## Question 1:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Moments about $x$-axis (or parallel axis) | M1 | Require all terms. Dimensionally correct. Condone sign errors |
| $3m \times 8 + m \times 0 + 2m \times -2 = 6m \times 2k$ | A1 | Correct unsimplified equation |
| Moments about $y$-axis (or parallel axis) | M1 | Require all terms. Dimensionally correct. Condone sign errors |
| $3m \times a + m \times -4 + 2m \times 5 = 6m \times k$ | A1 | Correct unsimplified equation |
| Eliminate $k$ and solve for $a$: $6a + 12 = 20$, $a = \frac{4}{3}$ | A1 | Or equivalent. 1.3 or better |
| **Total** | **(5)** | |

**1alt (Vector method):**

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Form vector equation in $k$ and $a$ | M1 | Require all terms. Dimensionally correct. Condone sign errors |
| $3m\binom{a}{8} + m\binom{-4}{0} + 2m\binom{5}{-2} = 6m\binom{k}{2k}$ | A1 | Or equivalent. Correct unsimplified equation |
| Use components from dimensionally correct equation to form two separate equations | M1 | Seen or implied. Condone processing errors |
| $\Rightarrow 3a + 6 = 6k$, $\ 20 = 12k$ | A1 | Pair of correct unsimplified equations |
| Eliminate $k$ and solve for $a$: $6a + 12 = 20$, $a = \frac{4}{3}$ | A1 | 1.3 or better |
| **Total** | **(5)** | |

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## Question 2a:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Use of $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ | M1 | Accept $\pm m(\mathbf{v} - \mathbf{u})$ |
| $\frac{3}{4}\mathbf{v} = \frac{3}{4}(4\mathbf{i} + \mathbf{j}) + (-6\mathbf{i} + 4\mathbf{j})\left(= -3\mathbf{i} + \frac{19}{4}\mathbf{j}\right)$ | A1 | Correct unsimplified equation |
| $\mathbf{v} = \frac{4}{3}\left(-3\mathbf{i} + \frac{19}{4}\mathbf{j}\right) = -4\mathbf{i} + \frac{19}{3}\mathbf{j}$ (m s$^{-1}$) | A1 | $-4\mathbf{i} + 6.3\mathbf{j}$ (m s$^{-1}$) or better. ISW. Accept as a column vector |
| **Total** | **(3)** | |

## Question 2b:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Change in direction | M1 | Use **u** and their **v** to find a relevant angle between the two velocities (e.g. not just $\tan^{-1}\frac{1}{4}$) |
| $= 180° - \tan^{-1}\frac{1}{4} - \tan^{-1}\frac{19}{12}$ or $\tan^{-1}4 + \tan^{-1}\frac{12}{19}$ | A1ft | Correct unsimplified. Follow their **v** $(180° - 14.036...° - 57.724...°)$ |
| $= 108°\ (108.2°)\ (1.89 \text{ radians})$ | A1 | $108.2°$ or better. Accept $252°$ |
| **Total** | **(3)** | |

**2b alt (Scalar product):**

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Using scalar product | M1 | Using vectors **u** and their **v** or equivalent |
| $\cos\theta = \dfrac{-16 + \frac{19}{3}}{\sqrt{17}\sqrt{16 + \left(\frac{19}{3}\right)^2}}$ | A1ft | Follow their **v** |
| $\theta = 108°\ (108.2°)$ | A1 | (1.89 radians) Accept $252°$ |
| **Total** | **(3)** | |

**2b alt (Cosine rule):**

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Use cosine rule in triangle with sides $\sqrt{17}$, $\dfrac{\sqrt{505}}{3}$, $\dfrac{8\sqrt{13}}{3}$ | M1 | Or equivalent |
| $\cos\theta = \dfrac{17 + \frac{505}{9} - \frac{64 \times 13}{9}}{2\sqrt{17}\cdot\frac{\sqrt{505}}{3}}\ (= -0.313)$ | A1 | |
| $\theta = 108°\ (108.2°)$ | A1 | (1.89 radians) Accept $252°$ |
| **Total** | **(3)** | **[6]** |

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