## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Slant height $= 5a$ | B1 | Slant height $= 5a$ seen anywhere (could be on diagram) |
| Correct masses: Hemisphere $= 2\pi(4a)^2 k\lambda$, Cone $= \pi \times 4a \times 5a \times \lambda$, Total $= 32\pi a^2 k\lambda + 20\pi a^2\lambda$ | B1 | Correct masses/mass ratio for hemisphere, cone and combined shape |
| Mass ratios: Hemisphere $= 8k$, Cone $= 5$, Total $= 8k+5$ | B1 | Correct distances seen. If using Alts 1 or 3 minus signs not needed here |
| Distances from vertex: Hemisphere $= 5a$, Cone $= 2a$, Total $= 4a$ | B1 | |
| $40ka + 10a = 4a(8k+5)$ | M1A1 | M1: Moments equation attempted with all 3 terms (2 terms if about $G$). Condone inconsistent mass dimensions. A1: Correct moments equation, all signs correct |
| $k = 1.25$ | A1 | Correct value for $k$: $1.25$, $\frac{5}{4}$, $1\frac{1}{4}$ or any equivalent fraction |

**ALT 1** - Moments about centre of mass $G$:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distances: $(-) a$, $2a$, $0$ | B1 | Accept $a$ provided minus appears in equation |
| $-8ka + 10a = 0$ | M1A1 | |

**ALT 2** - Moments about lowest point of hemisphere:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distances: $2a$, $5a$, $3a$ | B1 | |
| $16ka - 5a = a(8k+5)$ | M1A1 | |

**ALT 3** - Moments about centre of circular base:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distances: $2a$, $(-)a$, $a$ | B1 | |
| $16ka + 25a = 3a(8k+5)$ | M1A1 | |

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## Question 2a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ | | |
| $T\cos\theta = mg$ | M1A1 | M1: Resolving vertically. $T\cos\theta$ or $T\sin\theta$ accepted. A1: Correct equation |
| $\frac{4}{5}T = mg \rightarrow T = \frac{5mg}{4}$ | A1 | Correct tension |

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## Question 2b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = a + \frac{5a}{4} \times \frac{3}{5} = \frac{7a}{4}$ | B1 | Correct radius of motion seen explicitly or used in N2L |
| $T\sin\theta = m\omega^2\left(\frac{7a}{4}\right)$ | M1A1A1 | M1: Attempt at horizontal equation of motion. Allow either form of acceleration. $T\cos\theta$ or $T\sin\theta$ accepted. Allow if $r = \frac{3a}{4}$ used. A1: Correct LHS. A1: Correct RHS, acceleration must be in $r\omega^2$ form |
| $\frac{3}{5}\left(\frac{5mag}{4}\right) = \frac{7}{4}ma\omega^2$ | DM1 | Substitute trig and eliminate $T$ to find $\omega$ or $\omega^2$. Depends on first M mark in (b) |
| $\omega = \sqrt{\dfrac{3g}{7a}}$ | A1 | Correct value for $\omega$ (square root sign must cover all terms) |

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## Question 3a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^9$ | B1 | Identifying correct limits |
| $V = (\pi)\int_0^9\left(3-\sqrt{x}\right)^2 dx$ | M1 | Attempt at $(\pi)\int_0^9(3-\sqrt{x})^2\,dx$. $\pi$ not needed. Limits may be missing. Minimum accepted for squaring: $9 \pm k\sqrt{x} \pm x$. At least one term integrated (power increased) |
| $V = (\pi)\int_0^9\left(9 - 6\sqrt{x} + x\right)dx$ | | |
| $V = (\pi)\left[9x - 4x^{\frac{3}{2}} + \frac{x^2}{2}\right]_0^9$ | A1 | Correct integration. $\pi$ not needed but correct limits now needed |
| $V = (\pi)\left[81 - 108 + \frac{81}{2}\right]\ (-\pi[0])$ | | |
| $V = \dfrac{27}{2}\pi$ | A1* | **Given** result from fully correct working. $\pi$ must not just appear on final line without justification |

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## Question 3b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\pi)\int xy^2\,dx = (\pi)\int\left(9x - 6x^{\frac{3}{2}} + x^2\right)dx$ | M1 | $\pi$ not needed |
| $= (\pi)\left[\frac{9}{2}x^2 - \frac{12}{5}x^{\frac{5}{2}} + \frac{x^3}{3}\right]_0^9$ | A1 | Correct integration. $\pi$ not needed |
| $= (\pi)\left(\left[\frac{729}{2} - \frac{2916}{5} + \frac{729}{3}\right] - [0]\right)$ | A1 | Correct result from substitution of correct upper limit. Lower limit is 0 |
| $= \dfrac{243}{10}(\pi)$ | | |
| $\bar{x} = \dfrac{(\pi)\int xy^2\,dx}{(\pi)\int y^2\,dx}$ | DM1 | Use of $\bar{x} = \frac{(\pi)\int xy^2\,dx}{(\pi)\int y^2\,dx}$. $\pi$ must appear in both or neither. Depends on both previous M marks |
| $\bar{x} = \dfrac{\left(\dfrac{243}{10}\right)}{\left(\dfrac{27}{2}\right)} = 1.8$ | A1 | Accept any exact equivalent e.g. $\frac{18}{10}$, $\frac{9}{5}$, $1\frac{4}{5}$ etc |