| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 work-energy question with straightforward application of formulas. Part (a) requires calculating work against gravity and friction using W = Fd, while part (b) applies work-energy principle with given friction coefficient. The trigonometry is simple (sin α given directly), and all steps follow routine procedures taught in M2 with no novel problem-solving required. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.02b Calculate work: constant force, resolved component6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F_{\max} = \frac{1}{3} \times 2g\cos\alpha\ (= 5.90...)\) | M1 | Use of \(F = \mu R\). Seen or implied. Condone sine/cosine confusion. Condone \(g\) missing |
| WD against friction \(= 6 \times \textit{their}\ F_{\max}\) | M1 | \((= 35.4...\text{(J)})\) Seen or implied as part of the 4th M mark |
| PE gain \(= 2g \times 6 \times \sin\alpha\) \(\left(= 6\times\frac{42}{5} = 50.4\right)\) | M1 | Dimensionally correct. Condone sine/cosine confusion |
| Total WD = WD against friction + WD against gravity (gain in PE) | DM1 | Dependent on the 3 preceding M marks. Require both terms and no extras |
| Total WD \(= 85.8\text{(J)}\) or \(86\text{(J)}\) | A1 | 3 sf or 2 sf only. \(\left(8\sqrt{10}+36\right)\frac{g}{7}\) is A0 (incorrect units) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work-energy equation (KE gained = loss in GPE - WD against friction) | M1 | Must be using work-energy. Need all terms, no extras and dimensionally correct. Condone sign errors. Condone sine/cosine confusion |
| \(\frac{1}{2}\times 2v^2 = 2g\times 6\sin\alpha - 6\times\frac{2}{3}g\cos\alpha\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation. They must have started with correct expressions, but follow through on any calculation errors | |
| \(v = 3.87\left(\mathrm{ms}^{-1}\right)\ \textit{or}\ 3.9\left(\mathrm{ms}^{-1}\right)\) | A1 | 3 sf or 2 sf only |
| [4] |
## Question 3:
**Part 3a:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F_{\max} = \frac{1}{3} \times 2g\cos\alpha\ (= 5.90...)$ | M1 | Use of $F = \mu R$. Seen or implied. Condone sine/cosine confusion. Condone $g$ missing |
| WD against friction $= 6 \times \textit{their}\ F_{\max}$ | M1 | $(= 35.4...\text{(J)})$ Seen or implied as part of the 4th M mark |
| PE gain $= 2g \times 6 \times \sin\alpha$ $\left(= 6\times\frac{42}{5} = 50.4\right)$ | M1 | Dimensionally correct. Condone sine/cosine confusion |
| Total WD = WD against friction + WD against gravity (gain in PE) | DM1 | Dependent on the 3 preceding M marks. Require both terms and no extras |
| Total WD $= 85.8\text{(J)}$ or $86\text{(J)}$ | A1 | 3 sf or 2 sf only. $\left(8\sqrt{10}+36\right)\frac{g}{7}$ is A0 (incorrect units) |
| **[5]** | | |
**Part 3b:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation (KE gained = loss in GPE - WD against friction) | M1 | Must be using work-energy. Need all terms, no extras and dimensionally correct. Condone sign errors. Condone sine/cosine confusion |
| $\frac{1}{2}\times 2v^2 = 2g\times 6\sin\alpha - 6\times\frac{2}{3}g\cos\alpha$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation. They must have started with correct expressions, but follow through on any calculation errors |
| $v = 3.87\left(\mathrm{ms}^{-1}\right)\ \textit{or}\ 3.9\left(\mathrm{ms}^{-1}\right)$ | A1 | 3 sf or 2 sf only |
| **[4]** | | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f06fe7-4d9c-4009-8931-3ecbc31fa5e5-06_323_1043_255_513}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A rough ramp is fixed to horizontal ground.\\
The ramp is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 3 } { 7 }$\\
The line $A B$ is a line of greatest slope of the ramp, with $B$ above $A$ and $A B = 6 \mathrm {~m}$, as shown in Figure 1.
A block $P$ of mass 2 kg is pushed, with constant speed, in a straight line up the slope from $A$ to $B$. The force pushing $P$ acts parallel to $A B$.
The coefficient of friction between $P$ and the ramp is $\frac { 1 } { 3 }$\\
The block is modelled as a particle and air resistance is negligible.
\begin{enumerate}[label=(\alph*)]
\item Use the model to find the total work done in pushing the block from $A$ to $B$.
The block is now held at $B$ and released from rest.
\item Use the model and the work-energy principle to find the speed of the block at the instant it reaches $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2024 Q3 [9]}}