| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Horizontal force on slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics problem requiring resolution of forces parallel and perpendicular to an inclined plane with friction. While it involves multiple force components (weight, normal reaction, friction, horizontal force) and requires careful trigonometry with tan α = 3/4, it follows a routine procedure taught in all M1 courses. The 'on the point of sliding' condition (F = μR) is a standard concept, making this slightly easier than average overall. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(F = \mu R\) | B1 | Seen or implied |
| \((\searrow)\; R = 10\sin\alpha + 5g\cos\alpha\) | M1 A2 | M1 for resolving perpendicular to plane; A1A1 for correct equation (A1A0 if one error) |
| \((\nearrow)\; F = 5g\sin\alpha - 10\cos\alpha\) | M1 A2 | M1 for resolving parallel to plane; A1A1 for correct equation |
| \(\mu = \dfrac{g\sin\alpha - 2\cos\alpha}{2\sin\alpha + g\cos\alpha} = 0.47\) or \(0.473\) | M1 A1 (9) | Third M1 for eliminating \(R\) to get equation in \(\mu\) only; A1 for \(0.47\) or \(0.473\) |
# Question 4:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $F = \mu R$ | B1 | Seen or implied |
| $(\searrow)\; R = 10\sin\alpha + 5g\cos\alpha$ | M1 A2 | M1 for resolving perpendicular to plane; A1A1 for correct equation (A1A0 if one error) |
| $(\nearrow)\; F = 5g\sin\alpha - 10\cos\alpha$ | M1 A2 | M1 for resolving parallel to plane; A1A1 for correct equation |
| $\mu = \dfrac{g\sin\alpha - 2\cos\alpha}{2\sin\alpha + g\cos\alpha} = 0.47$ or $0.473$ | M1 A1 (9) | Third M1 for eliminating $R$ to get equation in $\mu$ only; A1 for $0.47$ or $0.473$ |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c809d34e-83db-4a16-a831-001f9f36b1c3-10_291_926_251_516}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle $P$ of mass 5 kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10 N . The plane is inclined to the horizontal at an angle $\alpha$ where $\tan \alpha = \frac { 3 } { 4 }$, as shown in Figure 1. The line of action of the force lies in the vertical plane containing $P$ and a line of greatest slope of the plane. The coefficient of friction between $P$ and the plane is $\mu$. Given that $P$ is on the point of sliding down the plane, find the value of $\mu$.
\includegraphics[max width=\textwidth, alt={}, center]{c809d34e-83db-4a16-a831-001f9f36b1c3-13_2460_72_311_27}\\
\hfill \mbox{\textit{Edexcel M1 2017 Q4 [9]}}