| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Energy methods for pulley systems |
| Difficulty | Standard +0.3 This is a standard M2 pulley system question using energy methods. Part (a) requires calculating potential energy changes for two particles (straightforward with given angle), and part (b) applies the work-energy principle including friction work. While it involves multiple components (PE changes, friction work, KE), these are routine M2 techniques with no novel insight required. The tan α = 3/4 simplifies calculations. Slightly easier than average due to clear structure and standard methodology. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02d 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 | Marks | Guidance |
| PE lost \(= 2mgh - mgh\sin\alpha \left(= \frac{7mgh}{5}\right)\) | M1 A1 | Two term expression for PE lost; condone sign errors and sin/cos confusion, but must be vertical distance moved. A1: both terms correct, \(\sin\alpha\) correct, need not be simplified. Allow 13.72\(mh\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Normal reaction \(R = mg\cos\alpha \left(= \frac{4mg}{5}\right)\) | B1 | Normal reaction between A and plane. Must use \(\cos\alpha\) but need not be substituted |
| Work-energy: \(\frac{1}{2}mv^2 + \frac{1}{2}\cdot 2mv^2 = \frac{7mgh}{5} - \frac{5}{8}\cdot\frac{4mg}{5}\cdot h\) | M1 A2,1,0 | M1(NB question specifies work & energy): PE lost = work done against friction plus KE gained; condone sign errors; must include KE of both particles. A1A1 all three elements correct (including signs); A1A0 two elements correct |
| \(\Rightarrow \frac{3}{2}mv^2 = \frac{9mgh}{10} \Rightarrow v^2 = \frac{3}{5}gh\) | A1 | \(V^2\) correct (NB \(kgh\) specified in Q) |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| PE lost $= 2mgh - mgh\sin\alpha \left(= \frac{7mgh}{5}\right)$ | M1 A1 | Two term expression for PE lost; condone sign errors and sin/cos confusion, but must be vertical distance moved. A1: both terms correct, $\sin\alpha$ correct, need not be simplified. Allow 13.72$mh$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Normal reaction $R = mg\cos\alpha \left(= \frac{4mg}{5}\right)$ | B1 | Normal reaction between A and plane. Must use $\cos\alpha$ but need not be substituted |
| Work-energy: $\frac{1}{2}mv^2 + \frac{1}{2}\cdot 2mv^2 = \frac{7mgh}{5} - \frac{5}{8}\cdot\frac{4mg}{5}\cdot h$ | M1 A2,1,0 | M1(NB question specifies work & energy): PE lost = work done against friction plus KE gained; condone sign errors; **must include KE of both particles**. A1A1 all three elements correct (including signs); A1A0 two elements correct |
| $\Rightarrow \frac{3}{2}mv^2 = \frac{9mgh}{10} \Rightarrow v^2 = \frac{3}{5}gh$ | A1 | $V^2$ correct (NB $kgh$ specified in Q) |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{778a0276-6738-40e6-90b2-a536ce5abe6a-06_330_1118_203_411}
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\caption{Figure 2}
\end{center}
\end{figure}
Two particles $A$ and $B$, of mass $m$ and $2 m$ respectively, are attached to the ends of a light inextensible string. The particle $A$ lies on a rough plane inclined at an angle $\alpha$ to the horizontal, where tan $\alpha = \frac { 3 } { 4 }$. The string passes over a small light smooth pulley $P$ fixed at the top of the plane. The particle $B$ hangs freely below $P$, as shown in Figure 2. The particles are released from rest with the string taut and the section of the string from $A$ to $P$ parallel to a line of greatest slope of the plane. The coefficient of friction between $A$ and the plane is $\frac { 5 } { 8 }$. When each particle has moved a distance $h , B$ has not reached the ground and $A$ has not reached $P$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the potential energy lost by the system when each particle has moved a distance $h$.
When each particle has moved a distance $h$, they are moving with speed $v$. Using the workenergy principle,
\item find an expression for $v ^ { 2 }$, giving your answer in the form $k g h$, where $k$ is a number.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q4 [7]}}