May/June 2025 Paper 41 Worked Answers (A-Level Maths 9709 AS)

13 questions · 50 marks · 75 minutes

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1. Question 1

   4 marksNewton's Second Law on Inclined Planes
   Step 1: Resolve perpendicular to the plane: $R = 15g\cos 25^\circ = 15 \times 10 \times \cos 25^\circ \approx 135.9$ N. Step 2: Calculate the friction force: $F = \mu R = 0.3 \times 135.9 \approx 40.8$ N. Step 3: Resolve along the plane (up positive): $T - F - 15g\sin 25^\circ = 15 \times 1.6$. Step 4: The component of weight down the plane is $15 \times 10 \times \sin 25^\circ \approx 63.4$ N. Step 5: $T = 24 + 40.8 + 63.4 = 128$ N (to 3 s.f.).
   Method:

   Resolve perpendicular to find $R$, calculate friction $F = \mu R$, then apply $F = ma$ along the plane to find $T$.
   Examiner tips

   • Always resolve perpendicular to the plane first to find the normal reaction before calculating friction
   • Draw a clear force diagram showing weight components, friction, tension, and normal reaction

2. Question 2

   4 marksEquilibrium of Coplanar Forces
   Step 1: Since $Q$ is perpendicular to $P$, the component of all forces in the direction of $P$ must be zero (as $Q$ has no component in the $P$-direction). Step 2: Resolve in the direction of $P$: $P + 6\sin 50^\circ - 12\sin 40^\circ = 0$. Step 3: $P = 12\sin 40^\circ - 6\sin 50^\circ = 12 \times 0.6428 - 6 \times 0.7660 = 7.7135 - 4.5963 = 3.12$ N (to 3 s.f.).
   Method:

   Use the fact that $Q \perp P$ means components in the $P$-direction sum to zero. Resolve and solve for $P$.
   Examiner tips

   • When the resultant is perpendicular to a force, the net component in that force's direction is zero
   • Be very careful about which angle goes with sine and which with cosine when resolving

3. Question 3a

   3 marksVelocity-Time Graphs and Kinematics
   Step 1: Find the speed at the end of the acceleration phase: $v = 2 + 0.4 \times 15 = 8$ m/s. Step 2: The constant-speed phase lasts $25$ s at $8$ m/s. Step 3: The deceleration phase starts at $t = 15 + 25 = 40$ s. The cyclist decelerates from $8$ m/s to rest. Using the area of the triangle on the velocity-time graph: $\dfrac{1}{2} \times 8 \times (T - 40) = 120$. Step 4: $4(T - 40) = 120$, so $T - 40 = 30$, giving $T = 70$ s.
   Method:

   Find maximum speed using $v = u + at$, then use triangle area formula for the deceleration phase to find the total time.
   Examiner tips

   • The area under a velocity-time graph represents displacement
   • For uniform deceleration to rest, the v-t graph is a triangle

4. Question 4a(i)

   1 marksPower and Driving Force
   Step 1: At constant speed, acceleration is zero, so the driving force equals the resistance: $D = 2000$ N. Step 2: Power $= D \times v = 2000 \times 25 = 50000$ W.
   Method:

   At constant speed, driving force = resistance. Then $P = Fv$.
   Examiner tips

   • At constant speed on a horizontal road, driving force = resistance
   • $P = Fv$ is the key formula connecting power, force, and velocity

5. Question 4a(ii)

   2 marksPower and Acceleration
   Step 1: Find the driving force: $D = \dfrac{P}{v} = \dfrac{60000}{20} = 3000$ N. Step 2: Apply Newton's second law: $D - R = ma$, so $3000 - 2000 = 20000a$. Step 3: $1000 = 20000a$, giving $a = 0.05$ m/s$^2$.
   Method:

   Calculate driving force $D = P/v$, then use $D - R = ma$ to find $a$.
   Examiner tips

   • Always find the driving force first using $D = P/v$
   • Then apply Newton's second law with net force = driving force minus resistance

