# Question 3:

## Part 3a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | Dimensionally correct equation. Condone sine/cosine confusion for their angle |
| $R\sin\theta = mg$ | A1 | Correct unsimplified equation |
| Horizontal equation of motion | M1 | Dimensionally correct equation. Condone sine/cosine confusion for their angle. Accept any correct form for the acceleration. |
| $R\cos\theta = ma\,(= mr\omega^2)$ | A1 | Correct unsimplified equation. Accept any correct form for the acceleration. |
| Solve for $a$ | DM1 | Eliminate $R$ and $\theta$ to solve for $a$. Dependent on both previous M marks. |
| $\frac{g}{a} = \tan\theta \Rightarrow a = \frac{4}{3}g$ | A1 | Correct only (or exact equivalent) |

## Part 3b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $8d\omega^2 = \frac{4}{3}g \Rightarrow \omega = \sqrt{\frac{g}{6d}}$ | M1 | Use $a = r\omega^2$ or $a = \frac{v^2}{r}$ to obtain $\omega$ or $v$ $\left(v = \sqrt{\frac{32dg}{3}}\right)$ |
| Use of $T = \frac{2\pi}{\omega}$ | M1 | Complete method to find $T$ |
| $T = 2\pi\sqrt{\frac{6d}{g}}$ oe OR $15\sqrt{\frac{d}{g}}$ or better | A1 | $15\sqrt{\frac{d}{g}}$ or better — must be in terms of $d$ and $g$. Accept $T = \frac{2\pi}{\sqrt{\frac{g}{6d}}}$ |

## Question 4a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about the $y$-axis: $\int(\rho)xy\,dx$ | M1 | Correct method for moments about the $x$-axis: $\int(\rho)xy\,dx$ or $\int(\rho)\frac{1}{2}x^2\,dy$. Integrand in one variable only. |
| $=(\rho)\int x\sqrt{36-\frac{9x^2}{4}}\,dx = k\left(36-\frac{9x^2}{4}\right)^{\frac{3}{2}}$ | M1 | Integrate to obtain $k\left(A-Bx^2\right)^{\frac{3}{2}}$. Ignore limits and/or constant of integration. |
| $=\left[-\frac{4}{27}\left(36-\frac{9x^2}{4}\right)^{\frac{3}{2}}(\rho)\right]_0^4 \;(=32(\rho))$ | A1 | Correct integration with correct limits seen or implied. |
| $\bar{x}=\frac{\int xy\,dx}{6\pi}$ | DM1 | Complete method to obtain $\bar{x}$. Dependent on **first** M1. |
| $\bar{x}=\frac{32}{6\pi}=\frac{16}{3\pi}$ * | A1* | Obtain given answer from correct working. |

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## Question 4b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about the $x$-axis: $\int\frac{1}{2}y^2(\rho)\,dx\left(=\frac{1}{2}(\rho)\int 36-\frac{9x^2}{4}\,dx\right)$ | M1 | Correct method for moments about the $y$-axis: $\int(\rho)xy\,dy$ or $\int(\rho)\frac{1}{2}y^2\,dx$. |
| $=\frac{1}{2}(\rho)\left[36x-\frac{3}{4}x^3\right]_0^4\left(=\frac{1}{2}(\rho)(144-48)=48(\rho)\right)$ | A1 | Correct integration with correct limits seen or implied. |
| $\bar{y}=\frac{\int\frac{1}{2}y^2\,dx}{6\pi}$ | DM1 | Complete method to obtain $\bar{y}$. Dependent on previous M1. |
| $=\frac{48}{6\pi}\left(=\frac{8}{\pi}\right)$ | A1 | Correct exact equivalent. |

