| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Energy methods on slope |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics question combining energy conservation and friction on a slope. Part (a) is simple energy conservation, part (b) requires resolving forces to find friction (standard F=μR), and part (c) applies work-energy principle. All techniques are routine for FM1 with no novel insight required, though the multi-part structure and Further Maths context place it slightly above average A-level difficulty. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Use the principle of conservation of mechanical energy and model | M1 | Correct number of terms, dimensionally correct, condone sign errors M0 if they use \(v^2 = u^2 + 2as\) |
| \(\frac{1}{2}mW^2 - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = mgh\) | A1 | Correct equation with at most one error |
| A1 | Correct equation | |
| \(W = \sqrt{4gh} = 2\sqrt{gh}\) | A1 | Either (need \(W = \)) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = mg \cos \theta\) | M1 | Condone sin/cos confusion and allow cos(\(\frac{4}{5}\)) etc |
| \(F = \frac{1}{3}R\) | M1 | \(F = \frac{1}{3}R\) |
| \(F = \frac{4}{15}mg\) | A1 | Accept \(0.27\) mg or better |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Use the work-energy principle and the model: | M1 | Correct number of terms, dimensionally correct, condone sign errors and sin/cos confusion and allow cos(\(\frac{4}{5}\)) etc |
| A to B: \(mgd \sin \theta - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = \frac{4}{15}mgd\) | A1ft | Correct equation with at most one error |
| or \(mg(h + d \sin \theta) - mgh - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = \frac{4}{15}mgd\) | A1ft | Correct equation ft on their answer to (b) (and (a) if they use A to the ground.) |
| OR | ||
| A to the ground: \(mg(h + d \sin \theta) - \frac{1}{2}m\left(\sqrt{4gh}\right)^2 = \frac{4}{15}mgd\) | ||
| Solve for \(d\) in terms of \(h\) or \(h\) in terms of \(d\) | M1 | Solve for \(d\), must have at least 3 terms, with two of them in \(d\) |
| \(d = 3h\) | A1 | cao |
| (5) |
## 3a
| Use the principle of conservation of mechanical energy and model | M1 | Correct number of terms, dimensionally correct, condone sign errors M0 if they use $v^2 = u^2 + 2as$ |
| $\frac{1}{2}mW^2 - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = mgh$ | A1 | Correct equation with at most one error |
| | A1 | Correct equation |
| $W = \sqrt{4gh} = 2\sqrt{gh}$ | A1 | Either (need $W = $) |
| | (4) | |
## 3b
| $R = mg \cos \theta$ | M1 | Condone sin/cos confusion and allow cos($\frac{4}{5}$) etc |
| $F = \frac{1}{3}R$ | M1 | $F = \frac{1}{3}R$ |
| $F = \frac{4}{15}mg$ | A1 | Accept $0.27$ mg or better |
| | (3) | |
## 3c
| Use the work-energy principle and the model: | M1 | Correct number of terms, dimensionally correct, condone sign errors and sin/cos confusion and allow cos($\frac{4}{5}$) etc |
| **A to B:** $mgd \sin \theta - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = \frac{4}{15}mgd$ | A1ft | Correct equation with at most one error |
| **or** $mg(h + d \sin \theta) - mgh - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = \frac{4}{15}mgd$ | A1ft | Correct equation ft on their answer to (b) (and (a) if they use A to the ground.) |
| **OR** | | |
| **A to the ground:** $mg(h + d \sin \theta) - \frac{1}{2}m\left(\sqrt{4gh}\right)^2 = \frac{4}{15}mgd$ | | |
| Solve for $d$ in terms of $h$ or $h$ in terms of $d$ | M1 | Solve for $d$, must have at least 3 terms, with two of them in $d$ |
| $d = 3h$ | A1 | cao |
| | (5) | |
**Total for Question 3: (12 marks)**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{885dd96e-ecaa-4a7f-acb4-f5cf636f491b-06_458_725_246_671}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the end elevation of a building which sits on horizontal ground. The side of the building is vertical and has height $h$.
A small stone of mass $m$ is at rest on the roof of the building at the point $A$. The stone slides from rest down a line of greatest slope of the roof and reaches the edge $B$ of the roof with speed $\sqrt { 2 g h }$
The stone then moves under gravity before hitting the ground with speed $W$.\\
In a model of the motion of the stone from $\boldsymbol { B }$ to the ground
\begin{itemize}
\item the stone is modelled as a particle
\item air resistance is ignored
\end{itemize}
Using the principle of conservation of mechanical energy and the model,
\begin{enumerate}[label=(\alph*)]
\item find $W$ in terms of $g$ and $h$.
In a model of the motion of the stone from $\boldsymbol { A }$ to $\boldsymbol { B }$
\begin{itemize}
\item the stone is modelled as a particle of mass $m$
\item air resistance is ignored
\item the roof of the building is modelled as a rough plane inclined to the horizontal at an angle $\theta$, where $\tan \theta = \frac { 3 } { 4 }$
\item the coefficient of friction between the stone and the roof is $\frac { 1 } { 3 }$
\item $A B = d$
\end{itemize}
Using this model,
\item find, in terms of $m$ and $g$, the magnitude of the frictional force acting on the stone as it slides down the roof,
\item use the work-energy principle to find $d$ in terms of $h$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 AS 2024 Q3 [12]}}