| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | String at angle to slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem involving resolving forces on a slope with a string at an angle. It requires routine application of Newton's second law in two perpendicular directions (parallel and perpendicular to slope), plus friction = μR. The question guides students through the steps (draw diagram, find R, find μ), making it slightly easier than average but still requiring competent force resolution. |
| Spec | 3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Correct force diagram showing \(R\), \(T\), \(F\), \(mg\) | B1 | Correct force diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(R+20\sin30°=6g\cos10°\) | M1 | Resolving perpendicular to the slope with 3 terms |
| \(R=6g\cos10°-20\sin30°\) | A1 | Correct equation |
| \(R=47.9\text{ N}\) (to 3 sf) | dM1 | Solving for \(R\) |
| A1 | Correct \(R\) from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(F=\mu R\) | M1 | Use of \(F=\mu R\) |
| \(6\times0.4=20\cos30°-6g\sin10°-\mu R\) | M1 | Resolving parallel to slope to get 4 term equation of motion |
| A1 | Correct equation | |
| \(\mu R=4.710\) | A1 | Correct \(F/\mu R\) |
| \(\mu=\frac{4.710}{47.91}=0.0983\) | dM1 | Solving for \(\mu\) |
| A1 | AWRT 0.098 |
## Question 8:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Correct force diagram showing $R$, $T$, $F$, $mg$ | B1 | Correct force diagram |
**Part (b)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $R+20\sin30°=6g\cos10°$ | M1 | Resolving perpendicular to the slope with 3 terms |
| $R=6g\cos10°-20\sin30°$ | A1 | Correct equation |
| $R=47.9\text{ N}$ (to 3 sf) | dM1 | Solving for $R$ |
| | A1 | Correct $R$ from correct working |
**Part (c)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $F=\mu R$ | M1 | Use of $F=\mu R$ |
| $6\times0.4=20\cos30°-6g\sin10°-\mu R$ | M1 | Resolving parallel to slope to get 4 term equation of motion |
| | A1 | Correct equation |
| $\mu R=4.710$ | A1 | Correct $F/\mu R$ |
| $\mu=\frac{4.710}{47.91}=0.0983$ | dM1 | Solving for $\mu$ |
| | A1 | AWRT 0.098 |8 A rough slope is inclined at an angle of $10 ^ { \circ }$ to the horizontal. A particle of mass 6 kg is on the slope. A string is attached to the particle and is at an angle of $30 ^ { \circ }$ to the slope. The tension in the string is 20 N . The diagram shows the slope, the particle and the string.\\
\includegraphics[max width=\textwidth, alt={}, center]{7e0585ea-062a-487c-8e39-37a4ed414ff8-6_259_684_518_676}
The particle moves up the slope with an acceleration of $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show the forces acting on the particle.
\item Show that the magnitude of the normal reaction force is 47.9 N , correct to three significant figures.
\item Find the coefficient of friction between the particle and the slope.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q8 [11]}}