9231/33 May 2026 — Worked Solutions (v2 Corrected)

Question 1

Finding $v$ in terms of $a$ and $g$

The string makes an angle of $60°$ with the vertical. The radius of the horizontal circle is $r = 2a\sin 60° = a\sqrt{3}$.

Resolving vertically: $T\cos 60° = mg$ $\Rightarrow$ $\dfrac{T}{2} = mg$ $\Rightarrow$ $T = 2mg$

Resolving horizontally (toward centre): $T\sin 60° = \dfrac{mv^2}{r}$

$2mg \cdot \dfrac{\sqrt{3}}{2} = \dfrac{mv^2}{a\sqrt{3}}$

$mg\sqrt{3} = \dfrac{mv^2}{a\sqrt{3}}$ $\Rightarrow$ $v^2 = 3ag$

$v = \sqrt{3ag}$

Tip: For a conical pendulum, always draw the triangle formed by the string, vertical, and horizontal radius. Resolve vertically for $T$, then use $T\sin\theta$ as the centripetal force.

Question 2 [Corrected v3 — energy equation fix]

Correction note (v3): The v2 answer contained an error in the Stage 2 energy equation. The equation mixed an overall displacement $(a+e)$ for GPE and friction with the Stage 1 endpoint KE $\frac{1}{2}mv_a^2$. The correct approach uses the overall energy balance from $O$ to $a+e$ with initial KE $= 0$, yielding a quadratic in $e$ whose solution simplifies to $e = \dfrac{a\sqrt{3}}{3}$.

(a) Distance travelled before first coming to rest

Stage 1: From $O$ to the point where the string becomes taut (distance $= a$).

The string is slack throughout this stage, so there is no tension and no elastic potential energy. Only gravity and friction act on $P$.

$$\frac{1}{2}mv_a^2 = mga\sin 30° - \mu mg a\cos 30°$$

$$\frac{1}{2}mv_a^2 = \frac{mga}{2} - \frac{mga\sqrt{3}}{6} = \frac{mga(3 - \sqrt{3})}{6}$$

Since $3 - \sqrt{3} \approx 1.27 > 0$, we have $v_a > 0$, confirming that $P$ is still moving when the string becomes taut.

Stage 2: From $x = a$ to the point of maximum extension at $x = a + e$, where $e$ is the extension of the string beyond its natural length.

Write the overall energy balance from $O$ to $a + e$:

$$\underbrace{0}_{\text{Initial KE}} = \underbrace{(-mg(a+e)\sin 30°)}_{\text{Final GPE}} + \underbrace{\frac{\lambda e^2}{2a}}_{\text{Final EPE}} + \underbrace{\mu mg(a+e)\cos 30°}_{\text{Friction loss}}$$

Rearranging: $mg(a+e)\sin 30° - \mu mg(a+e)\cos 30° = \dfrac{\lambda e^2}{2a}$

$$\frac{(a+e)(3-\sqrt{3})}{6} = \frac{e^2}{a}$$

This gives the quadratic equation $6e^2 - a(3-\sqrt{3})e - a^2(3-\sqrt{3}) = 0$.

The discriminant is $\Delta = a^2(3-\sqrt{3})^2 + 24a^2(3-\sqrt{3}) = a^2(84 - 30\sqrt{3})$.

Noting that $84 - 30\sqrt{3} = (5\sqrt{3}-3)^2$, we have $\sqrt{\Delta} = a(5\sqrt{3}-3)$.

$$e = \frac{a(3-\sqrt{3}) + a(5\sqrt{3}-3)}{12} = \frac{4a\sqrt{3}}{12} = \frac{a\sqrt{3}}{3}$$

Total distance from $O$:

$$a + e = a + \frac{a\sqrt{3}}{3} = \frac{a(3+\sqrt{3})}{3}$$

Distance $= \dfrac{a(3 + \sqrt{3})}{3} \approx 1.577\,a$

(b) Does $P$ remain at rest?

At the point of maximum extension, the extension is $e = \dfrac{a\sqrt{3}}{3}$.

