| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| 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 mechanics problem involving resolving forces on an inclined plane with friction. Part (a) requires straightforward resolution of forces in equilibrium (given the normal reaction, students resolve parallel and perpendicular to the slope), and part (b) is a routine application of F=ma with friction. The tan α = 3/4 setup is common, and all steps follow standard textbook methods with no novel insight required. Slightly easier than average due to being given the normal reaction directly. |
| Spec | 3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve vertically | M1 | Complete method to obtain equation in \(F\) only |
| \((\uparrow) F\sin\alpha + 68.6\cos\alpha = 5g\) OR \(-F\sin\alpha + 68.6\cos\alpha = 5g\) | A1 | Correct equation in \(F\) only, trig does not need substituting |
| \((\swarrow) 68.6 = X\sin\alpha + 5g\cos\alpha\) (leads to \(X=49\) with \(g=9.8\)) | Other possible equations where \(X\) would need eliminating | |
| \((\nearrow) F + X\cos\alpha = 5g\sin\alpha\) OR \(-F + X\cos\alpha = 5g\sin\alpha\) | ||
| \((\rightarrow) F\cos\alpha + X = 68.6\sin\alpha\) OR \(-F\cos\alpha + X = 68.6\sin\alpha\) | ||
| \(9.8\) N (or \(\frac{49}{5}\)) | A1 | cao, must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Down the plane | A1 | Allow "down" or "downwards" or arrow \(\swarrow\), must appear as written answer not just on diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion down the plane | M1 | Correct no. of terms, dimensionally correct |
| \(5g\sin\alpha - F = 5a\) | A1 | Allow \((-a)\) instead of \(a\) |
| Resolve perpendicular to plane | M1 | NB: M0 if \(R=68.6\) N used |
| \(R = 5g\cos\alpha\) | A1 | Correct equation |
| \(F = 0.5R\) seen | M1 | Could be seen on diagram |
| \(a = 1.96\) or \(2.0\) or \(2\ (\text{m s}^{-2})\) or \(\frac{1}{5}g\) | A1 | cao, must be positive |
## Question 2(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | Complete method to obtain equation in $F$ only |
| $(\uparrow) F\sin\alpha + 68.6\cos\alpha = 5g$ OR $-F\sin\alpha + 68.6\cos\alpha = 5g$ | A1 | Correct equation in $F$ only, trig does not need substituting |
| $(\swarrow) 68.6 = X\sin\alpha + 5g\cos\alpha$ (leads to $X=49$ with $g=9.8$) | | Other possible equations where $X$ would need eliminating |
| $(\nearrow) F + X\cos\alpha = 5g\sin\alpha$ OR $-F + X\cos\alpha = 5g\sin\alpha$ | | |
| $(\rightarrow) F\cos\alpha + X = 68.6\sin\alpha$ OR $-F\cos\alpha + X = 68.6\sin\alpha$ | | |
| $9.8$ N (or $\frac{49}{5}$) | A1 | cao, must be positive |
## Question 2(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Down the plane | A1 | Allow "down" or "downwards" or arrow $\swarrow$, must appear as written answer not just on diagram |
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion down the plane | M1 | Correct no. of terms, dimensionally correct |
| $5g\sin\alpha - F = 5a$ | A1 | Allow $(-a)$ instead of $a$ |
| Resolve perpendicular to plane | M1 | NB: M0 if $R=68.6$ N used |
| $R = 5g\cos\alpha$ | A1 | Correct equation |
| $F = 0.5R$ seen | M1 | Could be seen on diagram |
| $a = 1.96$ or $2.0$ or $2\ (\text{m s}^{-2})$ or $\frac{1}{5}g$ | A1 | cao, must be positive |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{414946db-64d7-44b8-801d-2c7805ee9cc6-04_282_627_246_721}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$\\
A small block $B$ of mass 5 kg is held in equilibrium on the plane by a horizontal force of magnitude $X$ newtons, as shown in Figure 1.
The force acts in a vertical plane which contains a line of greatest slope of the inclined plane.
The block $B$ is modelled as a particle.\\
The magnitude of the normal reaction of the plane on $B$ is 68.6 N .\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item find the magnitude of the frictional force acting on $B$,
\item state the direction of the frictional force acting on $B$.
The horizontal force of magnitude $X$ newtons is now removed and $B$ moves down the plane.
Given that the coefficient of friction between $B$ and the plane is 0.5
\end{enumerate}\item find the acceleration of $B$ down the plane.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2022 Q2 [10]}}