# Question 1:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Impulse-momentum equation | M1 | Dimensionally correct. Must be subtracting, but condone subtracting in the wrong order. |
| $\mathbf{J} = 0.5(-\mathbf{i}+6\mathbf{j}-4\mathbf{i}-3\mathbf{j})$ giving $\mathbf{J} = 0.5(-5\mathbf{i}+3\mathbf{j})$ | A1 | Correct unsimplified equation |
| Find magnitude of $\mathbf{J}$: | M1 | Correct application of Pythagoras to find the magnitude (from $\pm\mathbf{J}$) |
| $|\mathbf{J}|^2 = \frac{1}{4}(25+9)$, $\quad |\mathbf{J}| = \frac{\sqrt{34}}{2}$ (N s) | A1 | 2.9 or better (2.9154…) (from $\pm\mathbf{J}$) |
| **(4 marks)** | | |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct use of trig | M1 | Correct use of trig to find a relevant angle using $4\mathbf{i}+3\mathbf{j}$ and their $\mathbf{J}$; i.e. $\alpha°$ or $180°-\alpha°$. Allow $\left|\frac{\mathbf{a.b}}{|\mathbf{a}||\mathbf{b}|}\right|$ |
| $\alpha° = 180° - \tan^{-1}\frac{3}{4} - \tan^{-1}\frac{3}{5}$ or $\alpha° = \tan^{-1}\frac{4}{3} + \tan^{-1}\frac{5}{3}$ | A1ft | Correct unsimplified expression for the required angle. Follow their $\mathbf{J}$. A0 for $\left|\frac{\mathbf{a.b}}{|\mathbf{a}||\mathbf{b}|}\right|$. Do not ISW |
| $\alpha = 112$ | A1 | 110 or better (112.166…) or accept 247.8…° |
| **(3 marks)** | | |

## Part (b) — Alternate Method 1 (scalar product):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use scalar product of $\mu\mathbf{J}$ and $4\mathbf{i}+3\mathbf{j}$ to find the angle | M1 | |
| $\cos\alpha° = \frac{-20+9}{\sqrt{34}\times 5}$ | A1ft | |
| $\alpha = 112$ | A1 | |
| **(3 marks)** | | |

## Part (b) — Alternate Method 2 (cosine rule):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of cosine rule in triangle of momenta or equivalent | M1 | |
| $\alpha° = 180° - \cos^{-1}\left(\frac{34+25-37}{2\times5\times\sqrt{34}}\right)$ | A1ft | |
| $\alpha = 112$ | A1 | |
| **(3 marks)** | | |

**Total: 7 marks**