**(i)** $\sin 30° = 0.75\text{mg}x/1.2$ Extension is 0.8m | M1 A1 A1 B1 | For using Newton's second law with $a = 0$

PE loss = $\text{mg}(1.2 + 0.8\sin 30°)$ = (0.2mg) | B1 M1 | 

$\frac{1}{2}mv^2 = \text{mg} - 0.2\text{mg}$ | | 

**(ii)** EE gain = $0.75\text{mg}x(0.8)/(2 \times 1.2)$ = (0.2mg) | M1 A1 B1ft M1 | For an equation with terms representing PE, KE and EE in linear combination; It with $x$ or $d - 1.2$ replacing 0.8 in (ii)

Maximum speed is 3.96ms⁻¹ | A1 | 

**(iii)** PE loss = $\text{mg}(1.2 + x)\sin 30°$ or $\text{mgd}\sin 30°$ | M1 A1 | For using PE loss = EE gain to obtain a 3 term quadratic in $x$ or $d$; It with $x$ or $d - 1.2$ replacing 0.8 in (ii)

EE gain = $0.75\text{mg}x^2/(2 \times 1.2)$ or $0.75\text{mg}(d - 1.2)^2/(2 \times 1.2)$ | M1 | 

$[x^2 - 1.6x - 1.92 = 0, d^2 - 4d + 1.44 = 0]$ | A1 | 

Displacement is 3.6m | A1 | 

**Alternative for parts (ii) and (iii) for candidates who use Newton's second law and $a = v \frac{dv}{dx}$:**

In the following $x, y$ and $z$ represent displacement from equil. pos', extension, and distance OP respectively. | M1 | For using N2 with $a = v \frac{dv}{dx}$

$[mv \frac{dv}{dx} = \sin 30° - 0.75\text{mg}(0.8 + x)/1.2, mv \frac{dv}{dx} = \text{mg}\sin 30° - 0.75\text{mg}(1.2 - y)/1.2]$ | | 

$v^2/2 = 5gx/16 + C$ or | A1 M1 | For using $v^2(-0.8) = v^2(0)$ or $v^2(1.2) = 2(g\sin 30°)1.2$ as appropriate

$v^2/2 = gy/2 - 5gy^2/16 + C$ or | | 

$v^2/2 = 5gz/2 - 5gz^2/16 + C$ | | 

$[C = 0.6g + 5g(-0.8)^2/16$ or $C = 0.6g - 5g(1.2/4) + 5g(1.2)^2/16$ or $C = 0.6g - 5g(1.2/4) + 5g(1.2)^2/16]$ | M1 | For using $v_{\max}^2 = v^2(0)$ or $v^2(0.8)$ or $v^2(2)$ as appropriate

$v^2 = (-5x/8 + 1.6)g$ or $v^2 = (y - 5y^2/8 + 1.2)g$ or $v^2 = (5z/2 - 5z^2/8 - 0.9)g$ | | 

$[v_{\max}^2 = 1.6g$ or $0.8g - 0.4g + 1.2g$ or $5g - 2.5g - 0.9g]$ | M1 | For solving $v = 0$

Maximum speed is 3.96ms⁻¹ | A1 | 

**(iii)** $[5x^2 - 12.8 = 0 \Rightarrow x = 1.6, 5y^2 - 8y - 9.6 = 0 \Rightarrow y = 2.4, 5z^2 - 20z + 7.2 = 0 \Rightarrow z = 3.6]$ | M1 A1 | 

Displacement is 3.6m | A1 | 

**Alternative for parts (ii) and (iii) for candidates who use Newton's second law and SHM analysis:**

$[m\ddot{x} = \text{mg}\sin 30° - 0.75\text{mg}(0.8 + x)/1.2 \Rightarrow \ddot{x} = -\omega^2x; v^2 = \omega^2(a^2 - x^2)]$ | M1 A1 M1 | For using N2 with $v^2 = \omega^2(a^2 - x^2)$

$v^2 = 5g(a^2 - x^2)/8$ | | 

$v^2 = 5g(2.56 - x^2)/8$ | A1 M1 | For using $v_{\max}^2 = v^2(0)$

$[v_{\max}^2 = 5g \times 2.56 + 8]$ | | 

Maximum speed is 3.96ms⁻¹ | A1 | 

**(iii)** $[2.56 - x^2 = 0 \Rightarrow x = 1.6]$ | M1 | For solving $v = 0$

Displacement is 3.6m | A1 | 

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