# Question 3:

## Part 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Large disc: $\pi R^2$, distance 0; Small disc: $\pi r^2$, distance $R-r$; Template: $\pi R^2 - \pi r^2$, distance $\pm kr$ | B1 | Correct area ratios and distances seen or implied. Allow $+kr$ or $-kr$ |
| Moments about axis through $X$ | M1 | Or moments about a parallel axis. Need all terms but condone sign errors. Do not need to see the zero term. Dimensionally consistent. Could be part of a vector equation |
| $(0 \times \pi R^2) - \pi r^2 \times (R-r) = (\pi R^2 - \pi r^2) \times (-kr)$ Moments about the left-hand end of the diameter through $X$ and $Y$ gives: $\pi R^2 \cdot R - \pi r^2(2R - r) = \pi(R^2 - r^2)(R - kr)$ | A1 | Correct unsimplified equation. Do not need to see the zero term. Must be using $-kr$ (unless they have changed the sign on the left-hand side) |
| $r = \frac{k}{1-k}R$ | A1* | Obtain **given answer** from correct working e.g. via $\frac{r}{R+r} = k$. If they use $\bar{x}$ in place of $\pm kr$ and never substitute $\pm kr$ they can score B0M1A0A0 |

## Part 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 < \frac{k}{1-k}R < R$ | M1 | Use of correct inequality |
| $(0 <) k < (1-k) \Rightarrow (0 <) k < \frac{1}{2}$ | A1 | Correct only. Only need the right-hand value. A0 with an incorrect left hand value |

## Part 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = \frac{4}{9} \Rightarrow r = \frac{4}{5}R$ | B1 | Seen or implied (this mark could be implied by the correct expression for $\tan\alpha$ in terms of $k$) |
| $\tan\alpha = \frac{R}{kr} \left(= \frac{R}{\frac{4}{9} \times \frac{4}{5}R} = \frac{45}{16}\right)$ $\left(\tan\alpha = \frac{1-k}{k^2}\right)$ | M1 | Correct use of trig in a correct triangle. Available for finding $90 - \alpha$ |
| $\alpha = 70°$ | A1 | Or better (70.426…). Accept 109.6, 250.4 and 289.6 |

## Part 3(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about an axis through $P$ | M1 | Dimensionally consistent, condone sign errors and missing $g$ throughout. The equation should be of the form $M_1gR = Mg \times$ a distance (in $r$ or $R$). Moments about any other axis requires use of the forces acting at $P$ |
| M(P), $M_1gR = Mg \times \frac{4}{9}r$ or $M_1gR = Mg \times \frac{16}{45}R$ | A1 | Correct unsimplified equation in $r$ and/or $R$ |
| $M_1 = \frac{16}{45}M$ | A1 | Accept $0.36\, M$ or better |

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