# Question 5:

**Part (a):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}MuL = \frac{1}{3}M(2L)^2\omega$ | M1 A2 | M1 for conservation of angular momentum; A1 and A2 for each side |
| $\omega = \frac{3u}{8L}$ | A1 (4) | Printed answer |

**Part (b):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3Mg}{2} - Mg = ML\omega^2$ | M1 A1 | First M1 for resolving inwards; First A1 for correct equation |
| $\frac{Mg}{2} = ML\left(\frac{3u}{8L}\right)^2$ | DM1 | Second DM1 (dep on previous M) for substituting for $\omega$ |
| $u = \sqrt{\frac{32gL}{9}}$ | A1 (4) | Second A1 for answer |

**Part (c):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $M(A):\ 0 = I\ddot{\theta} \Rightarrow \ddot{\theta} = 0$ | M1 | M1 for moments equation showing $\ddot{\theta} = 0$ |
| $(\rightarrow)\ X = ML\ddot{\theta} = 0$ | A1 (2) | A1 for resolving horizontally to give zero |

**Part (d):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}\left(\frac{4ML^2}{3}\right)\omega^2 = MgL(1 - \cos\alpha)$ | M1 A1 A1 | First M1 for energy equation; First A1 for KE in terms of $M$, $L$, $\omega$; Second A1 for PE |
| $\frac{1}{3} = (1 - \cos\alpha)$ | DM1 | Second DM1 (dep on 1st M) for substituting for $\omega$ and $u$ |
| $\alpha = \arccos\!\left(\frac{2}{3}\right) = 48°$ or better | A1 (5) | Third A1 for $48°$ or better (0.84 rad or better) |

## Question 6a:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $m = m_0(1+kx)$, $\quad \frac{\mathrm{d}m}{\mathrm{d}t} = m_0 kv$ | M1A1 | M1 for differentiating $m$ wrt $t$, using $v = \frac{\mathrm{d}x}{\mathrm{d}t}$; A1 for correct expression for $\frac{\mathrm{d}m}{\mathrm{d}t}$ in terms of $v$ |
| $(m+\delta m)(v+\delta v) - mv = -mg\delta t$ | M1 | Second M1 for impulse-momentum principle |
| $\frac{\mathrm{d}v}{\mathrm{d}t} + \frac{v}{m}\frac{\mathrm{d}m}{\mathrm{d}t} = -g$ | A1 | Second A1 for correct diff eqn in $m$, $v$ and $t$ |
| $\frac{\mathrm{d}v}{\mathrm{d}t} + \frac{vm_0 kv}{m_0(1+kx)} = -g$ | M1 | Third M1 for sub for $m$ and $\frac{\mathrm{d}m}{\mathrm{d}t}$ |
| $v\frac{\mathrm{d}v}{\mathrm{d}x} + \frac{kv^2}{(1+kx)} = -g$ | M1 | Fourth M1 for changing $\frac{\mathrm{d}v}{\mathrm{d}t}$ to $v\frac{\mathrm{d}v}{\mathrm{d}x}$ to $\frac{1}{2}\frac{\mathrm{d}(v^2)}{\mathrm{d}x}$ |
| $\frac{\mathrm{d}(v^2)}{\mathrm{d}x} + \frac{2kv^2}{(1+kx)} = -2g$ (printed answer) | A1 | Third A1 for printed answer | 

**Total: 7 marks**

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## Question 6b:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $x=0, v^2=2gh \Rightarrow 2gh = A - \frac{2g}{3k} \Rightarrow A = 2gh + \frac{2g}{3k}$ | M1A1 | M1 for using initial conditions; A1 for correct expression for $A$ |
| $v=0 \Rightarrow \frac{3kA}{2g} = (1+kH)^3$ | M1 | Second M1 for putting $v=0$ |
| $3kh+1 = (1+kH)^3$ | M1 | Third M1 for solving for $H$ in terms of $h$ only |
| $\frac{3h}{7} = H$ | A1 | Second A1 for $3h/7$ |

**Total: 5 marks**

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