| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | October |
| Marks | 17 |
| 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 involving forces on a slope with friction. Part (a) requires resolving perpendicular to the plane, (b) uses limiting friction when on the point of moving, and (c) applies equations of motion. While it has multiple parts and requires careful resolution of the horizontal force, these are routine techniques for M1 with no novel insight required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments3.02d Constant acceleration: SUVAT formulae3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03i Normal reaction force3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R(\perp \text{ plane}): R = 0.5g\cos 30° + 5\sin 30°\) | M1A1A1 | M1 for resolution perpendicular to plane with usual rules; first and second A1 for correct equation, \(-1\) each error |
| \(R = 6.743\ldots = 6.7\) or \(6.74\) N | A1 | Must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R(\parallel \text{ plane}): F = 5\cos 30° - 0.5g\sin 30°\ (= 1.880\ldots)\) | M1A1A1 | First M1 for resolution parallel to plane; first and second A1 for correct equation, \(-1\) each error |
| \(\mu = \frac{F}{R} = \frac{1.880}{6.743} = 0.27880\ldots = 0.28\) or \(0.279\) | M1A1 | Second M1 for use of \(\mu = \frac{F}{R}\); A1 for 0.28 or 0.279 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| NL2: \(0.5g\sin 30° - F' = 0.5a\) | M1A1 | First M1 for equation of motion parallel to plane |
| \(R(\perp \text{ plane}): R' = 0.5g\cos 30°\ (= 4.2435\ldots)\) | M1A1 | Second M1 for resolution perpendicular to plane |
| Use of \(F' = \mu R' = 0.2787\ldots \times R'\ (= 1.18345\ldots)\) and solve for \(a\) | DM1 | Dependent on both previous M marks |
| \(a = 2.53\ldots\) m s\(^{-2}\) | A1 | A1 for 2.53 or better; if \(v\) wrong but \(v = 3.9\) allow \(a = 2.5\) or \(2.54\) |
| \(v^2 = 2as = 2 \times 2.533 \times 3\) | M1 | Fourth M1 independent but must have used equation of motion to find \(a\); M0 if \(a\) is negative down the plane |
| \(v = 3.9\) or \(3.90\) ms\(^{-1}\) | A1 | (8) [17] |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R(\perp \text{ plane}): R = 0.5g\cos 30° + 5\sin 30°$ | M1A1A1 | M1 for resolution perpendicular to plane with usual rules; first and second A1 for correct equation, $-1$ each error |
| $R = 6.743\ldots = 6.7$ or $6.74$ N | A1 | Must be positive | (4) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R(\parallel \text{ plane}): F = 5\cos 30° - 0.5g\sin 30°\ (= 1.880\ldots)$ | M1A1A1 | First M1 for resolution parallel to plane; first and second A1 for correct equation, $-1$ each error |
| $\mu = \frac{F}{R} = \frac{1.880}{6.743} = 0.27880\ldots = 0.28$ or $0.279$ | M1A1 | Second M1 for use of $\mu = \frac{F}{R}$; A1 for 0.28 or 0.279 | (5) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| NL2: $0.5g\sin 30° - F' = 0.5a$ | M1A1 | First M1 for equation of motion parallel to plane |
| $R(\perp \text{ plane}): R' = 0.5g\cos 30°\ (= 4.2435\ldots)$ | M1A1 | Second M1 for resolution perpendicular to plane |
| Use of $F' = \mu R' = 0.2787\ldots \times R'\ (= 1.18345\ldots)$ and solve for $a$ | DM1 | Dependent on both previous M marks |
| $a = 2.53\ldots$ m s$^{-2}$ | A1 | A1 for 2.53 or better; if $v$ wrong but $v = 3.9$ allow $a = 2.5$ or $2.54$ |
| $v^2 = 2as = 2 \times 2.533 \times 3$ | M1 | Fourth M1 independent but must have used equation of motion to find $a$; M0 if $a$ is negative down the plane |
| $v = 3.9$ or $3.90$ ms$^{-1}$ | A1 | (8) **[17]** |
**SC for 8c:** If 5N force not removed, max score M1A0M1A0DM1A0M0A0 with usual rules for M marks.8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5f2d38d9-b719-4205-8cb0-caa959afc46f-28_268_634_292_657}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A rough plane is inclined at $30 ^ { \circ }$ to the horizontal. A particle $P$ of mass 0.5 kg is held at rest on the plane by a horizontal force of magnitude 5 N , as shown in Figure 4. The force acts in a vertical plane containing a line of greatest slope of the inclined plane. The particle is on the point of moving up the plane.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the normal reaction of the plane on $P$.
\item Find the coefficient of friction between $P$ and the plane.
The force of magnitude 5 N is now removed and $P$ accelerates from rest down the plane.
\item Find the speed of $P$ after it has travelled 3 m down the plane.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q8 [17]}}