1

Resolve vertically:   $T\cos 60° = mg$   $\Rightarrow$   $T = 2mg$ M1
Radius of circle   $r = 2a\sin 60° = a\sqrt{3}$ M1
Resolve horizontally:   $T\sin 60° = \dfrac{mv^2}{r}$ M1

$2mg \cdot \dfrac{\sqrt{3}}{2} = \dfrac{mv^2}{a\sqrt{3}}$ AG
$mg\sqrt{3} = \dfrac{mv^2}{a\sqrt{3}}$   $\Rightarrow$   $v^2 = 3ag$ A1

(a)   [5 marks]

Stage 1: From $O$ to the point where the string becomes taut (distance $a$). The string is slack, so no tension and no EPE.

$\frac{1}{2}mv_a^2 = mga\sin 30° - \mu mga\cos 30°$ M1
$\frac{1}{2}mv_a^2 = \dfrac{mga}{2} - \dfrac{mga\sqrt{3}}{6} = \dfrac{mga(3 - \sqrt{3})}{6}$ A1

Since $3 - \sqrt{3} > 0$, the particle has speed $v_a > 0$ at $x = a$ and continues into stage 2.

Stage 2: From $x = a$ to maximum extension at $x = a + e$. Let extension $= e$.

Energy balance (overall, from $O$ to $a + e$):
Initial energy $=$ Final energy $+$ Energy lost to friction

$0 = -mg(a+e)\sin 30° + \dfrac{\lambda e^2}{2a} + \mu mg(a+e)\cos 30°$ M1
$mg(a+e)\sin 30° - \mu mg(a+e)\cos 30° = \dfrac{mge^2}{a}$
$\dfrac{(a+e)(3 - \sqrt{3})}{6} = \dfrac{e^2}{a}$ A1

This gives $6e^2 = a(a+e)(3-\sqrt{3})$, i.e. $6e^2 - a(3-\sqrt{3})e - a^2(3-\sqrt{3}) = 0$.

Solving the quadratic: $e = \dfrac{a(3-\sqrt{3}) + a\sqrt{84 - 30\sqrt{3}}}{12}$

Since $84 - 30\sqrt{3} = (5\sqrt{3}-3)^2$, we have $\sqrt{84-30\sqrt{3}} = 5\sqrt{3}-3$, giving
$e = \dfrac{a(3-\sqrt{3}+5\sqrt{3}-3)}{12} = \dfrac{4a\sqrt{3}}{12} = \dfrac{a\sqrt{3}}{3}$ A1

Total distance from $O = a + e = a + \dfrac{a\sqrt{3}}{3} = \dfrac{a(3+\sqrt{3})}{3}$ A1

(b)   [1 mark]

At maximum extension, 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 $= mg\sin 30° = \dfrac{mg}{2}$.
Limiting friction $= \mu mg\cos 30° = \dfrac{mg\sqrt{3}}{6}$.

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

Since $\dfrac{5\sqrt{3}}{6} \approx 1.44 > 0.5$, we have $T + F_{\max} > mg\sin 30°$. M1A1

(a)   [3 marks]

Place $AB$ on the $x$-axis: $A = (0, 0)$, $B = (4a, 0)$, $C = (2a, 2a\sqrt{3})$.
Midpoint of $BC$: $M = (3a, a\sqrt{3})$.

Centroid of triangle: $G = \!\left(2a, \dfrac{2a\sqrt{3}}{3}\right)$ M1

Centre of mass of combined object (total mass $2m$):
$\bar{y} = \dfrac{m \cdot \dfrac{2a\sqrt{3}}{3} + m \cdot a\sqrt{3}}{2m}$ M1
$= \dfrac{2a\sqrt{3}/3 + a\sqrt{3}}{2} = \dfrac{5a\sqrt{3}/3}{2} = \dfrac{5a\sqrt{3}}{6}$ A1

(b)   [4 marks]

$\bar{x} = \dfrac{m \cdot 2a + m \cdot 3a}{2m} = \dfrac{5a}{2}$ M1
Centre of mass $G'$ at $\!\left(\dfrac{5a}{2}, \dfrac{5a\sqrt{3}}{6}\right)$.

When suspended from $A = (0,0)$, the line $AG'$ is vertical (downward).
Direction of $AG'$: $\left(\dfrac{5a}{2}, \dfrac{5a\sqrt{3}}{6}\right) \propto (3, \sqrt{3})$ M1
Direction of $AC$: $(2a, 2a\sqrt{3}) \propto (1, \sqrt{3})$ M1

Angle between $AC$ and vertical (direction $(3, \sqrt{3})$):
$\cos\varphi = \dfrac{(1)(3) + (\sqrt{3})(\sqrt{3})}{2 \times \sqrt{9 + 3}} = \dfrac{3 + 3}{2\sqrt{12}} = \dfrac{6}{4\sqrt{3}} = \dfrac{\sqrt{3}}{2}$ A1
$\varphi = 30°$

(a)   [3 marks]

Normal reaction $R$ acts along $OP$ (toward centre).
Resolve vertically:   $R\cos\theta = mg$   $\Rightarrow$   $R = \dfrac{mg}{\cos\theta}$ M1
Resolve horizontally:   $R\sin\theta = \dfrac{mv^2}{a\sin\theta}$ M1
$mg\tan\theta = \dfrac{mv^2}{a\sin\theta}$   $\Rightarrow$   $v^2 = ag\sin^2\!\theta \sec\theta$

