| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2020 |
| Session | January |
| Marks | 6 |
| 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 straightforward application of the work-energy principle to a particle on an inclined plane with friction. Students must identify forces (weight component, friction), calculate work done by each force over the given distance, and apply the work-energy equation. While it requires careful bookkeeping of energy terms and resolving forces, it follows a standard M2 template with no novel problem-solving required. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Work energy equation: \(KE\text{ lost} = WD + PE\text{ gain}\) | M1 | Must be using work-energy. Require all terms. Dimensionally correct. Allow their WD, but must be WD not \(F\). Condone sine/cosine confusion and sign errors |
| \(\frac{1}{2}\times 2\times 16 = WD + 2g\times 2.5\sin\theta\) | A1 | Unsimplified equation with at most one error |
| \((WD = 9)\) | A1 | Correct unsimplified equation. NB: \(16 = WD + 7\) seen scores 3 marks |
| Use of \(F = \mu\times 2g\cos\theta\) | B1 | \((F = \mu\times 19.398...)\) Allow \(\pm\). This mark is available if they use a *suvat* approach |
| Use of Work done \(= 2.5F\) | B1 | Allow \(\pm\) |
| \(9 = 2.5\times\mu\times 2g\cos\theta \Rightarrow \mu = 0.19\) | A1 | Or 0.186. Max 3 sf. Not \(\frac{3\sqrt{3}}{28}\) |
## Question 2:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Work energy equation: $KE\text{ lost} = WD + PE\text{ gain}$ | M1 | Must be using work-energy. Require all terms. Dimensionally correct. Allow their WD, but must be WD not $F$. Condone sine/cosine confusion and sign errors |
| $\frac{1}{2}\times 2\times 16 = WD + 2g\times 2.5\sin\theta$ | A1 | Unsimplified equation with at most one error |
| $(WD = 9)$ | A1 | Correct unsimplified equation. NB: $16 = WD + 7$ **seen** scores 3 marks |
| Use of $F = \mu\times 2g\cos\theta$ | B1 | $(F = \mu\times 19.398...)$ Allow $\pm$. This mark is available if they use a *suvat* approach |
| Use of Work done $= 2.5F$ | B1 | Allow $\pm$ |
| $9 = 2.5\times\mu\times 2g\cos\theta \Rightarrow \mu = 0.19$ | A1 | Or 0.186. Max 3 sf. Not $\frac{3\sqrt{3}}{28}$ |
**Total: [6]**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c16c17b6-2c24-4939-b3b5-63cd63646b76-04_239_796_246_577}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A rough straight ramp is fixed to horizontal ground. The ramp is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 7 }$. The points $A$ and $B$ are on a line of greatest slope of the ramp with $A B = 2.5 \mathrm {~m}$ and $B$ above $A$, as shown in Figure 1. A package of mass 2 kg is projected up the ramp from $A$ with speed $4 \mathrm {~ms} ^ { - 1 }$ and first comes to instantaneous rest at $B$. The coefficient of friction between the package and the ramp is $\mu$. The package is modelled as a particle.
Use the work-energy principle to find the value of $\mu$.\\
(6)\\
\hfill \mbox{\textit{Edexcel M2 2020 Q2 [6]}}