# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $j$-component of $B$ unchanged: $-3\mathbf{j}$ retained | B1 | $j$-component preserved (smooth spheres, line of centres along $\mathbf{i}$) |
| Conservation of momentum in $\mathbf{i}$ direction: $4(4) + 2(-2) = 4v_A + 2v_B$ | M1 | CLM along line of centres |
| $12 = 4v_A + 2v_{Bi}$ | A1 | Correct momentum equation |
| Direction of $B$ changed by $90°$: original $\mathbf{i}$-component $= -2$, so after collision $\mathbf{i}$-component $\perp$ original direction requires $\frac{9}{2}$ | M1 | Using $90°$ condition: $(-2)\cdot v_{Bi} + (-3)(-3) = 0$ giving $v_{Bi} = \frac{9}{2}$ |
| Velocity of $B = (\frac{9}{2}\mathbf{i} - 3\mathbf{j})\ \text{ms}^{-1}$ | A1 | Shown |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| From momentum: $4v_{Ai} = 12 - 2(\frac{9}{2}) = 3$, so $v_{Ai} = \frac{3}{4}$ | M1 A1 | Finding velocity of $A$ after collision |
| $e = \frac{v_{Bi} - v_{Ai}}{u_{Ai} - u_{Bi}} = \frac{\frac{9}{2} - \frac{3}{4}}{4-(-2)}$ | M1 A1 | Correct Newton's law of restitution |
| $e = \frac{\frac{15}{4}}{6} = \frac{15}{24} = \frac{5}{8}$ | A1 | Correct value |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse on $B$ = $m_B(v_B - u_B) = 2\left[(\frac{9}{2}\mathbf{i}-3\mathbf{j})-(-2\mathbf{i}-3\mathbf{j})\right]$ | M1 | Impulse = change in momentum |
| $= 2(\frac{13}{2}\mathbf{i}) = 13\mathbf{i}$ | A1 | Correct value |
| Units: $\text{N s}$ | B1 | Correct units stated |

I can see these are answer space pages (pages 17-20) for Questions 6 and 7 of what appears to be an AQA mechanics exam (P59194/Jun13/MM03). However, these pages only contain **blank answer spaces** for students to write in — they do not contain any mark scheme content.

To get the mark scheme for this paper, you would need to:

1. Search the **AQA website** (aqa.org.uk) for the mark scheme for **P59194/Jun13/MM03**
2. Search for **"AQA MM03 June 2013 Mark Scheme"**
3. Check revision resource sites that host past paper mark schemes

Would you like me to help you **work through the solutions** to Question 7 (the aircraft/helicopter interception problem) shown on page 18? I can solve each part showing full working:

- **(a)(i)** Finding the bearing for $v_A = 200$ to intercept $H$
- **(a)(ii)** Finding the time to intercept
- **(b)** Finding the bearing for closest approach when $v_A = 150$

Just let me know and I'll work through the mathematics in full.