| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Standard +0.3 This is a standard M3 equilibrium problem requiring resolution of forces in two directions and application of Hooke's law (T = λx/l). The setup is straightforward with tan α = 3/4 giving sin α = 3/5 and cos α = 4/5. Students resolve horizontally (T = F sin α) and vertically (F cos α = mg), eliminate F to find T = 3mg/4, then use Hooke's law to find extension and hence length. While it requires multiple steps, it follows a well-practiced routine with no novel insight needed, making it slightly easier than average. |
| Spec | 6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F\cos\alpha = mg\) | M1, A1 | Resolve vertically or horizontally, correct no. of terms, condone sign errors and sin/cos confusion |
| \(F\sin\alpha = T\) | A1 | Correct horizontal equation (A2 for \(T=mg\tan\alpha\) from triangle of forces) |
| \(T=\frac{2mgx}{l}\) or \(T=\frac{2mg(AB-l)}{l}\) | M1 | Hooke's Law. Must be an extension not \(AB\) |
| \(\frac{3}{4}mg=\frac{2mgx}{l}\) | dM1 | Substitute trig to produce equation in \(x\) (and \(l\)) only, dependent on previous M's and having two equations |
| \(AB=\frac{11l}{8}\) | A1 | Cao. Accept \(1.375l\), \(1.4l\), \(1.38l\) |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F\cos\alpha = mg$ | M1, A1 | Resolve vertically or horizontally, correct no. of terms, condone sign errors and sin/cos confusion |
| $F\sin\alpha = T$ | A1 | Correct horizontal equation (A2 for $T=mg\tan\alpha$ from triangle of forces) |
| $T=\frac{2mgx}{l}$ or $T=\frac{2mg(AB-l)}{l}$ | M1 | Hooke's Law. Must be an extension not $AB$ |
| $\frac{3}{4}mg=\frac{2mgx}{l}$ | dM1 | Substitute trig to produce equation in $x$ (and $l$) only, dependent on previous M's and having two equations |
| $AB=\frac{11l}{8}$ | A1 | Cao. Accept $1.375l$, $1.4l$, $1.38l$ |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-04_252_842_285_609}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A light elastic string AB has natural length I and modulus of elasticity 2 mg .\\
The end A of the elastic string is attached to a fixed point. The other end B is attached to a particle of mass m . The particle is held in equilibrium, with the elastic string taut and horizontal, by a force of magnitude F . The line of action of the force and the elastic string lie in the same vertical plane. The direction of the force makes an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$, with the upward vertical, as shown in Figure 2.\\
Find, in terms of I , the length AB .\\
\includegraphics[max width=\textwidth, alt={}, center]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-04_2264_53_311_1981}
\hfill \mbox{\textit{Edexcel M3 2023 Q2 [6]}}