# Question 1:

## Part 1(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The rods are uniform and the axes of symmetry intersect at midpoint of $AC$ | B1 | Any equivalent clear justification. Needs to mention uniformity and symmetry and the midpoint of $AC$ |

**Total: (1)**

## Part 1(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use moments e.g. $M(A): (2aW + aW + 3aW = 4aT_B + aW)$ | M1 | Form ANY moments equation. Require all terms. Dimensionally correct. Condone sign errors. |
| e.g. $M(A): 5W \cdot 2a\cos 60° = 4aT_B$ or $M(B): 3a \times 5W = 4aT_A$ | A1 | Correct unsimplified (including trig) equation e.g. $M(G): T_A \cdot 2a\cos 60° = T_B(4a - 2a\cos 60°)$ or $T_B(4a\cos^2 30°)$ |
| Resolving vertically: $T_A + T_B = 5W$ | M1 | Form a second equation in $T_A$ and/or $T_B$, solve for $T_A$ and $T_B$ |
| $\Rightarrow T_A = \dfrac{15W}{4}, \quad T_B = \dfrac{5W}{4}$ | A1 | Both tensions correct. If answers reversed, allow M marks. |

**Total: (4)**

## Part 1(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_A$ will be the larger, so first to exceed $6W$, use $T_A = 6W$ (e.g. by $M(B)$ but may use two equations) to form equation in $k$ only | M1 | Realise first to break is rope at $A$, complete method using $T_A = 6W$. Require all terms. Dimensionally correct. Condone sign errors. M0 if $T_B = 6W$ used (gives $k = 9.5$) |
| $6W \times 4a = 5W \times 3a + kW \times 6a + 2kW \times 2a$ | A1 | Unsimplified equation or inequality in $k$ only with at most one error |
| $24aW = 15aW + 10kaW$ | A1 | Correct unsimplified equation or inequality in $k$ only |
| $k = 0.9$ | A1 | Correct only. Decimal or fraction. |

**Total: (4)**

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