Substitute $v^2 = \dfrac{3ag}{2}$:
$\dfrac{3}{2} = \dfrac{\sin^2\theta}{\cos\theta} = \dfrac{1 - \cos^2\!\theta}{\cos\theta}$
$3\cos\theta = 2 - 2\cos^2\!\theta$   $\Rightarrow$   $2\cos^2\!\theta + 3\cos\theta - 2 = 0$
$\cos\theta = \dfrac{-3 + \sqrt{9 + 16}}{4} = \dfrac{-3 + 5}{4} = \dfrac{1}{2}$ A1
$\theta = 60°$   AG

(b)   [2 marks]

$R = \dfrac{mg}{\cos 60°} = \dfrac{mg}{1/2} = 2mg$ M1A1

(c)   [2 marks]

$r = a\sin 60° = \dfrac{a\sqrt{3}}{2}$ M1A1

(a)   [4 marks]

Going up: $m\dfrac{dv}{dt} = -mg - 0.5v$
$\dfrac{dv}{dt} + 0.5v = -10$ M1

Integrating factor: $e^{0.5t}$
$\dfrac{d}{dt}\!\left[ve^{0.5t}\right] = -10e^{0.5t}$ M1
$ve^{0.5t} = -20e^{0.5t} + C$
At $t = 0$: $15 = -20 + C$   $\Rightarrow$   $C = 35$ A1

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

At rest: $35e^{-0.5T} = 20$   $\Rightarrow$   $e^{-0.5T} = \dfrac{4}{7}$ M1
$T = 2\ln\dfrac{7}{4}$   AG

(b)   [4 marks]

$x = \displaystyle\int_0^T v\,dt = \int_0^T (35e^{-0.5t} - 20)\,dt$ M1
$= \left[-70e^{-0.5t} - 20t\right]_0^T$ M1
$= -70e^{-0.5T} - 20T + 70$
$= 70\!\left(1 - \dfrac{4}{7}\right) - 20T = 30 - 20T$ A1

$= 30 - 20 \times 2\ln\dfrac{7}{4} = 30 - 40\ln\dfrac{7}{4}$ A1

(a)   [4 marks]

Let the line of centres be the $x$-axis (from $A$ to $B$).
Smooth spheres → impulse along line of centres only.
$y$-components of velocity unchanged: $A_y = 2u\sin\alpha$, $B_y = 0$ M1

Along $x$-direction:
Conservation of momentum: $3m(2u\cos\alpha) = 3m\,v_{Ax} + m\,v_{Bx}$ M1
Restitution: $v_{Bx} - v_{Ax} = -\dfrac{1}{3}(0 - 2u\cos\alpha) = \dfrac{2u\cos\alpha}{3}$ M1

Solving: $v_{Bx} = 2u\cos\alpha$,   $v_{Ax} = \dfrac{4u\cos\alpha}{3}$
$\tan\varphi = \dfrac{A_y}{A_x} = \dfrac{2u\sin\alpha}{\dfrac{4u\cos\alpha}{3}} = \dfrac{3\sin\alpha}{2\cos\alpha} = \dfrac{3}{2}\tan\alpha$ A1   AG

(b)   [4 marks]

$v_B = \sqrt{v_{Bx}^2 + v_{By}^2} = \sqrt{(2u\cos\alpha)^2 + 0^2} = 2u\cos\alpha$ M1A1

(a)   [2 marks]

$OA = \dfrac{U^2\sin 2\alpha}{g} = \dfrac{20^2 \sin 120°}{10} = \dfrac{400 \times \frac{\sqrt{3}}{2}}{10} = 20\sqrt{3}$ M1A1

(b)   [2 marks]

At $A$: $v_x = 20\cos 60° = 10$,   $v_y = -20\sin 60° = -10\sqrt{3}$ M1
After rebound: $v'_x = v_x = 10$,   $v'_y = e \times 10\sqrt{3} = 5\sqrt{3}$ M1
Time of second flight: $T' = \dfrac{2v'_y}{g} = \dfrac{10\sqrt{3}}{10} = \sqrt{3} = eT_1$
$AB = v'_x \times T' = 10\sqrt{3} = e \times 20\sqrt{3} = e \cdot OA$ A1   AG

(c)   [3 marks]

Speed after rebound = $\sqrt{v_x'^2 + v_y'^2} = \sqrt{10^2 + (5\sqrt{3})^2} = \sqrt{100 + 75}$ M1
$= \sqrt{175} = 5\sqrt{7}$ A1A1

(d)   [4 marks]

Greatest height after rebound (vertical motion only):
$0 = v_y'^2 - 2gh$   $\Rightarrow$   $h = \dfrac{v_y'^2}{2g} = \dfrac{(5\sqrt{3})^2}{20} = \dfrac{75}{20} = \dfrac{15}{4}$ M1A1

Total time = time of first flight + time to reach max height after rebound
$= \dfrac{2U\sin\alpha}{g} + \dfrac{v_y'}{g} = \dfrac{20\sqrt{3}}{10} + \dfrac{5\sqrt{3}}{10} = 2\sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{5\sqrt{3}}{2}$ M1A1