## Question 5:

**Moments about vertical axis (AF):**

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{Mg}{2}\times\frac{1}{2}a+\frac{Mg}{2}\times 1.5a+3akMg=Mg(1+k)\bar{x}$ | M1 | Requires all terms, dimensionally correct but condone $g$ missing |
| Above equation correct | A2 | -1 each error; accept with $M$ and/or $g$ not seen |
| $\bar{x}=\frac{1+3k}{1+k}a$ | — | — |

**Moments about horizontal axis (AB or FE):**

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{Mg}{2}\times 1.5a+\frac{Mg}{2}\times 3.5a+4akMg=Mg(1+k)\bar{y}$ | M1 | Requires all terms, dimensionally correct but condone $g$ missing |
| Above equation correct | A2 | -1 each error; accept with $M$ and/or $g$ not seen; do not penalise repeated errors |
| $\bar{y}=\frac{2.5+4k}{1+k}a$ | — | — |

*Note: Working with axes through F gives $\bar{x}=\frac{1+3k}{1+k}a$ and $\bar{y}=\frac{1.5}{1+k}a$*

*SR: A candidate working with a mixture of mass and mass ratio can score 4/6: M1A0A0M1A2*

**Finding $\tan\theta$:**

| Working | Mark | Guidance |
|---------|------|----------|
| Use of $\tan\theta$ with their distances from $AF$ and $AB$ | M1 | Must be considering whole system; allow for inverted ratio |
| $\tan\theta=\frac{M+3kM}{2.5M+4kM}\left(=\frac{4}{7}\right)$ | A1 | Or exact equivalent |

**Solving for k:**

| Working | Mark | Guidance |
|---------|------|----------|
| Equate their $\tan\theta$ to $\frac{4}{7}$ and solve for $k$ | M1 | — |
| $7M+21kM=10M+16kM$ | — | — |
| $k=\frac{3}{5}$ | A1 | cso |
| | **(10)** | |

**Alternative (L-shape first):**

| Working | Mark | Guidance |
|---------|------|----------|
| $\bar{x}=a$ and $\bar{y}=\frac{5}{2}a$ or $\frac{3}{2}a$ | M1A2 | M1 requires all terms, dimensionally correct, condone $g/M$ missing; A1 for each correct |
| Combine with the particle | M1A2 | M1 requires all terms, dimensionally correct, condone $g$ missing; A1 for each correct |

**Geometrical approach:**

| Working | Mark | Guidance |
|---------|------|----------|
| Centre of mass at $\left(a,\frac{3}{2}a\right)$ relative to $F$ | M1A2 | — |
| Use of $\tan\theta$ to find $d_1$ or $d_2$ | M1 | — |
| $\left(\frac{5}{2}a\right)\tan\theta - a = \frac{3}{7}a$ | A1 | Correct for their centre of mass |
| $d_1=\left(\frac{3}{7}a\right)\cos\theta$ | A1 | Correct for their centre of mass |
| $3a-4a\tan\theta=\frac{5}{7}a$ | M1 | Use of $\tan\theta$ to find second distance |
| $d_2=\frac{5}{7}a\cos\theta$ | A1 | — |
| Moments about $A$: $Md_1=kMd_2$ | M1 | — |
| $\frac{3}{7}a\cos\theta=k\times\frac{5}{7}a\cos\theta\Rightarrow k=\frac{3}{5}$ | A1 | |
| | **(10)** | |

---