Question 2:

(i) $1.2mg\cos\theta = mg$ or $T\cos\theta = mg$ | M1A1

(ii) $\cos\theta = \frac{1}{1.2}$, $\theta = 33.55...$ (accept 34, 33.6 or better) | A1

$1.2mg\sin\theta = mr\omega^2$ or $T\sin\theta = mr\omega^2$ | M1A1

$1.2mg\sin\theta = ml\sin\theta\omega^2$ | A1

$\omega = \sqrt{\frac{1.2 \times 9.8}{1.2l}} = \sqrt{\frac{9.8}{l}}$, $58.8 = l\omega^2$, $l = 0.2$ m | dM1A1 (8)

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Question 3:

(a) $mv\frac{dv}{dx} = mg\sin 30° - \frac{1}{2}mx\omega^2$ | M1A1A1

$\frac{1}{2}v^2 = xg\sin 30° - \frac{1}{6}x^3\omega^2 + c$ | dM1A1ft

$\frac{1}{2}x^3\omega^2 - v^2 = 3g\sin 30° - \frac{9}{2}$ | dM1

$v = 4.5166...$ m s$^{-1}$ | A1cso (7)

(b) $v_0 = x\omega^2 = 6g\sin 30°$, $x = 0$ | M1A1 (2)

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Question 4:

(a) Ratio of masses: $2a : 2a : 4a : 2a = 11a^2$ | M1A1

Distances: $0$, $2a$, $4a$, $y$ = $\frac{1}{2}$ | B1

$0 + 2a + 84a + 2a^2 = 11y$ | M1A1ft

$y = \frac{25}{11}a = 2.272...a$ | A1 (6)

(b) $\tan\theta = \frac{a}{11} - \frac{25}{a}$ or $\tan\theta = \frac{11a}{25a}$ | M1A1ft

$\theta = 23.749...°$ Accept 24° or better | A1 (3)

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Question 5:

(a) $\frac{1}{2}mv^2 = m \times 7ag + mgasin\theta$ | M1A1A1

$v^2 = 7ag + 2ag\sin\theta - ag(7 + 2\sin\theta)$ * | A1 (4)

(b) At top $v^2 = 5ag$ | M1A1

$R = mg - m\frac{v^2}{a}$ or $m\frac{v^2}{a} - mg$ | M1A1

$R = 4mg$ or substitute for $v^2$ | dM1

$R > 0$ $\Rightarrow$ complete circles | A1cso (6)

(c) Max $v$ at lowest point | M1

$\sin\theta = 1$ $\Rightarrow$ $v^2 = 9ag$ | A1 (2)

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Question 6:

(a) $\frac{0.20}{8} = \frac{0.40}{16}$ * | M1A1

Length of string = 1 m or 100 cm | A1cso (3)

(b) $T = \frac{0.6}{0.4} = 24$ (or use half string) | M1A1

$2T\cos\theta = F$ | M1

$F = 2 \times \frac{4}{5} \times 192 - \frac{2}{5} \times 38.4$ or $F = 224 - 38.4 = 38$ | A1 (4)

(c) $\text{Initial EPE} = \frac{160.62}{2 \times 0.4 \times 5}$, Final EPE $= \frac{160.22}{2 \times 0.4 \times 5}$ | B1 (either)

$\frac{160.62 - 160.22}{20.4 \times 5} = 0.3v^2$ | M1A1A1

$0.3v^2 = 40(0.62^2 - 0.22^2)$ | dM1A1cso

$v = 6.531...$ Accept 6.5 m s$^{-1}$ or better or exact value $8$ m s$^{-1}$ (6)

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Question 7:

(a) $\frac{45e + 201e}{1.8 + 1.2}$ or $\frac{45(1-e) + 20e}{1.8 + 1.2}$ | M1A1A1

$e = 0.4$ or $e = 0.6$ * | A1cso (4)

(b) $AO = 2.2$ m | M1A1

$\frac{45y - 1.8 + 20(2.8 - y)}{1.8 + 1.2}$ | A1

$54y + 97.2 - 100.8 - 36y = 0$ | A1

$y = 2.2$ m * | A1cso (5)

(c) $\frac{20 \times 0.6x}{1.2} - \frac{45 \times 0.4x}{1.8} = 0.6x$ | B1

$\frac{625}{9} \times 0.6x = x$ SHM | M1A1ft

$\omega^2 = \frac{625}{9}$ or $\omega = \frac{25}{3}$ | A1

$x = 0.4\cos\left(\frac{25}{3}t\right) + 0.5$ | M1A1

Time from C to D: $t = \frac{3}{25}\cos^{-1}(0.8)$ | M1A1ft

Speed at D: $v^2 = \frac{625}{9}(0.5^2 - 0.4^2)$ or use $v = a\sin\theta$ | A1

$v = \frac{3}{2.5} \times 0.3$ | M1

Time from D to A: $\frac{1.8}{2.5}$ | M1

Total time: $\frac{3}{25}\cos^{-1}(0.8) + \frac{3}{25}\cos^{-1}(0.8) + 1.01977...$ s (accept 1.0 or better) | A1cao (8)

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