| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | Specimen |
| 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 requires multiple steps (friction work, PE changes for both particles, applying work-energy principle), all techniques are routine for this topic. The given sin α = 3/5 simplifies calculations, and the structure clearly guides students through parts (a), (b), (c). Slightly easier than average due to the scaffolding and standard setup. |
| Spec | 3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve perpendicular to plane: \(R = 2g\cos\alpha\) | B1 | |
| Use \(F = \mu R\): \(F = \dfrac{1}{4} \times 2g \times \dfrac{4}{5}\left(= \dfrac{2g}{5}\right)\) | M1 | with \(\dfrac{1}{4}\) and their \(R\) (3.92) |
| Work done: \(\text{WD} = 2.5 \times F\) | dM1 | For their \(F\) |
| \(= 2.5 \times \dfrac{2g}{5} = 9.8\) (J) | A1 | Accept \(g\) |
| (4) | If candidate found total work done but correct terms/processes for finding WD against friction visible, give B1M1DM1A0 (3/4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | 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 \(7g\) |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| KE gained \(+\) WD \(=\) loss in GPE | M1 | The question requires use of work-energy. Alternative methods score 0/4. Requires all terms but condone sign errors (must be considering both particles) |
| \(\dfrac{1}{2} \times 4v^2 + \dfrac{1}{2} \times 2v^2 + (\text{their (a)}) = (\text{their (b)})\) | A2 | Correct unsimplified. \(-1\) each error |
| \(3v^2 = 6g\) | ||
| \(v = \sqrt{2g} = 4.43\ (\text{m s}^{-1})\) | A1 | or 4.4. Accept \(\sqrt{2g}\) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equations of motion for each particle leading to \(T = \dfrac{12g}{5} = 23.52\) followed by W-E equation for \(P\): \(2.5T = \dfrac{1}{2} \times 2v^2 + 2g \times 2.5\sin\alpha + (a)\) | M1A2 | Equations of motion for each particle leading to \(T = \dfrac{12g}{5} = 23.52\) followed by W-E equation for \(Q\): \(\dfrac{1}{2} \times 4v^2 + 2.5T = 4g \times 2.5\) |
| \(v = \sqrt{2g} = 4.43\ (\text{m s}^{-1})\) | A1 | |
| (11 marks) | Use of \(\alpha = 36.9\) gives correct answers to 3 sf. Use of \(\alpha = 37\) gives correct answers to 2 sf only; A0 if 3 sf given. |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to plane: $R = 2g\cos\alpha$ | B1 | |
| Use $F = \mu R$: $F = \dfrac{1}{4} \times 2g \times \dfrac{4}{5}\left(= \dfrac{2g}{5}\right)$ | M1 | with $\dfrac{1}{4}$ and their $R$ (3.92) |
| Work done: $\text{WD} = 2.5 \times F$ | dM1 | For their $F$ |
| $= 2.5 \times \dfrac{2g}{5} = 9.8$ (J) | A1 | Accept $g$ |
| | **(4)** | If candidate found total work done but correct terms/processes for finding WD against friction visible, give B1M1DM1A0 (3/4) |
## Question 4(b):
| Answer/Working | Mark | 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 $7g$ |
| | **(3)** | |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| KE gained $+$ WD $=$ loss in GPE | M1 | The question requires use of work-energy. Alternative methods score 0/4. Requires all terms but condone sign errors (must be considering both particles) |
| $\dfrac{1}{2} \times 4v^2 + \dfrac{1}{2} \times 2v^2 + (\text{their (a)}) = (\text{their (b)})$ | A2 | Correct unsimplified. $-1$ each error |
| $3v^2 = 6g$ | | |
| $v = \sqrt{2g} = 4.43\ (\text{m s}^{-1})$ | A1 | or 4.4. Accept $\sqrt{2g}$ |
| | **(4)** | |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equations of motion for each particle leading to $T = \dfrac{12g}{5} = 23.52$ followed by W-E equation for $P$: $2.5T = \dfrac{1}{2} \times 2v^2 + 2g \times 2.5\sin\alpha + (a)$ | M1A2 | Equations of motion for each particle leading to $T = \dfrac{12g}{5} = 23.52$ followed by W-E equation for $Q$: $\dfrac{1}{2} \times 4v^2 + 2.5T = 4g \times 2.5$ |
| $v = \sqrt{2g} = 4.43\ (\text{m s}^{-1})$ | A1 | |
| | **(11 marks)** | Use of $\alpha = 36.9$ gives correct answers to 3 sf. Use of $\alpha = 37$ gives correct answers to 2 sf only; A0 if 3 sf given. |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f30ed5b8-880e-42de-860e-d1538fa68f11-12_540_1116_251_342}
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\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 $B$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q4 [11]}}