| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | String at angle to slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question involving resolving forces on an inclined plane with friction and an angled applied force. Part (a) requires resolving forces parallel and perpendicular to the plane to find acceleration, part (b) uses standard kinematics (v² = u² + 2as), and part (c) tests understanding of limiting friction when the particle comes to rest. While it has multiple parts and requires careful resolution of the 40° force, it follows a completely standard template for M1 slope problems with no novel insight required—slightly easier than the average A-level question due to its routine nature. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03e 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 |
| Perpendicular to plane: \(R + 18\sin 40° = 2g\cos 30°\) | M1 A1 | Correct number of terms, forces resolved perp to plane, condone sign errors and sin/cos confusion, forces and angles paired correctly. A1 for correct unsimplified equation. |
| Parallel to plane: \(18\cos 40° - F - 2g\sin 30° = 2a\) (or \(-2a\)) | M1 A1 A1 | M1: correct number of terms. A1: at most one error. A1: fully correct unsimplified equation. |
| \(F = 0.3R\) | B1 | Use of \(F = 0.3R\) |
| \(18\cos 40° - 0.3(2g\cos 30° - 18\sin 40°) - 2g\sin 30° = 2a\) | dM1 | Eliminate \(F\) and \(R\) to form equation in \(a\), dependent on two M's |
| \(a = 1.18\) or \(1.2\ (\text{m s}^{-2})\) | A1 cao | Must be 2 or 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = 2^2 + 2(\text{'1.18'})5\) | M1 A1ft | Complete method to form equation in \(v\) or \(v^2\). Follow through on their value for \(a\). |
| \(v = 3.98\) or \(4.0\) or \(4\ (\text{m s}^{-1})\) | A1 cao | Must be positive. Note \(a = 1.2\) leads to \(v = 4\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 2g\cos 30°\ \left(= g\sqrt{3}\right)\) | B1 | Correct expression or value for new \(R\) |
| Friction \(= 0.3 \times 2g\cos 30°\) OR \(0.3 \times 2g\sin 30°\) | M1 | Find max friction. M0 if previous \(R\) is used. |
| Compare friction with weight component parallel to plane e.g. \(2g\sin 30° - 0.3(2g\cos 30°)\ (=2a)\) OR \(0.3(2g\cos 30°) - 2g\sin 30°\ (=2a)\) | dM1 | Correct comparison between max friction and weight component (force parallel to slope), dependent on previous M |
| \((a) > 0\) OR \((a) < 0\), concludes \(P\) will not remain at rest | A1 | Correct statement from fully correct working. |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Perpendicular to plane: $R + 18\sin 40° = 2g\cos 30°$ | M1 A1 | Correct number of terms, forces resolved perp to plane, condone sign errors and sin/cos confusion, forces and angles paired correctly. A1 for correct unsimplified equation. |
| Parallel to plane: $18\cos 40° - F - 2g\sin 30° = 2a$ (or $-2a$) | M1 A1 A1 | M1: correct number of terms. A1: at most one error. A1: fully correct unsimplified equation. |
| $F = 0.3R$ | B1 | Use of $F = 0.3R$ |
| $18\cos 40° - 0.3(2g\cos 30° - 18\sin 40°) - 2g\sin 30° = 2a$ | dM1 | Eliminate $F$ and $R$ to form equation in $a$, dependent on two M's |
| $a = 1.18$ or $1.2\ (\text{m s}^{-2})$ | A1 cao | Must be 2 or 3 sf |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = 2^2 + 2(\text{'1.18'})5$ | M1 A1ft | Complete method to form equation in $v$ or $v^2$. Follow through on their value for $a$. |
| $v = 3.98$ or $4.0$ or $4\ (\text{m s}^{-1})$ | A1 cao | Must be positive. Note $a = 1.2$ leads to $v = 4$. |
N.B. For (a) and (b), penalise over accurate answers ONCE only.
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 2g\cos 30°\ \left(= g\sqrt{3}\right)$ | B1 | Correct expression or value for **new** $R$ |
| Friction $= 0.3 \times 2g\cos 30°$ OR $0.3 \times 2g\sin 30°$ | M1 | Find max friction. M0 if previous $R$ is used. |
| Compare friction with weight component parallel to plane e.g. $2g\sin 30° - 0.3(2g\cos 30°)\ (=2a)$ **OR** $0.3(2g\cos 30°) - 2g\sin 30°\ (=2a)$ | dM1 | Correct comparison between max friction and weight component (force parallel to slope), dependent on previous M |
| $(a) > 0$ **OR** $(a) < 0$, concludes $P$ will not remain at rest | A1 | Correct statement from fully correct working. |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{84c0eead-0a87-4d87-b33d-794a94bb466c-24_545_764_285_651}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A parcel of mass 2 kg is pulled up a rough inclined plane by the action of a constant force.
The force has magnitude 18 N and acts at an angle of $40 ^ { \circ }$ to the plane.\\
The line of action of the force lies in a vertical plane containing a line of greatest slope of the inclined plane.
The plane is inclined at an angle of $30 ^ { \circ }$ to the horizontal, as shown in Figure 5.\\
The coefficient of friction between the plane and the parcel is 0.3\\
The parcel is modelled as a particle $P$
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$
The points $A$ and $B$ lie on a line of greatest slope of the plane, where $A B = 5 \mathrm {~m}$ and $B$ is above $A$. Particle $P$ passes through $A$ with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction $A B$.
\item Find the speed of $P$ as it passes through $B$.
The force of 18 N is removed at the instant $P$ passes through $B$. As a result, $P$ comes to rest at the point $C$.
\item Determine whether $P$ will remain at rest at $C$. You must show all stages of your working clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q8 [15]}}