# Question 6:

## Part 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| In equilibrium: $mg+4mg\dfrac{l-e}{l}=4mg\dfrac{e}{l}$ | M1, A1, A1 | Need all three forces. Dimensionally correct. Unsimplified with at most one error. Correct unsimplified equation. |
| $5l=8e \Rightarrow e=\dfrac{5l}{8}$, $AE=l+\dfrac{5l}{8}=\dfrac{13l}{8}$ * | A1* | Obtain given answer. Must see $AE=$ |

**ALT1:** $mg+4mg\dfrac{(2l-AE)}{l}=4mg\dfrac{(AE-l)}{l}$ giving $AE=\dfrac{13l}{8}$ *

**ALT2:** $mg+4mg\dfrac{(\frac{l}{2}-e)}{l}=4mg\dfrac{(\frac{l}{2}+e)}{l}$, $e=\dfrac{l}{8}$, $AE=l+\dfrac{l}{2}+\dfrac{l}{8}=\dfrac{13l}{8}$ *

**[4]**

## Part 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion: $4mg\dfrac{\frac{5l}{8}+x}{l}-4mg\dfrac{\frac{3l}{8}-x}{l}-mg=-m\ddot{x}$ | M1, A1, A1 | All terms. Dimensionally correct. Condone use of $a$ for acceleration. Unsimplified with at most one error. Note: $x$ measured vertically down. |
| $\Rightarrow -m\ddot{x}=\dfrac{8mg}{l}x,\ \ddot{x}=-\dfrac{8g}{l}x$ * | A1* | Obtain given answer. Must use $\ddot{x}$ |

**[4]**

## Part 6c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $v^2=\omega^2(a^2-x^2)$ with $a=\dfrac{3l}{8}$ | M1 | Or use of equivalent correct formula |
| $=\dfrac{8g}{l}\left(\dfrac{9}{64}l^2-\dfrac{1}{64}l^2\right)$ | A1 | Correct unsimplified expression for $v$ or $v^2$ |
| $v=\sqrt{gl}$ | A1 | Correct only |

**[3]**

## Part 6d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\dfrac{3l}{8}\cos\omega t$ | B1 | Use of relevant formula with correct amplitude: $x=\frac{3l}{8}\cos\omega t$ or $x=\frac{3l}{8}\sin\omega t$ |
| Use of $-\dfrac{l}{8}=\dfrac{3l}{8}\cos\omega t$; or $\dfrac{l}{8}=\dfrac{3l}{8}\sin\omega t$ and correct use of $\frac{1}{2}\times\frac{2\pi}{\omega}$; or $\dfrac{l}{8}=\dfrac{3l}{8}\cos\omega t$ and $\pi-\cos^{-1}\!\left(\frac{1}{3}\right)$ | M1 | Complete method: $t=\frac{1}{\omega}\cos^{-1}\!\left(-\frac{1}{3}\right)$; or $t=\frac{1}{\omega}\sin^{-1}\!\left(\frac{1}{3}\right)$ with $\frac{1}{2}$ period; or $t=\frac{1}{\omega}\cos^{-1}\!\left(\frac{1}{3}\right)$ with $\pi$ |
| Required time $=\dfrac{2}{\omega}\cos^{-1}\!\left(-\dfrac{1}{3}\right)=\sqrt{\dfrac{l}{2g}}\cos^{-1}\!\left(-\dfrac{1}{3}\right)$ or $\dfrac{\pi}{\omega}+\dfrac{2}{\omega}\sin^{-1}\!\left(\dfrac{1}{3}\right)=\sqrt{\dfrac{l}{8g}}\left(\pi+2\sin^{-1}\!\left(\dfrac{1}{3}\right)\right)$ or $\dfrac{2}{\omega}\left[\pi-\cos^{-1}\!\left(\dfrac{1}{3}\right)\right]=\sqrt{\dfrac{l}{2g}}\left[\pi-\cos^{-1}\!\left(\dfrac{1}{3}\right)\right]$ | A1 | Accept $1.91\sqrt{\frac{l}{2g}}$, $1.35\sqrt{\frac{l}{g}}$, $0.43\sqrt{l}$, $3.82\sqrt{\frac{l}{8g}}$; $\cos^{-1}\!\left(-\frac{1}{3}\right)=1.91...$ |

