| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question requiring resolution of forces in perpendicular directions and Newton's second law. Part (a) uses the condition that perpendicular components cancel (ship moves along centre line), giving a simple equation to solve for T. Part (b) applies F=ma with the net force along the centre line. The steps are routine and methodical with no conceptual challenges beyond standard M1 techniques. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Perpendicular to centre line, forces balance: \(100000\sin 25° = T\sin 20°\) | M1 | Resolving perpendicular to centre line |
| \(T = \frac{100000\sin 25°}{sin 20°}\) | A1 | Correct equation |
| \(T = 123000 \text{ N}\) (123 270...) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Net force along centre line \(= 100000\cos 25° + T\cos 20° - 20000\) | M1 | Resolving along centre line |
| \(= 100000\cos 25° + 123270\cos 20° - 20000\) | A1 | Correct substitution of \(T\) |
| Net force \(= 186000 \text{ N}\) (approx) | A1 | Correct net force |
| \(a = \frac{F}{m} = \frac{186000}{500000}\) | M1 | Use of \(F = ma\), mass = 500 000 kg |
| \(a = 0.373 \text{ ms}^{-2}\) | A1 | Correct answer |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Perpendicular to centre line, forces balance: $100000\sin 25° = T\sin 20°$ | M1 | Resolving perpendicular to centre line |
| $T = \frac{100000\sin 25°}{sin 20°}$ | A1 | Correct equation |
| $T = 123000 \text{ N}$ (123 270...) | A1 | Correct value |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Net force along centre line $= 100000\cos 25° + T\cos 20° - 20000$ | M1 | Resolving along centre line |
| $= 100000\cos 25° + 123270\cos 20° - 20000$ | A1 | Correct substitution of $T$ |
| Net force $= 186000 \text{ N}$ (approx) | A1 | Correct net force |
| $a = \frac{F}{m} = \frac{186000}{500000}$ | M1 | Use of $F = ma$, mass = 500 000 kg |
| $a = 0.373 \text{ ms}^{-2}$ | A1 | Correct answer |
I can see these are exam answer space pages (pages 7-11) from what appears to be a Mechanics paper (P/Jun15/MM1B). The pages shown are **answer space pages only** - they don't contain mark scheme content. They show:
- Answer space for Question 3 (page 7)
- Question 4 with parts (a), (b), (c) and answer space (pages 8-9)
- Question 5 with parts (a)-(e) and answer space (pages 10-11)
The **question text** is visible for Q4 and Q5, but **no mark scheme** is present in these images. Mark schemes are separate documents.
Here is what I can extract from the **question text** to help you structure solutions:
---3 A ship has a mass of 500 tonnes. Two tugs are used to pull the ship using cables that are horizontal. One tug exerts a force of 100000 N at an angle of $25 ^ { \circ }$ to the centre line of the ship. The other tug exerts a force of $T \mathrm {~N}$ at an angle of $20 ^ { \circ }$ to the centre line of the ship. The diagram shows the ship and forces as viewed from above.\\
\includegraphics[max width=\textwidth, alt={}, center]{01338c87-302c-420f-a473-39b0796ccaed-06_279_844_539_664}
The ship accelerates in a straight line along its centre line.
\begin{enumerate}[label=(\alph*)]
\item $\quad$ Find $T$.
\item A resistance force of magnitude 20000 N directly opposes the motion of the ship. Find the acceleration of the ship.\\[0pt]
[4 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{01338c87-302c-420f-a473-39b0796ccaed-06_1419_1714_1288_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2015 Q3 [7]}}