6. Question 4b

   3 marksPower on an Incline
   Step 1: At constant speed, the net force is zero. The driving force is $D = \dfrac{P}{v} = \dfrac{105000}{15} = 7000$ N. Step 2: Resolve along the slope: $D - R - mg\sin\alpha = 0$, so $7000 - 2000 - 20000 \times 10 \times \sin\alpha = 0$. Step 3: $5000 = 200000\sin\alpha$, giving $\sin\alpha = 0.025$. Step 4: $\alpha = \sin^{-1}(0.025) = 1.43^\circ$.
   Method:

   Find $D = P/v$, then use $D - R - mg\sin\alpha = 0$ at constant speed to find $\alpha$.
   Examiner tips

   • At constant speed on a slope, the driving force must overcome both the resistance and the weight component down the slope
   • For small angles, $\sin\alpha \approx \alpha$ in radians, but always use exact calculation

7. Question 5a

   3 marksMomentum and Kinetic Energy
   Step 1: From momentum: $mu = 6$, so $u = \dfrac{6}{m}$. Step 2: From kinetic energy: $\dfrac{1}{2}mu^2 = 36$. Step 3: Substitute $u = \dfrac{6}{m}$: $\dfrac{1}{2}m \times \dfrac{36}{m^2} = 36$, giving $\dfrac{18}{m} = 36$. Step 4: $m = 0.5$ kg. Then $u = \dfrac{6}{0.5} = 12$ m/s.
   Method:

   Form simultaneous equations from momentum and KE, divide to find $u$, then substitute back for $m$.
   Examiner tips

   • Dividing KE by momentum gives $\frac{1}{2}u$, which is a quick way to find $u$
   • Always check: $mu = 0.5 \times 12 = 6$ and $\frac{1}{2}(0.5)(144) = 36$

8. Question 5b

   4 marksCollision and Loss of Kinetic Energy
   Step 1: Conservation of momentum: $0.4v = 0.4w + 1.2 \times 3w = 0.4w + 3.6w = 4w$. So $v = 10w$. Step 2: Loss of KE $= \dfrac{1}{2}(0.4)v^2 - \dfrac{1}{2}(0.4)w^2 - \dfrac{1}{2}(1.2)(3w)^2 = 14.4$. Step 3: Substitute $v = 10w$: $0.2(100w^2) - 0.2w^2 - 0.6(9w^2) = 14.4$. Step 4: $20w^2 - 0.2w^2 - 5.4w^2 = 14.4$, so $14.4w^2 = 14.4$, giving $w = 1$ m/s. Step 5: $v = 10w = 10$ m/s.
   Method:

   Use conservation of momentum to express $v$ in terms of $w$, then substitute into the KE loss equation to find $w$ and hence $v$.
   Examiner tips

   • Always use conservation of momentum first to reduce the number of unknowns
   • Be careful with the KE formula: $\frac{1}{2}mv^2$ for each particle

9. Question 6a

   5 marksConnected Particles on Inclined Planes
   Step 1: For particle $Q$ (moving down its plane): $0.8g\sin 40^\circ - T = 0.8a$. So $0.8 \times 10 \times 0.6428 - T = 0.8a$, giving $5.142 - T = 0.8a$. Step 2: For particle $P$ (moving up its plane): $T - 0.4g\sin\theta = 0.4a$. So $T - 0.4 \times 10 \times 0.3 = 0.4a$, giving $T - 1.2 = 0.4a$. Step 3: Add the two equations: $5.142 - 1.2 = 1.2a$, so $3.942 = 1.2a$. Step 4: $a = 3.29$ m/s$^2$ (to 3 s.f.).
   Method:

   Write $F = ma$ for each particle along its plane, then add the equations to eliminate $T$ and find $a$.
   Examiner tips

   • For connected particles, the acceleration is the same for both and the tension is the same throughout the string
   • Draw separate force diagrams for each particle

