9709/41 Oct/Nov 2025 Worked Answers (A-Level Maths 9709 AS)

13 questions · 50 marks · 75 minutes

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1. Question 1a

   2 marksDriving Force (Newton's Second Law)
   Step 1: Apply Newton's second law along the direction of motion: $D - R = ma$, where $D$ is the driving force, $R = 400$ N is the resistance, $m = 1200$ kg, and $a = 0.3$ m/s$^2$. Step 2: Substitute: $D - 400 = 1200 \times 0.3 = 360$. Step 3: Solve: $D = 360 + 400 = 760$ N.
   Method:

   Apply $F_{\text{net}} = ma$ where $F_{\text{net}} = D - R$, then solve for $D$.
   Examiner tips

   • Always draw a force diagram showing all horizontal forces
   • The driving force must overcome resistance AND provide the resultant force for acceleration

2. Question 1b

   2 marksPower
   Step 1: Power is given by $P = Fv$, where $F$ is the driving force and $v$ is the speed. Step 2: Substitute: $P = 760 \times 18 = 13680$ W.
   Method:

   Apply $P = Fv$ directly using the driving force and the instantaneous speed.
   Examiner tips

   • Power = driving force times velocity, not net force times velocity
   • Ensure consistent units: Newtons and m/s give Watts

3. Question 2a

   3 marksConservation of Momentum
   Step 1: P could either continue forward at $1$ m/s or rebound at $1$ m/s after the collision. Step 2: Case 1 (P continues at $1$ m/s forward): $4(6) = 4(1) + 8v$, so $24 = 4 + 8v$, giving $v = 2.5$ m/s. Step 3: Case 2 (P rebounds at $1$ m/s): $4(6) = 4(-1) + 8v$, so $24 = -4 + 8v$, giving $v = 3.5$ m/s. Step 4: The two possible speeds of Q are $2.5$ m/s and $3.5$ m/s. The larger is $3.5$ m/s.
   Method:

   Apply conservation of momentum twice: once with P continuing forward and once with P rebounding.
   Examiner tips

   • Always consider both directions for the particle whose speed (not velocity) is given after a collision
   • Check physical validity: Q must move in the original direction of P

4. Question 2b

   2 marksKinetic Energy Loss in Collisions
   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.
   Method:

   Compute $\frac{1}{2}mv^2$ for each particle before and after, then subtract to find the loss.
   Examiner tips

   • Kinetic energy is a scalar: use speed (not velocity) when computing $\frac{1}{2}mv^2$
   • Momentum is always conserved in a collision, but kinetic energy is only conserved in a perfectly elastic collision

5. Question 3

   6 marksResultant of Coplanar Forces
   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.).
   Method:

   Resolve each force into horizontal and vertical components, sum each set of components, then use Pythagoras.
   Examiner tips

   • Draw a clear diagram showing all forces with their angles measured from the horizontal
   • Be careful with signs: forces pointing left or downward contribute negative components

6. Question 4

   6 marksEnergy Method for Connected Particles
   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. $2m + 6 = 8m - 14.4 - 9.6$. $2m + 6 = 8m - 24$. $6m = 30$. $m = 5$.
   Method:

   Apply the work-energy principle to the whole system: KE gain = PE loss by A - PE gain by B - work against resistance.
   Examiner tips

   • In energy problems with connected particles, both particles have the same speed and travel the same distance
   • Remember: the vertical height gained by B is $d\sin\theta$, not $d$

7. Question 5

   6 marksFriction on Inclined Plane
   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: $\alpha = \tan^{-1}(2.75) = 70.0°$.
   Method:

   Resolve forces parallel and perpendicular to the plane, apply $F = \mu R$, cancel $mg$, and solve for $\alpha$.
   Examiner tips

   • Always resolve the applied force into components parallel and perpendicular to the inclined plane
   • At limiting equilibrium, $F = \mu R$ with friction opposing the direction of potential motion

8. Question 6

   6 marksCollision on Inclined Plane
   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.
   Method:

   Find A's pre-collision speed using $v = at$, apply conservation of momentum to find A's post-collision velocity, then use $v^2 = u^2 + 2as$ for motion up the plane.
   Examiner tips

   • Be careful with signs: define a positive direction and stick to it throughout
   • After the collision, A decelerates at the same rate as before (smooth plane)

9. Question 7a

   2 marksVariable Acceleration
   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$^2$.
   Method:

   Expand $v$, differentiate to get $a$, then substitute $t = 3$.
   Examiner tips

   • Expand the brackets before differentiating - it is simpler than using the product rule
   • Acceleration is the derivative of velocity with respect to time

10. Question 7b

    5 marksDistance with Variable Velocity
    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.
    Method:

    Find when $v = 0$ to locate direction changes, integrate over each sub-interval, then sum the absolute displacements.
    Examiner tips

    • Total distance is not the same as displacement - you must account for changes of direction
    • Find zeros of $v$ to determine when the particle reverses

11. Question 7c

    2 marksReturn to Origin
    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.
    Method:

    Integrate velocity to find displacement as a function of $t$, set $s = 0$, and show no positive solution exists (negative discriminant).
    Examiner tips

    • $v = 0$ means the particle is instantaneously at rest, NOT that it is at the origin
    • To check return to O, examine the displacement $s(t) = \int_0^t v\,du$, not the velocity

12. Question 8a

    2 marksTension in String (Connected Particles)
    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.).
    Method:

    Apply $F = ma$ to particle B, using the horizontal component of tension as the net horizontal force.
    Examiner tips

    • When a string is at an angle, resolve the tension into horizontal and vertical components
    • Apply Newton's second law to the lighter particle - it involves fewer forces

13. Question 8b

    6 marksCoefficient of Friction
    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.).
    Method:

    Resolve forces on A vertically to find $R_A$, then horizontally to find friction $F$, then compute $\mu = F/R_A$.
    Examiner tips

    • Be careful: the applied force and string tension both have vertical components that affect the normal reaction
    • The applied force lifts A (reducing $R_A$) while the string tension pulls A down (increasing $R_A$)