| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Connected particles pulley energy method |
| Difficulty | Standard +0.3 This is a standard M2 connected particles problem using work-energy methods. While it involves multiple steps (friction work, PE changes for both particles, applying work-energy principle), all techniques are routine for M2: resolving forces on an incline, calculating friction work, and applying conservation of energy. The given sin α = 3/5 simplifies calculations. Slightly easier than average due to straightforward setup and clear structure. |
| Spec | 3.03o Advanced connected particles: and pulleys6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Resolve perpendicular to the plane: \(R = 2g\cos\alpha\) | B1 | |
| Use \(F = \mu R\): \(F = \frac{1}{4} \times 2g \times \frac{4}{5}\left(= \frac{2g}{5}\right)\) | M1 | with \(\frac{1}{4}\) and their \(R\) (3.92) |
| Work done: \(WD = 2.5 \times F\) | DM1 | For their \(F\) |
| \(= 2.5 \times \frac{2g}{5} = 9.8\) (J) | A1 | Accept \(g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Change in PE: \(\pm(4g \times 2.5 - 2g \times 2.5\sin\alpha)\) | M1 | Requires one gaining and one losing. Condone trig confusion |
| \(= \pm(4g \times 2.5 - 2g \times 1.5)\) | A1 | \(\pm\) (correct unsimplified) |
| PE lost \(= 7g = 68.6\) (J) | A1 | or 69 (J). Accept \(7g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| KE gained \(+\) WD \(=\) loss in GPE | M1 | The question requires the use of work-energy. Alternative methods score 0/4. Requires all terms but condone sign errors (must be considering both particles) |
| \(\frac{1}{2} \times 4v^2 + \frac{1}{2} \times 2v^2 + (\text{their (a)}) = (\text{their (b)})\) | A2 | Correct unsimplified. \(-1\) each error |
| \(3v^2 = 6g\) | ||
| \(v = \sqrt{2g} = 4.43\) (m s\(^{-1}\)) | A1 | or 4.4. Accept \(\sqrt{2g}\) |
# Question 4:
## Part 4a:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Resolve perpendicular to the plane: $R = 2g\cos\alpha$ | B1 | |
| Use $F = \mu R$: $F = \frac{1}{4} \times 2g \times \frac{4}{5}\left(= \frac{2g}{5}\right)$ | M1 | with $\frac{1}{4}$ and their $R$ (3.92) |
| Work done: $WD = 2.5 \times F$ | DM1 | For their $F$ |
| $= 2.5 \times \frac{2g}{5} = 9.8$ (J) | A1 | Accept $g$ |
## Part 4b:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Change in PE: $\pm(4g \times 2.5 - 2g \times 2.5\sin\alpha)$ | M1 | Requires one gaining and one losing. Condone trig confusion |
| $= \pm(4g \times 2.5 - 2g \times 1.5)$ | A1 | $\pm$ (correct unsimplified) |
| PE lost $= 7g = 68.6$ (J) | A1 | or 69 (J). Accept $7g$ |
## Part 4c:
| Working/Answer | Marks | Guidance |
|---|---|---|
| KE gained $+$ WD $=$ loss in GPE | M1 | The question requires the use of work-energy. Alternative methods score 0/4. Requires all terms but condone sign errors (must be considering both particles) |
| $\frac{1}{2} \times 4v^2 + \frac{1}{2} \times 2v^2 + (\text{their (a)}) = (\text{their (b)})$ | A2 | Correct unsimplified. $-1$ each error |
| $3v^2 = 6g$ | | |
| $v = \sqrt{2g} = 4.43$ (m s$^{-1}$) | A1 | or 4.4. Accept $\sqrt{2g}$ |
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4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e6d100ff-dd4a-4591-a0a3-81761773045e-07_544_1264_251_338}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Two particles $P$ and $Q$, of mass 2 kg and 4 kg respectively, are connected by a light inextensible string. Initially $P$ is held at rest at the point $A$ on a rough fixed plane inclined at $\alpha$ to the horizontal ground, where $\sin \alpha = \frac { 3 } { 5 }$. The string passes over a small smooth pulley fixed at the top of the plane. The particle $Q$ hangs freely below the pulley and 2.5 m above the ground, as shown in Figure 1. The part of the string from $P$ to the pulley lies along a line of greatest slope of the plane. The system is released from rest with the string taut. At the instant when $Q$ hits the ground, $P$ is at the point $B$ on the plane. The coefficient of friction between $P$ and the plane is $\frac { 1 } { 4 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the work done against friction as $P$ moves from $A$ to $B$.
\item Find the total potential energy lost by the system as $P$ moves from $A$ to $B$.
\item Find, using the work-energy principle, the speed of $P$ as it passes through $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q4 [11]}}