Tension $T = \dfrac{\lambda e}{a} = \dfrac{2mg \cdot a\sqrt{3}/3}{a} = \dfrac{2mg\sqrt{3}}{3}$ (up the plane)

Weight component down the plane $= mg\sin 30° = \dfrac{mg}{2}$

Maximum available friction $= \mu mg\cos 30° = \dfrac{mg\sqrt{3}}{6}$ (acts up the plane, opposing the tendency to slide down)

Check: $T + F_{\max} = \dfrac{2mg\sqrt{3}}{3} + \dfrac{mg\sqrt{3}}{6} = \dfrac{5mg\sqrt{3}}{6} \approx 1.44\,mg$

Compare with $mg\sin 30° = 0.5\,mg$:

$\dfrac{5\sqrt{3}}{6} \approx 1.44 > 0.5$ $\Rightarrow$ $T + F_{\max} > mg\sin 30°$

Yes, $P$ remains at rest.

Yes, $P$ remains at rest because $T + F_{\max} > mg\sin 30°$.

Note: The friction needed to maintain equilibrium is $\dfrac{mg}{2} - \dfrac{2mg\sqrt{3}}{3} = \dfrac{mg(3 - 4\sqrt{3})}{6}$. Since this is negative ($4\sqrt{3} \approx 6.93 > 3$), the tension alone already exceeds the weight component, so friction actually acts down the plane to balance the forces. Either way, equilibrium is maintained.

Question 3

(a) Distance of centre of mass from $AB$

Set up coordinates with $A = (0, 0)$, $B = (4a, 0)$, $C = (2a, 2a\sqrt{3})$.

Midpoint of $BC$: $M = \left(\dfrac{4a + 2a}{2}, \dfrac{0 + 2a\sqrt{3}}{2}\right) = (3a, a\sqrt{3})$

Centroid of $\triangle ABC$: $G = \left(\dfrac{0 + 4a + 2a}{3}, \dfrac{0 + 0 + 2a\sqrt{3}}{3}\right) = \left(2a, \dfrac{2a\sqrt{3}}{3}\right)$

Combined CoM (lamina mass $m$ + particle mass $m$, total $2m$):

$$\bar{y} = \frac{m \cdot \frac{2a\sqrt{3}}{3} + m \cdot a\sqrt{3}}{2m} = \frac{\frac{2a\sqrt{3}}{3} + a\sqrt{3}}{2} = \frac{\frac{5a\sqrt{3}}{3}}{2} = \frac{5a\sqrt{3}}{6}$$

Distance from $AB = \dfrac{5a\sqrt{3}}{6}$

(b) Angle $AC$ makes with downward vertical when suspended from $A$

$\bar{x} = \dfrac{m \cdot 2a + m \cdot 3a}{2m} = \dfrac{5a}{2}$

CoM at $G' = \!\left(\dfrac{5a}{2}, \dfrac{5a\sqrt{3}}{6}\right)$.

When suspended from $A = (0,0)$, the line $AG'$ is vertical (downward direction).

Direction of $AG' = \left(\dfrac{5a}{2}, \dfrac{5a\sqrt{3}}{6}\right) \propto (3, \sqrt{3})$

Direction of $AC = (2a, 2a\sqrt{3}) \propto (1, \sqrt{3})$

Angle between $AC$ and the vertical:

$$\cos\varphi = \frac{(1)(3) + (\sqrt{3})(\sqrt{3})}{\sqrt{1+3}\;\cdot\;\sqrt{9+3}} = \frac{3 + 3}{2\sqrt{12}} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$$

$\varphi = 30°$

Angle $= 30°$

Tip: When a lamina is suspended from a point, the centre of mass hangs directly below the point of suspension. The angle a side makes with the vertical equals the angle between the side's direction and the direction from the suspension point to the CoM.

Question 4

(a) Showing $\theta = 60°$

The normal reaction $R$ acts along $OP$ (toward the centre $O$).

