A Level Mathematics 9709/41 October/November 2025 Past Paper with Answers

Question 1

A car of mass $1200$ kg travels along a horizontal road against a resistance of $400$ N. The car has speed $18$ m/s and acceleration $0.3$ m/s$^2$. What is the driving force of the car?

1. $760$ N
2. $360$ N
3. $400$ N
4. $1160$ N

Question 2

A car travelling at $18$ m/s has a driving force of $760$ N. What is the power output of the engine at this instant?

1. $13680$ W
2. $42.2$ W
3. $778$ W
4. $7600$ W

Question 3

Particle P of mass $4$ kg moves at $6$ m/s towards particle Q of mass $8$ kg, which is stationary. After the collision, P has speed $1$ m/s. How many possible speeds does Q have after the collision, and what is the larger of these speeds?

1. Two possible speeds; the larger is $3.5$ m/s
2. Two possible speeds; the larger is $4$ m/s
3. One possible speed: $2.5$ m/s
4. Two possible speeds; the larger is $2.5$ m/s

Question 4 · Answer A

Step 1: KE before the collision $= \dfrac{1}{2}(4)(6^2) = \dfrac{1}{2}(4)(36) = 72$ J. Step 2: KE after the collision $= \dfrac{1}{2}(4)(1^2) + \dfrac{1}{2}(8)(2.5^2) = 2 + 25 = 27$ J. Step 3: KE lost $= 72 - 27 = 45$ J.

Question 5 · Answer A

Step 1: Resolve each force into horizontal ($X$) and vertical ($Y$) components.

• $50$ N upward: $X = 0$, $Y = 50$.
• $30$ N at $30°$ above positive $x$: $X = 30\cos30° = 25.98$, $Y = 30\sin30° = 15$.
• $65$ N at $45°$ below positive $x$: $X = 65\cos45° = 45.96$, $Y = -65\sin45° = -45.96$.
• $40$ N at $60°$ below negative $x$: $X = -40\cos60° = -20$, $Y = -40\sin60° = -34.64$.

Step 2: Sum components: $X_{\text{total}} = 0 + 25.98 + 45.96 - 20 = 51.94$. $Y_{\text{total}} = 50 + 15 - 45.96 - 34.64 = -15.60$. Step 3: Resultant magnitude: $R = \sqrt{51.94^2 + 15.60^2} = \sqrt{2697.8 + 243.4} = \sqrt{2941.2} = 54.2$ N (3 s.f.).

Question 6 · Answer A

Step 1: Both A and B travel $0.8$ m. A descends $0.8$ m. B moves $0.8$ m up the plane, gaining vertical height $0.8 \times 0.6 = 0.48$ m. Step 2: KE gained by the system $= \dfrac{1}{2}(m + 3)(2^2) = 2(m + 3) = 2m + 6$. Step 3: PE lost by A $= m \times 10 \times 0.8 = 8m$. PE gained by B $= 3 \times 10 \times 0.48 = 14.4$. Work done against resistance $= 12 \times 0.8 = 9.6$. Step 4: Energy equation: KE gained = PE lost by A $-$ PE gained by B $-$ work against resistance. $2m + 6 = 8m - 14.4 - 9.6$. $2m + 6 = 8m - 24$. $6m = 30$. $m = 5$.

Question 7 · Answer A

Step 1: The applied force $6mg$ at angle $\alpha$ to the line of greatest slope has component $6mg\cos\alpha$ along the plane (up the slope) and $6mg\sin\alpha$ perpendicular to the plane (into the surface). Step 2: Perpendicular to plane: $R = mg\cos\alpha + 6mg\sin\alpha$. Step 3: Parallel to plane (equilibrium at limiting friction): $6mg\cos\alpha + mg\sin\alpha = 0.5R$. Step 4: Substitute $R$: $6\cos\alpha + \sin\alpha = 0.5(\cos\alpha + 6\sin\alpha)$. Step 5: Expand: $6\cos\alpha + \sin\alpha = 0.5\cos\alpha + 3\sin\alpha$. Step 6: Rearrange: $5.5\cos\alpha = 2\sin\alpha$, so $\tan\alpha = 2.75$. Step 7: $\alpha = \tan^{-1}(2.75) = 70.0°$.

