| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Particle on inclined plane - force parallel to slope |
| Difficulty | Standard +0.3 This is a standard M1 friction problem requiring resolution of forces parallel and perpendicular to the plane in two equilibrium scenarios, then solving simultaneous equations. The setup is clearly defined with limiting friction in both directions, making it slightly easier than average but requiring careful algebraic manipulation of the two cases. |
| Spec | 3.03e Resolve forces: two dimensions3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolve perp to plane: \(R = mg\cos\alpha\) | M1 A1 | First resolution, correct no. of terms, condone sign errors and sin/cos confusion; if cos(4/5) used, treat as A error but allow recovery |
| Resolve parallel (up slope): \(mg\sin\alpha + F = 2P\) | M1 | Second resolution, correct no. of terms |
| \(mg\sin\alpha - F = P\) | A1 | Correct equation (A0 if different \(R\)'s or \(F\)'s used) |
| Use \(F = \mu R\) | M1 | |
| Substitute for trig, eliminate \(P\) and \(F\), solve for \(\mu\) | M1 | |
| \(\mu = 0.25\) | A1 | cao; if \(g\) consistently omitted, max M1A0M1A0A0M1M1A1 |
| Alternative via horizontal/vertical: | ||
| \((\rightarrow)\ 2P\cos\alpha = R\sin\alpha + F\cos\alpha\) | M1 A1 | |
| \((\rightarrow)\ P\cos\alpha = R\sin\alpha - F\cos\alpha\) | M1 A1 | |
| \((1)+(2)\): \(3P\cos\alpha = 2R\sin\alpha\) | ||
| \((1)-(2)\): \(P\cos\alpha = 2F\cos\alpha\) | ||
| Divide: \(\frac{1}{3} = \frac{F}{R}\cot\alpha\) | A1 | |
| Use \(F = \mu R\); substitute for trig and solve for \(\mu\) | M1 M1 | |
| \(\mu = 0.25\) | A1 |
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolve perp to plane: $R = mg\cos\alpha$ | M1 A1 | First resolution, correct no. of terms, condone sign errors and sin/cos confusion; if cos(4/5) used, treat as A error but allow recovery |
| Resolve parallel (up slope): $mg\sin\alpha + F = 2P$ | M1 | Second resolution, correct no. of terms |
| $mg\sin\alpha - F = P$ | A1 | Correct equation (A0 if different $R$'s or $F$'s used) |
| Use $F = \mu R$ | M1 | |
| Substitute for trig, eliminate $P$ and $F$, solve for $\mu$ | M1 | |
| $\mu = 0.25$ | A1 | cao; if $g$ consistently omitted, max M1A0M1A0A0M1M1A1 |
| **Alternative via horizontal/vertical:** | | |
| $(\rightarrow)\ 2P\cos\alpha = R\sin\alpha + F\cos\alpha$ | M1 A1 | |
| $(\rightarrow)\ P\cos\alpha = R\sin\alpha - F\cos\alpha$ | M1 A1 | |
| $(1)+(2)$: $3P\cos\alpha = 2R\sin\alpha$ | | |
| $(1)-(2)$: $P\cos\alpha = 2F\cos\alpha$ | | |
| Divide: $\frac{1}{3} = \frac{F}{R}\cot\alpha$ | A1 | |
| Use $F = \mu R$; substitute for trig and solve for $\mu$ | M1 M1 | |
| $\mu = 0.25$ | A1 | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1bdc84b-c8a1-4e7c-a2ba-48b40c6a6d36-14_209_511_246_721}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A particle of mass $m$ rests in equilibrium on a fixed rough plane under the action of a force of magnitude $X$. The force acts up a line of greatest slope of the plane, as shown in Figure 3.
The plane is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$\\
The coefficient of friction between the particle and the plane is $\mu$.
\begin{itemize}
\item When $X = 2 P$, the particle is on the point of sliding up the plane.
\item When $X = P$, the particle is on the point of sliding down the plane.
\end{itemize}
Find the value of $\mu$.\\
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel M1 2022 Q5 [8]}}