Resolving vertically: $R\cos\theta = mg$ … (1)

Resolving horizontally: $R\sin\theta = \dfrac{mv^2}{a\sin\theta}$ … (2)

From (1): $R = \dfrac{mg}{\cos\theta}$. Substituting into (2):

$$\frac{mg\sin\theta}{\cos\theta} = \frac{mv^2}{a\sin\theta} \quad\Rightarrow\quad v^2 = ag\,\frac{\sin^2\!\theta}{\cos\theta}$$

Given $v^2 = \dfrac{3ag}{2}$:

$$\frac{3}{2} = \frac{\sin^2\!\theta}{\cos\theta} = \frac{1 - \cos^2\!\theta}{\cos\theta}$$

$3\cos\theta = 2 - 2\cos^2\!\theta$ $\Rightarrow$ $2\cos^2\!\theta + 3\cos\theta - 2 = 0$

$\cos\theta = \dfrac{-3 \pm \sqrt{9 + 16}}{4} = \dfrac{-3 \pm 5}{4}$

Taking the positive root: $\cos\theta = \dfrac{1}{2}$, so $\theta = 60°$.

$\theta = 60°$ (shown)

(b) Normal reaction

$R = \dfrac{mg}{\cos 60°} = \dfrac{mg}{1/2} = 2mg$

$R = 2mg$

$r = a\sin 60° = \dfrac{a\sqrt{3}}{2}$

$r = \dfrac{a\sqrt{3}}{2}$

Question 5

(a) Time to come to rest

Going up, both gravity and resistance act downward:

$$1 \times \frac{dv}{dt} = -10 - 0.5v$$

$$\frac{dv}{dt} + 0.5v = -10$$

Integrating factor: $e^{0.5t}$

$$\frac{d}{dt}\!\left[ve^{0.5t}\right] = -10e^{0.5t}$$

$$ve^{0.5t} = -20e^{0.5t} + C$$

At $t = 0$: $v = 15$, so $15 = -20 + C \Rightarrow C = 35$

$$v = 35e^{-0.5t} - 20$$

At rest ($v = 0$): $35e^{-0.5T} = 20$

$e^{-0.5T} = \dfrac{4}{7}$ $\Rightarrow$ $-0.5T = \ln\dfrac{4}{7}$

$T = 2\ln\dfrac{7}{4}$

$T = 2\ln\dfrac{7}{4} \approx 1.12$ s

(b) Greatest height

$$x = \int_0^T v\,dt = \int_0^T (35e^{-0.5t} - 20)\,dt$$

$$= \left[-70e^{-0.5t} - 20t\right]_0^T$$

$$= -70e^{-0.5T} - 20T - (-70)$$

$$= 70\!\left(1 - \frac{4}{7}\right) - 20T$$

$$= 70 \times \frac{3}{7} - 20T = 30 - 20T$$

$$= 30 - 20 \times 2\ln\frac{7}{4} = 30 - 40\ln\frac{7}{4}$$

Greatest height $= 30 - 40\ln\dfrac{7}{4} \approx 7.62$ m

Tip: For resistive motion, the ODE $\dfrac{dv}{dt} + kv = c$ always has solution $v = \dfrac{c}{k} + \left(u - \dfrac{c}{k}\right)e^{-kt}$. Here $c = -g = -10$, $k = 0.5$, giving terminal velocity $= c/k = -20$ m/s (the negative sign confirms it acts downward).

Question 6

(a) Showing $\tan\varphi = \frac{3}{2}\tan\alpha$

Take $x$-axis along line of centres (from $A$ to $B$).

Since the spheres are smooth, the impulse acts only along the line of centres ($x$-direction). Therefore the $y$-components of velocity are unchanged:

$A_y^{\text{after}} = 2u\sin\alpha$, $B_y^{\text{after}} = 0$

Along $x$ (conservation of momentum):

$$3m(2u\cos\alpha) = 3m\,v_{Ax} + m\,v_{Bx} \quad\Rightarrow\quad 6u\cos\alpha = 3v_{Ax} + v_{Bx} \quad\text{...(1)}$$

Newton's restitution law:

$$v_{Bx} - v_{Ax} = -\frac{1}{3}\!\left(0 - 2u\cos\alpha\right) = \frac{2u\cos\alpha}{3} \quad\text{...(2)}$$

From (2): $v_{Ax} = v_{Bx} - \dfrac{2u\cos\alpha}{3}$. Sub into (1):

