| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Block on horizontal plane motion |
| Difficulty | Moderate -0.8 This is a straightforward M1 friction question requiring standard application of Newton's laws and friction formulae (F ≤ μR). All parts follow directly from basic principles: resolving vertically for R, calculating maximum friction, recognizing equilibrium when P < Fmax, and using F=ma when P > Fmax. No problem-solving insight needed, just routine application of learned procedures. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03r Friction: concept and vector form3.03s Contact force components: normal and frictional3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram showing: Weight \(W = 98\) N (down), Normal reaction \(R\) (up), Applied force \(P\) (horizontal right), Friction \(F\) (horizontal left) | B1 | All four forces correctly labelled |
| Answer | Marks |
|---|---|
| \(R = mg = 10 \times 9.8 = 98\) N | B1 |
| Answer | Marks |
|---|---|
| \(F_{max} = \mu R = 0.5 \times 98 = 49\) N | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Friction \(= 30\) N | B1 | Since \(P < F_{max}\), block remains at rest so friction equals \(P\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P - F = ma\) | M1 | Newton's second law applied |
| \(80 - 49 = 10a\) | A1 | Correct equation with \(F = 49\) N |
| \(a = \frac{31}{10} = 3.1\) m s\(^{-2}\) | A1 |
# Question 2:
**(a)**
Diagram showing: Weight $W = 98$ N (down), Normal reaction $R$ (up), Applied force $P$ (horizontal right), Friction $F$ (horizontal left) | B1 | All four forces correctly labelled
**(b)(i)**
$R = mg = 10 \times 9.8 = 98$ N | B1 |
**(b)(ii)**
$F_{max} = \mu R = 0.5 \times 98 = 49$ N | B1 |
**(b)(iii)**
Friction $= 30$ N | B1 | Since $P < F_{max}$, block remains at rest so friction equals $P$
**(c)**
$P - F = ma$ | M1 | Newton's second law applied
$80 - 49 = 10a$ | A1 | Correct equation with $F = 49$ N
$a = \frac{31}{10} = 3.1$ m s$^{-2}$ | A1 |
---2 A block, of mass 10 kg , is at rest on a rough horizontal surface, when a horizontal force, of magnitude $P$ newtons, is applied to the block, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-04_108_962_461_539}
The coefficient of friction between the block and the surface is 0.5 .
\begin{enumerate}[label=(\alph*)]
\item Draw and label a diagram to show all the forces acting on the block.
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the magnitude of the normal reaction force acting on the block.
\item Find the maximum possible magnitude of the friction force between the block and the surface.
\item Given that $P = 30$, state the magnitude of the friction force acting on the block.
\end{enumerate}\item Given that $P = 80$, find the acceleration of the block.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-05_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2010 Q2 [7]}}