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## Question 4c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct use of trigonometry | M1 | Correct use of trigonometry to find a relevant angle. |
| $\tan\theta^\circ=\frac{\text{their }\bar{y}}{4-\frac{16}{3\pi}}\left(=\frac{6}{3\pi-4}\right)$ | A1ft | Correct unsimplified expression for $\tan\theta$ or its reciprocal. Follow their $\bar{y}$. |
| $\theta=47.9$ (48 or better) | A1 | 48 or better (47.8823...). 0.836 radians is A0. |

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## Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a\omega^2=18$ | B1 | Use the model to state correct equation for greatest acceleration. Accept $\pm18$. |
| Use $v^2=\omega^2(a^2-x^2)$ | M1 | Use the model to form a second equation in $a$ and $\omega$. |
| $2.4^2=\omega^2(a^2-0.3^2)$ | A1 | Correct unsimplified equation with $x$ and $v$ substituted. |
| Form equation in $a$ only and solve for $a$, e.g. $\frac{a}{a^2-0.09}=\frac{18\times25}{144}$ $(18a^2-5.76a-1.62=0)$ | M1 | Solve the simultaneous equations to obtain $a$. |
| $\Rightarrow a=0.5$ * | A1* | Correct only from **correct** working. Must have used positive $\omega^2$. |

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## Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Solve for $\omega$ and use max speed $=a\omega$ | M1 | Complete method using the model to obtain greatest speed. E.g. $v=\sqrt{\omega^2(0.5^2-0^2)}$ |
| Greatest speed $=0.5\times6=3\,(\text{ms}^{-1})$ | A1 | Correct only. |

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## Question 5c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|v\|=a\omega\sin\omega t$ **OR** $v^2=\omega^2(a^2-x^2)$ and $x=a\sin\omega t$ | B1ft | Set up a correct model to find time when speed is 2. ft their $\omega$. (If starting from $O$ then $\|v\|=a\omega\cos\omega t$) |
| $\Rightarrow2=3\sin6t\;(t=0.1216...)$ **OR** $\frac{\sqrt{5}}{6}=0.5\sin6t\;(t=0.14017...)$ | M1 | Solves their equation(s) to obtain a critical value for $t$. (condone degrees: 41.8°/48.2°) |
| $S=\frac{2\pi}{6}-4t$ **OR** $S=4\times0.14017...$ | M1 | Correct method to obtain $S$; must be in radians. |
| $=0.5607...$ | A1 | 0.56 or better. |

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## Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Mass}=\int_0^6\pi\cdot2^2\lambda(x+2)\,dx$ | M1 | Correct method for total mass. |
| $=4\pi\lambda\left[\frac{x^2}{2}+2x\right]_0^6$ | A1 | Correct integration with limits seen or implied. |
| $=4\pi\lambda\left(\frac{36}{2}+12\right)=120\lambda\pi\,(\text{kg})$ * | A1* | Obtain given answer from correct working. |

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## Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment about $y$-axis $=\int_0^6 4\pi x\lambda(x+2)\,dx$ | M1 | Correct method for moments about $y$-axis (condone missing $\pi$ and $\lambda$). |
| $=4\pi\lambda\left[\frac{x^3}{3}+x^2\right]_0^6\left(=4\lambda\pi(72+36)=432\lambda\pi\right)$ | A1 | Correct integration with correct limits. |
| Distance from $O=\frac{\text{their }432\lambda\pi}{120\lambda\pi}$ | DM1 | |
| $=\frac{432}{120}=3.6\,(\text{cm})$ * | A1* | Obtain given answer from correct working. |

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## Question 6c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\frac{3}{8}\times3$ | B1 | |
| Moments about a diameter of the base | M1 | |
| $120\lambda\pi\times3.6+\left(6+\frac{3}{8}\times3\right)\times\frac{2}{3}\pi(3)^3\lambda=(120+18)\pi\lambda d$ | A1, A1 | |
| $\left(d=\frac{747}{184}=4.0597...\right)$ | | |
| $\tan\alpha^\circ=\frac{2}{\text{their }d}$ | M1 | |
| $\alpha=26.2$ | A1 | |