10. Question 6b

    4 marksEnergy Method with Friction
    Step 1: When $Q$ moves $2.5$ m down its plane, $P$ moves $2.5$ m up its plane (inextensible string). Both reach the same speed $v$. Step 2: GPE lost by $Q$: $0.8 \times 10 \times 2.5 \times \sin 30^\circ = 0.8 \times 10 \times 2.5 \times 0.5 = 10$ J. Step 3: GPE gained by $P$: $0.4 \times 10 \times 2.5 \times 0.3 = 3$ J. Step 4: By the work-energy principle: $\dfrac{1}{2}(0.4 + 0.8)v^2 = 10 - 3 - 1.6 = 5.4$. Step 5: $0.6v^2 = 5.4$, so $v^2 = 9$, giving $v = 3$ m/s.
    Method:

    Apply the work-energy principle: total KE gained = GPE lost by Q minus GPE gained by P minus work done against friction.
    Examiner tips

    • The work-energy principle accounts for all energy transfers: KE gained, GPE changes, and work done against friction
    • Both particles have the same speed since they are connected by an inextensible string

11. Question 7a

    6 marksVariable Acceleration Kinematics
    Step 1: Differentiate to find velocity: $v = \dfrac{ds}{dt} = 4 \times \dfrac{3}{2}t^{1/2} - 8 = 6t^{1/2} - 8$. Step 2: Set $v = 0$: $6t^{1/2} - 8 = 0$, so $t^{1/2} = \dfrac{8}{6} = \dfrac{4}{3}$. Step 3: Square both sides: $t = \dfrac{16}{9}$ s.
    Method:

    Differentiate $s$ to find $v$, set $v = 0$, solve for $t^{1/2}$, then square to get $t$.
    Examiner tips

    • The derivative $\frac{d}{dt}(t^{3/2}) = \frac{3}{2}t^{1/2}$ by the power rule
    • After solving $t^{1/2} = k$, square to find $t = k^2$

12. Question 7a_ii

    4 marksTotal Distance with Variable Acceleration
    Step 1: Evaluate $s$ at key times. $s(0) = 0$. At $t = \dfrac{16}{9}$: $\left(\dfrac{16}{9}\right)^{3/2} = \dfrac{16}{9} \times \dfrac{4}{3} = \dfrac{64}{27}$. So $s\!\left(\dfrac{16}{9}\right) = 4 \times \dfrac{64}{27} - 8 \times \dfrac{16}{9} = \dfrac{256}{27} - \dfrac{384}{27} = -\dfrac{128}{27}$. Step 2: $s(9) = 4(9)^{3/2} - 8(9) = 4 \times 27 - 72 = 36$. Step 3: The particle moves from $s = 0$ to $s = -\dfrac{128}{27}$ (distance $= \dfrac{128}{27}$), then reverses direction and moves to $s = 36$ (distance $= 36 + \dfrac{128}{27} = \dfrac{1100}{27}$). Step 4: Total distance $= \dfrac{128}{27} + \dfrac{1100}{27} = \dfrac{1228}{27} \approx 45.5$ m.
    Method:

    Find displacement at $t = 0$, $t = 16/9$ (where $v = 0$), and $t = 9$. Total distance is the sum of absolute displacement changes over each interval.
    Examiner tips

    • Total distance requires splitting the journey at points where the particle changes direction
    • Distance is always positive; take absolute values of displacement changes

13. Question 7b

    7 marksIntegration for Displacement
    Step 1: Integrate acceleration to find velocity: $v = \int (0.6 - 0.4t)\,dt = 0.6t - 0.2t^2 + C$. Step 2: At $t = 0$, $v = 6$, so $C = 6$. Therefore $v = 0.6t - 0.2t^2 + 6$. Step 3: Set $v = -8$: $0.6t - 0.2t^2 + 6 = -8$, giving $-0.2t^2 + 0.6t + 14 = 0$. Step 4: Multiply by $-5$: $t^2 - 3t - 70 = 0$. Factorise: $(t - 10)(t + 7) = 0$. Step 5: Since $t > 0$, $t = 10$ s.
    Method:

    Integrate acceleration $a = 0.6 - 0.4t$ to get $v$, use $v(0) = 6$ to find the constant, set $v = -8$, and solve the resulting quadratic.
    Examiner tips

    • When integrating acceleration, do not forget the constant of integration determined by the initial velocity
    • When solving a quadratic for time, reject the negative root