| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Particle on inclined plane - force parallel to slope |
| Difficulty | Moderate -0.8 This is a standard M1 friction problem with straightforward application of equilibrium equations. Part (a) requires resolving perpendicular to the plane (one equation), part (b) uses F=μR and resolving parallel to the plane, and part (c) checks if friction is sufficient to prevent sliding down. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(240 = \frac{1}{2}(u+34)10\) | M1 A1 | Complete method to produce equation in \(u\) only; correct equation (\(u^2 - 48u + 476 = 0\) oe is possible) |
| \(u = 14\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(34 = 14 + 10a \Rightarrow a = 2\) | M1 A1 | Equation in \(a\) only (M0 if \(v=34\) when \(s=120\) used) |
| \(120 = 14t + \frac{1}{2} \times 2 \times t^2\) | M1 A1 | 3-term quadratic in \(t\) only, allow sign errors; must have found value of \(a\) |
| \(t^2 + 14t - 120 = 0\) | ||
| Solving: \(t = -20\) or \(6\) | DM1 | Dependent on previous M1 |
| \(t = 6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(34 = 14 + 10a \Rightarrow a = 2\) | M1 A1 | |
| \(v^2 = 14^2 + 2 \times 2 \times 120 \Rightarrow v = 26\) | M1 A1 | Complete method to obtain equation in \(t\) only, allow sign errors |
| AND \(26 = 14 + 2t\) | ||
| \(t = 6\) | DM1 A1 |
## Question 4:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $240 = \frac{1}{2}(u+34)10$ | M1 A1 | Complete method to produce equation in $u$ only; correct equation ($u^2 - 48u + 476 = 0$ oe is possible) |
| $u = 14$ | A1 | |
### Part (b) — EITHER method
| Answer/Working | Marks | Guidance |
|---|---|---|
| $34 = 14 + 10a \Rightarrow a = 2$ | M1 A1 | Equation in $a$ only (M0 if $v=34$ when $s=120$ used) |
| $120 = 14t + \frac{1}{2} \times 2 \times t^2$ | M1 A1 | 3-term quadratic in $t$ only, allow sign errors; must have found value of $a$ |
| $t^2 + 14t - 120 = 0$ | | |
| Solving: $t = -20$ or $6$ | DM1 | Dependent on previous M1 |
| $t = 6$ | A1 | |
### Part (b) — OR method
| Answer/Working | Marks | Guidance |
|---|---|---|
| $34 = 14 + 10a \Rightarrow a = 2$ | M1 A1 | |
| $v^2 = 14^2 + 2 \times 2 \times 120 \Rightarrow v = 26$ | M1 A1 | Complete method to obtain equation in $t$ only, allow sign errors |
| AND $26 = 14 + 2t$ | | |
| $t = 6$ | DM1 A1 | |
---4.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{94d9432d-1723-4549-ad5e-d4be0f5fd083-007_330_675_287_644}
\end{center}
\end{figure}
A particle $P$ of mass 2.5 kg rests in equilibrium on a rough plane under the action of a force of magnitude $X$ newtons acting up a line of greatest slope of the plane, as shown in Figure 3. The plane is inclined at $20 ^ { \circ }$ to the horizontal. The coefficient of friction between $P$ and the plane is 0.4 . The particle is in limiting equilibrium and is on the point of moving up the plane. Calculate
\begin{enumerate}[label=(\alph*)]
\item the normal reaction of the plane on $P$,
\item the value of $X$.
The force of magnitude $X$ newtons is now removed.
\item Show that $P$ remains in equilibrium on the plane.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q4}}