A Level Mathematics 9709/41 May/June 2025 Past Paper with Answers

Question 1

A block of mass $15$ kg is pulled by a rope up a rough plane inclined at $25^\circ$ to the horizontal. The coefficient of friction between the block and the plane is $0.3$ and the block has acceleration $1.6$ m/s$^2$ directed up the plane. Take $g = 10$ m/s$^2$. Find the tension in the rope, which is parallel to the plane.

1. $128$ N
2. $87.4$ N
3. $106$ N
4. $63.4$ N

Question 2

Three coplanar forces $P$ N, $6$ N and $12$ N act at a point $O$. The $6$ N force acts at $50^\circ$ above the horizontal (directed to the upper right) and the $12$ N force acts at $40^\circ$ below the horizontal (directed to the lower left). The resultant of all three forces is $Q$ N and it is perpendicular to $P$. By resolving in the direction of $P$, find the value of $P$.

1. $3.12$ N
2. $6.16$ N
3. $3.72$ N
4. $1.41$ N

Question 3

A cyclist starts from point $O$ with speed $2$ m/s and accelerates uniformly at $0.4$ m/s$^2$ for $15$ s. The cyclist then travels at constant speed for $25$ s before decelerating uniformly to rest at point $P$. The distance covered during the deceleration phase is $120$ m. Find the total time for the journey from $O$ to $P$.

1. $70$ s
2. $55$ s
3. $64$ s
4. $80$ s

Question 4 · Answer A

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.

Question 5 · Answer A

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$^2$.

Question 6 · Answer A

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: $\alpha = \sin^{-1}(0.025) = 1.43^\circ$.

Question 7 · Answer A

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.

Question 8 · Answer A

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.

Question 9 · Answer A

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.).

Question 10 · Answer A

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.

Question 11 · Answer A

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.

Question 12 · Answer A

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.

Question 13 · Answer A

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.