Edexcel M2 2016 January — Question 4 11 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2016
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeConnected particles pulley energy method
DifficultyStandard +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.
Spec3.03o 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

4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e6d100ff-dd4a-4591-a0a3-81761773045e-07_544_1264_251_338} \captionsetup{labelformat=empty} \caption{Figure 1}
\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 \(\frac { 1 } { 4 }\).
  1. Find the work done against friction as \(P\) moves from \(A\) to \(B\).
  2. Find the total potential energy lost by the system as \(P\) moves from \(A\) to \(B\).
  3. Find, using the work-energy principle, the speed of \(P\) as it passes through \(B\).

Question 4:
Part 4a:
AnswerMarks Guidance
Working/AnswerMarks 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:
AnswerMarks Guidance
Working/AnswerMarks 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:
AnswerMarks Guidance
Working/AnswerMarks Guidance
KE gained \(+\) WD \(=\) loss in GPEM1 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]}}