**[3]**

**(Total: 14)**

## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Conservation of mechanical energy (all terms required) | M1 | All terms required. Dimensionally correct. $\cos\theta = \frac{5}{13}$, $\sin\theta = \frac{12}{13}$ |
| $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg(r + r\cos\theta)$ | A1 | Correct unsimplified equation |
| $v^2 = u^2 - \frac{36}{13}gr$ * | A1* | Obtain given answer from correct working |
| **[3]** | | |

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## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion | M1 | All terms required. Dimensionally correct. Condone sign errors and sin/cos confusion. Condone use of $R=0$ |
| $R + mg\cos\theta = \frac{mv^2}{r}$ | A1 | Correct unsimplified equation. Condone (strict) inequality the right way round |
| Use $R \geq 0$ and solve for $u^2$ | M1 | Complete method to obtain $u^2$. Condone use of $R=0$ or $R>0$ |
| $\frac{mv^2}{r} - mg\cos\theta \geq 0$ * $\Rightarrow u^2 - \frac{36}{13}gr \cdot \frac{5}{13}gr$, $u^2 \geq \frac{41}{13}gr$ | A1* | Obtain given answer from correct working. Must have stated $R \geq 0$. If no reference to $R$, max mark is M1A1M1A0* |
| **[4]** | | |

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## Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $BC = 2r\sin\theta = \frac{24}{13}r$ | B1 | Or equivalent $BC = 1.846\ldots r$ |
| Relevant vertical motion e.g. time to return to level of $BC$ | M1 | Complete method vertically using *suvat* |
| $t = \frac{2v\sin\theta}{g} = \frac{24v}{13g}$ | A1 | Correct unsimplified expression for time. Accept $\frac{24}{13g} \times 4\sqrt{\frac{gr}{13}}$, $\frac{24}{13}\sqrt{\frac{16r}{13g}}$, $\frac{96}{13}\sqrt{\frac{r}{13g}}$, $0.65\sqrt{r}$ |
| Relevant horizontal motion e.g. distance travelled by $P$ | M1 | Complete method horizontally |
| $= (v\cos\theta)t = v^2 \times \frac{120}{169g}$ | A1 | Correct unsimplified expression for distance. $0.87r$, $\frac{1920}{2197}r$, $0.0892gr$ |
| $= \frac{16gr}{13} \times \frac{120}{169g} = \frac{160r}{169} \times \frac{12}{13} < 2r \times \frac{12}{13}$ | A1* | Obtain given conclusion from correct working |

**ALT 1 (for last 3 marks):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal: time $T$ required to travel length $BC$ | M1 | Complete method horizontally |
| $2r\sin\theta = v\cos\theta \times T$; $T = \frac{2r\frac{12}{13}}{4\sqrt{\frac{gr}{13}} \times \frac{5}{13}} = 1.38\sqrt{r}$ | A1 | Correct unsimplified expression for $T$ |
| $t < T$ since $0.654\sqrt{r} < 1.38\sqrt{r}$, hence falls into the bowl * | A1* | Obtain given conclusion from correct working |

**ALT 2 (for last 3 marks):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal: speed $V$ required to reach $C$ | M1 | Complete method horizontally |
| $-V\sin\theta = V\sin\theta - g\cdot\frac{2r\sin\theta}{V\cos\theta}$ $\Rightarrow V = \sqrt{\frac{gr}{\cos\theta}} = \sqrt{\frac{13gr}{5}}$ | A1 | Correct unsimplified expression for $V$ |
| $v < V$ since $\sqrt{\frac{13gr}{5}} < \sqrt{\frac{16gr}{13}}$, hence falls into the bowl * | A1* | Obtain given conclusion from correct working |

| SC: If range formula $\text{Range} = \frac{2v^2\sin\theta\cos\theta}{g}$ is quoted correctly | M1A1M1A1 | |
| **[6]** | | |
| **Total: [13]** | | |