| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – accelerating (finding acceleration or applied force) |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics problem involving resolving forces, friction, and Newton's second law. Part (a) uses equilibrium conditions with straightforward vertical/horizontal resolution, while part (b) applies F=ma with known friction coefficient. The calculations are routine with no conceptual surprises, making it slightly easier than average for M1 content. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolving vertically: \(R + 60\sin 30° = 20g\) | M1 | Correct vertical resolution |
| \(R = 20(9.8) - 60(0.5)\) | A1 | Correct equation |
| \(R = 196 - 30 = 166\) N | A1 | Correct answer shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Constant speed \(\Rightarrow\) friction \(=\) horizontal component of tension | M1 | Correct equilibrium condition |
| \(F = 60\cos 30°\) | A1 | Correct expression |
| \(F = \mu R\): \(\mu = \frac{60\cos 30°}{166}\) | M1 | Use of \(F=\mu R\) |
| \(\mu = \frac{51.96...}{166} = 0.313\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| New normal reaction: \(R' = 20g - T\sin 30°\) | M1 | Resolving vertically with new \(T\) |
| Friction: \(F' = \mu R' = 0.313(196 - 0.5T)\) | A1 | Correct friction expression |
| Resolving horizontally: \(T\cos 30° - F' = 20(0.8)\) | M1 | Newton's second law horizontally |
| \(T\cos 30° - 0.313(196-0.5T) = 16\) | A1 | Correct equation |
| \(0.866T - 61.3 + 0.1565T = 16\) | DM1 | Solving for \(T\) |
| \(T = 75.5\) N | A1 | Correct answer |
# Question 8:
## Part (a)(i)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolving vertically: $R + 60\sin 30° = 20g$ | M1 | Correct vertical resolution |
| $R = 20(9.8) - 60(0.5)$ | A1 | Correct equation |
| $R = 196 - 30 = 166$ N | A1 | Correct answer shown |
## Part (a)(ii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Constant speed $\Rightarrow$ friction $=$ horizontal component of tension | M1 | Correct equilibrium condition |
| $F = 60\cos 30°$ | A1 | Correct expression |
| $F = \mu R$: $\mu = \frac{60\cos 30°}{166}$ | M1 | Use of $F=\mu R$ |
| $\mu = \frac{51.96...}{166} = 0.313$ | A1 | Correct value |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| New normal reaction: $R' = 20g - T\sin 30°$ | M1 | Resolving vertically with new $T$ |
| Friction: $F' = \mu R' = 0.313(196 - 0.5T)$ | A1 | Correct friction expression |
| Resolving horizontally: $T\cos 30° - F' = 20(0.8)$ | M1 | Newton's second law horizontally |
| $T\cos 30° - 0.313(196-0.5T) = 16$ | A1 | Correct equation |
| $0.866T - 61.3 + 0.1565T = 16$ | DM1 | Solving for $T$ |
| $T = 75.5$ N | A1 | Correct answer |
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To get mark scheme content, you would need to share the actual mark scheme PDF/images for this paper. AQA mark schemes are published separately from question papers and are available on the AQA website.8 The diagram shows a block, of mass 20 kg , being pulled along a rough horizontal surface by a rope inclined at an angle of $30 ^ { \circ }$ to the horizontal.\\
\includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-16_323_1194_411_424}
The coefficient of friction between the block and the surface is $\mu$. Model the block as a particle which slides on the surface.
\begin{enumerate}[label=(\alph*)]
\item If the tension in the rope is 60 newtons, the block moves at a constant speed.
\begin{enumerate}[label=(\roman*)]
\item Show that the magnitude of the normal reaction force acting on the block is 166 N .
\item Find $\mu$.
\end{enumerate}\item If the rope remains at the same angle and the block accelerates at $0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, find the tension in the rope.
\includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-18_2488_1719_219_150}\\
\includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-19_2488_1719_219_150}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q8 [12]}}