| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with inextensible strings or gravity only |
| Difficulty | Challenging +1.2 This is a standard M4 potential energy question requiring systematic application of PE = mgh, geometric relationships (using cosine rule or coordinates), differentiation for equilibrium, and second derivative test for stability. While it involves multiple steps and careful geometry, it follows a well-established template for this topic with no novel insights required. The algebra is moderately involved but routine for Further Maths students. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| The length of string from \(P\) to \(R\) is \(\sqrt{x^2 + d^2}\) | B1 | Correct expression for PR |
| The length of string from \(P\) to the mass \(3m\) = total string length \(- \sqrt{x^2+d^2}\) | M1 | Using total string length |
| Height of \(3m\) below \(P\) = total string \(- \sqrt{x^2+d^2}\) | A1 | Correct height expression |
| PE of system \(= -3mg(\text{total} - \sqrt{x^2+d^2}) - mgx\) \(= 3mg\sqrt{x^2+d^2} - mgx + \text{constant}\) | A1 | Fully correct derivation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dx} = \frac{3mgx}{\sqrt{x^2+d^2}} - mg = 0\) | M1 | Differentiating and setting to zero |
| \(3x = \sqrt{x^2 + d^2}\) | M1 | Rearranging |
| \(9x^2 = x^2 + d^2 \Rightarrow x = \frac{d}{2\sqrt{2}}\) | A1 | Correct value of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2V}{dx^2} = \frac{3mg \cdot d^2}{(x^2+d^2)^{3/2}}\) | M1 | Correct second derivative |
| Substituting equilibrium value | M1 | Substitution |
| \(\frac{d^2V}{dx^2} > 0\), therefore stable equilibrium | A1 | Correct conclusion |
## Question 4:
**(a)** Show potential energy expression
| The length of string from $P$ to $R$ is $\sqrt{x^2 + d^2}$ | B1 | Correct expression for PR |
| The length of string from $P$ to the mass $3m$ = total string length $- \sqrt{x^2+d^2}$ | M1 | Using total string length |
| Height of $3m$ below $P$ = total string $- \sqrt{x^2+d^2}$ | A1 | Correct height expression |
| PE of system $= -3mg(\text{total} - \sqrt{x^2+d^2}) - mgx$ $= 3mg\sqrt{x^2+d^2} - mgx + \text{constant}$ | A1 | Fully correct derivation |
**(b)** Find $x$ at equilibrium
| $\frac{dV}{dx} = \frac{3mgx}{\sqrt{x^2+d^2}} - mg = 0$ | M1 | Differentiating and setting to zero |
| $3x = \sqrt{x^2 + d^2}$ | M1 | Rearranging |
| $9x^2 = x^2 + d^2 \Rightarrow x = \frac{d}{2\sqrt{2}}$ | A1 | Correct value of $x$ |
**(c)** Stability
| $\frac{d^2V}{dx^2} = \frac{3mg \cdot d^2}{(x^2+d^2)^{3/2}}$ | M1 | Correct second derivative |
| Substituting equilibrium value | M1 | Substitution |
| $\frac{d^2V}{dx^2} > 0$, therefore **stable** equilibrium | A1 | Correct conclusion |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cf941854-3a33-4d9d-9fa0-ce9a63227599-08_479_807_246_571}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A light inextensible string of length $2 a$ has one end attached to a fixed point $A$. The other end of the string is attached to a particle $P$ of mass $m$. A second light inextensible string of length $L$, where $L > \frac { 12 a } { 5 }$, has one of its ends attached to $P$ and passes over a small smooth peg fixed at a point $B$. The line $A B$ is horizontal and $A B = 2 a$. The other end of the second string is attached to a particle of mass $\frac { 7 } { 20 } m$, which hangs vertically below $B$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy of the system, when the angle $P A B = 2 \theta$, is
$$\frac { 1 } { 5 } m g a ( 7 \sin \theta - 10 \sin 2 \theta ) + \text { constant. }$$
\item Show that there is only one value of $\cos \theta$ for which the system is in equilibrium and find this value.
\item Determine the stability of the position of equilibrium.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 Q4 [10]}}