Question 8 · Answer A

Step 1: Acceleration of A down the plane: $a = g\sin\theta = 10 \times 0.5 = 5$ m/s$^2$. Step 2: Speed of A just before collision: $v = u + at = 0 + 5 \times 1 = 5$ m/s (down the plane). Step 3: Take up the plane as positive. Conservation of momentum: $4(3) + 2(-5) = 4(0) + 2u$. $12 - 10 = 0 + 2u$, so $u = 1$ m/s (up the plane). Step 4: A now moves up the plane at $1$ m/s with deceleration $5$ m/s$^2$. Using $v^2 = u^2 + 2as$: $0 = 1^2 + 2(-5)s$, so $s = \dfrac{1}{10} = 0.1$ m.

Question 9 · Answer A

Step 1: Expand: $v = \dfrac{1}{2}(3t^2 - 14t + 8)$. Step 2: Differentiate: $a = \dfrac{dv}{dt} = \dfrac{1}{2}(6t - 14) = 3t - 7$. Step 3: At $t = 3$: $a = 3(3) - 7 = 9 - 7 = 2$ m/s$^2$.

Question 10 · Answer A

Step 1: $v = 0$ when $t = 1$ and $t = 3$. The particle changes direction at these times. Step 2: Check signs: at $t = 0$, $v = (-1)(-3) = 3 > 0$. At $t = 2$, $v = (1)(-1) = -1 < 0$. At $t = 3.5$, $v = (2.5)(0.5) = 1.25 > 0$. Step 3: Expand: $v = t^2 - 4t + 3$. Integrate: $\int v\,dt = \dfrac{t^3}{3} - 2t^2 + 3t$. Step 4: $s(0 \to 1) = \dfrac{1}{3} - 2 + 3 - 0 = \dfrac{4}{3}$. Step 5: $s(1 \to 3) = \left(9 - 18 + 9\right) - \dfrac{4}{3} = 0 - \dfrac{4}{3} = -\dfrac{4}{3}$. Step 6: $s(3 \to 4) = \left(\dfrac{64}{3} - 32 + 12\right) - 0 = \dfrac{64}{3} - 20 = \dfrac{4}{3}$. Step 7: Total distance $= \dfrac{4}{3} + \dfrac{4}{3} + \dfrac{4}{3} = 4$ m.

Question 11 · Answer A

Step 1: Expand: $v = \dfrac{1}{3}(2t^2 - 11t + 15)$. Step 2: Integrate: $s(t) = \dfrac{1}{3}\left(\dfrac{2t^3}{3} - \dfrac{11t^2}{2} + 15t\right) = \dfrac{2t^3}{9} - \dfrac{11t^2}{6} + 5t$. Step 3: Set $s = 0$: $t\left(\dfrac{2t^2}{9} - \dfrac{11t}{6} + 5\right) = 0$. For $t > 0$, solve $\dfrac{2t^2}{9} - \dfrac{11t}{6} + 5 = 0$. Step 4: Multiply by $18$: $4t^2 - 33t + 90 = 0$. Discriminant $= 33^2 - 4(4)(90) = 1089 - 1440 = -351 < 0$. Step 5: Since the discriminant is negative, there are no real solutions for $t > 0$. The particle does not return to O.

Question 12 · Answer A

Step 1: Consider particle B alone. The only horizontal force on B is the horizontal component of the tension: $T\cos15°$. Step 2: Apply Newton's second law to B: $T\cos15° = 1 \times 1 = 1$. Step 3: $T = \dfrac{1}{\cos15°} = \dfrac{1}{0.9659} = 1.035 \approx 1.04$ N (3 s.f.).

Question 13 · Answer A

Step 1: For block A, resolve vertically. The applied force ($10$ N at $30°$) lifts A, and the string tension ($1.04$ N at $15°$ below horizontal) pulls A down. $R_A + 10\sin30° = 3g + T\sin15°$. $R_A = 30 + 1.04\sin15° - 10\sin30° = 30 + 0.269 - 5 = 25.27$ N. Step 2: For block A, resolve horizontally. $10\cos30° - T\cos15° - \mu R_A = 3 \times 1$. $8.660 - 1.004 - 25.27\mu = 3$. $4.656 = 25.27\mu$. Step 3: $\mu = \dfrac{4.656}{25.27} = 0.1843 \approx 0.184$ (3 s.f.).