$6u\cos\alpha = 3v_{Bx} - 2u\cos\alpha + v_{Bx} = 4v_{Bx} - 2u\cos\alpha$

$4v_{Bx} = 8u\cos\alpha$ $\Rightarrow$ $v_{Bx} = 2u\cos\alpha$

$v_{Ax} = 2u\cos\alpha - \dfrac{2u\cos\alpha}{3} = \dfrac{4u\cos\alpha}{3}$

Now $\tan\varphi = \dfrac{A_y}{A_x} = \dfrac{2u\sin\alpha}{\frac{4u\cos\alpha}{3}} = \dfrac{3\sin\alpha}{2\cos\alpha} = \dfrac{3}{2}\tan\alpha$

$\tan\varphi = \dfrac{3}{2}\tan\alpha$ (shown)

(b) Speed of $B$ after collision

$B$ moves along the line of centres ($B_y = 0$), so:

$$v_B = v_{Bx} = 2u\cos\alpha$$

$v_B = 2u\cos\alpha$

Note: Because $B$ was initially at rest and the impulse is along the line of centres, $B$'s velocity after collision is purely along the line of centres. $A$'s perpendicular velocity component is unchanged, but its parallel component is reduced.

Question 7 [Corrected v2 — $AB = e \cdot OA$]

Correction note: The original version stated $OB = e \cdot OA$, but $OB = OA + AB = (1+e)\,OA$. The correct relationship is $AB = e \cdot OA$, i.e. the second range equals $e$ times the first range. The question text and answers have been corrected accordingly.

(a) Finding $OA$

Using the range formula: $OA = \dfrac{U^2\sin 2\alpha}{g}$

$= \dfrac{20^2 \times \sin 120°}{10} = \dfrac{400 \times \frac{\sqrt{3}}{2}}{10} = 20\sqrt{3}$

$OA = 20\sqrt{3} \approx 34.6$ m

(b) Showing $AB = e \cdot OA$

At $A$ (first landing): $v_x = U\cos\alpha = 10$, $v_y = -U\sin\alpha = -10\sqrt{3}$

After rebound: $v'_x = v_x = 10$ (unchanged, smooth horizontal ground)

$v'_y = e \times U\sin\alpha = \dfrac{1}{2} \times 10\sqrt{3} = 5\sqrt{3}$ (upward)

Time of first flight: $T_1 = \dfrac{2U\sin\alpha}{g} = \dfrac{20\sqrt{3}}{10} = 2\sqrt{3}$

Time of second flight: $T_2 = \dfrac{2v'_y}{g} = \dfrac{10\sqrt{3}}{10} = \sqrt{3} = \dfrac{1}{2}T_1 = eT_1$

$AB = v'_x \times T_2 = 10\sqrt{3}$

$e \cdot OA = \dfrac{1}{2} \times 20\sqrt{3} = 10\sqrt{3}$

Therefore $AB = e \cdot OA$. $\blacksquare$

$AB = e \cdot OA$ (shown)

$v = \sqrt{v_x'^2 + v_y'^2} = \sqrt{10^2 + (5\sqrt{3})^2} = \sqrt{100 + 75} = \sqrt{175} = 5\sqrt{7}$

$v = 5\sqrt{7} \approx 13.2$ m/s

(d) Greatest height after rebound & total time

Greatest height (using $v^2 = u^2 - 2gs$ for vertical motion):

$0 = (5\sqrt{3})^2 - 2 \times 10 \times h$

$h = \dfrac{75}{20} = \dfrac{15}{4} = 3.75$ m

Total time from projection to max height after rebound:

$= T_1 + t_{\text{rise}} = \dfrac{2U\sin\alpha}{g} + \dfrac{v'_y}{g}$

$= 2\sqrt{3} + \dfrac{5\sqrt{3}}{10} = 2\sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{4\sqrt{3} + \sqrt{3}}{2} = \dfrac{5\sqrt{3}}{2}$

Greatest height $= \dfrac{15}{4}$ m (3.75 m); Total time $= \dfrac{5\sqrt{3}}{2} \approx 4.33$ s

Tip: The result $AB = e \cdot OA$ for a smooth horizontal rebound is a useful general result. Each successive range is multiplied by $e$, so if the particle bounces $n$ times, the successive ranges form a geometric sequence with common ratio $e$.

These are worked solutions for the predicted paper. They represent one valid method for each question.
v2 corrected: Q2 two-stage model; Q7(b) $AB = e \cdot OA$.
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