| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Vehicle on slope with resistance |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question involving resolving forces on a slope at constant velocity (equilibrium), then applying Newton's second law with a different angle. Part (b) is essentially given (just verify P = mg sin 4°), part (c)(i) is a standard F=ma calculation, and part (c)(ii) requires a simple qualitative comment about resistance varying with speed. Below average difficulty due to minimal problem-solving required and standard bookwork application. |
| Spec | 3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03m Equilibrium: sum of resolved forces = 03.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Correct diagram with arrows for \(P\), \(R\), \(mg\) | B1 | Total: 1 — Must not use \(F\) instead of \(P\). Condone resistance instead of \(P\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(P = 100 \times 9.8 \sin 4°\) | M1 | Resolving weight (must see 100) |
| M1 | Using \(\sin 4°\) or \(\cos 86°\) | |
| \(= 68.4\) | A1 | Total: 3 — AG Correct \(P\) from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(100a = 100 \times 9.8\sin 5° - 100 \times 9.8\sin 4°\) | M1 | Three term equation of motion |
| A1 | Weight resolved correctly | |
| A1 | Correct equation | |
| \(a = \frac{100 \times 9.8\sin 5° - 100 \times 9.8\sin 4°}{100} = 0.171\) | A1 | Total: 4 — Correct \(a\). (Accept 0.170 or 0.17) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| You would expect \(P\) to vary with the speed of the car | B1 | Total: 1 — Correct explanation |
## Question 6:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Correct diagram with arrows for $P$, $R$, $mg$ | B1 | **Total: 1** — Must not use $F$ instead of $P$. Condone resistance instead of $P$ |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $P = 100 \times 9.8 \sin 4°$ | M1 | Resolving weight (must see 100) |
| | M1 | Using $\sin 4°$ or $\cos 86°$ |
| $= 68.4$ | A1 | **Total: 3** — **AG** Correct $P$ from correct working |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $100a = 100 \times 9.8\sin 5° - 100 \times 9.8\sin 4°$ | M1 | Three term equation of motion |
| | A1 | Weight resolved correctly |
| | A1 | Correct equation |
| $a = \frac{100 \times 9.8\sin 5° - 100 \times 9.8\sin 4°}{100} = 0.171$ | A1 | **Total: 4** — Correct $a$. (Accept 0.170 or 0.17) |
### Part (d)
| Working | Marks | Guidance |
|---------|-------|----------|
| You would expect $P$ to vary with the speed of the car | B1 | **Total: 1** — Correct explanation |6 A trolley, of mass 100 kg , rolls at a constant speed along a straight line down a slope inclined at an angle of $4 ^ { \circ }$ to the horizontal.
Assume that a constant resistance force, of magnitude $P$ newtons, acts on the trolley as it moves. Model the trolley as a particle.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show the forces acting on the trolley.
\item Show that $P = 68.4 \mathrm {~N}$, correct to three significant figures.
\item \begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the trolley if it rolls down a slope inclined at $5 ^ { \circ }$ to the horizontal and experiences the same constant force of magnitude $P$ that you found in part (b).
\item Make one criticism of the assumption that the resistance force on the trolley is constant.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q6 [9]}}