicular and parallel to the plane. Use $F_{\text{fric}} = \mu R$ at the point of slipping. Divide through to express $\mu$ in terms of $\tan\theta$.
    Examiner tips

    • The horizontal force has components both along and perpendicular to the incline
    • On the point of slipping down: friction acts up the slope and $F = \mu R$

15. Question 7b

    2 marksRange of Values (Friction Inequality)
    Step 1: For $\mu > 0$: both numerator and denominator must have the same sign. Step 2: The denominator $5 + 2\tan\theta > 0$ for $0^\circ < \theta < 90^\circ$ (since $\tan\theta > 0$). Step 3: So the numerator must be positive: $5\tan\theta - 2 > 0$, giving $\tan\theta > \dfrac{2}{5}$. Step 4: $\theta > \tan^{-1}\left(\dfrac{2}{5}\right) = 21.8^\circ$. Step 5: Therefore $21.8^\circ < \theta < 90^\circ$.
    Method:

    Since the denominator is positive for acute $\theta$, set the numerator $> 0$ and solve for $\theta$.
    Examiner tips

    • Since $\theta$ is acute, $\tan\theta > 0$ and the denominator is always positive
    • Only the numerator determines the sign of $\mu$