A Level Mathematics 9709/43 October/November 2025 Past Paper with Answers

Question 1

A runner moves along a straight track. The velocity-time graph consists of straight-line segments: from $(0, 0)$ to $(4, 6)$, from $(4, 6)$ to $(10, 0)$, from $(10, 0)$ to $(14, -6)$, from $(14, -6)$ to $(20, -6)$, from $(20, -6)$ to $(24, 0)$, from $(24, 0)$ to $(28, 6)$. What is the deceleration at $t = 8$?

1. $1$ m/s$^2$
2. $0.75$ m/s$^2$
3. $1.5$ m/s$^2$
4. $6$ m/s$^2$

Question 2

A runner moves along a straight track. The velocity-time graph consists of straight-line segments: from $(0, 0)$ to $(4, 6)$, from $(4, 6)$ to $(10, 0)$, from $(10, 0)$ to $(14, -6)$, from $(14, -6)$ to $(20, -6)$, from $(20, -6)$ to $(24, 0)$, from $(24, 0)$ to $(28, 6)$. What is the total distance travelled from $t = 0$ to $t = 28$?

1. $102$ m
2. $18$ m
3. $66$ m
4. $114$ m

Question 3

A car has maximum power $60$ kW and mass $500$ kg. The resistance to motion is $cv$ N, where $v$ m/s is the speed and $c$ is a constant. When the car travels at $30$ m/s on a horizontal road with maximum power, the acceleration is $0.4$ m/s$^2$. What is the value of $c$?

1. $60$
2. $2000$
3. $67$
4. $53$

Question 4 · Answer A

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.

Question 5 · Answer A

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.

Question 6 · Answer A

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.

Question 7 · Answer A

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.

Question 8 · Answer A

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$^2$.

Question 9 · Answer A

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.

Question 10 · Answer A

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.

Question 11 · Answer A

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.

Question 12 · Answer A

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 $5$.

Question 13 · Answer A

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 $m = 7$.

Question 14 · Answer A

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 $16\cos\theta$: $\mu = \dfrac{5\tan\theta - 2}{5 + 2\tan\theta}$.

Question 15 · Answer A

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 $21.8^\circ < \theta < 90^\circ$.