| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Standard +0.3 This is a standard M3 elastic string equilibrium problem requiring resolution of forces in two directions and application of Hooke's law. The given angle (tan α = 4/3 means sin α = 4/5, cos α = 3/5) makes calculations straightforward, and part (a) is a 'show that' which guides students to the answer. Slightly easier than average due to the structured nature and standard technique application. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| \[T = \frac{\lambda a}{4a}\] | B1 | Award for correct expression for tension in terms of \(\lambda\) and \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T\cos\alpha = mg\) | M1A1 | Resolve vertically. Must be dimensionally correct with correct number of terms, condone sign errors and sin/cos confusion. Eg: Parallel to string: \(T = kmg\sin\alpha + mg\cos\alpha\); Triangle of forces: \(T = \sqrt{(mg)^2 + (kmg)^2}\) |
| \(\frac{\lambda a}{4a} \times \frac{3}{5} = mg \Rightarrow \lambda = \frac{20mg}{3}\) | A1* | (4) Given answer from complete correct working. Must include a line of working before reaching given answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T\sin\alpha = kmg\) | M1A1 | Resolve horizontally. Must be dimensionally correct, condone sign errors and sin/cos confusion. Eg: Perp to string: \(mg\sin\alpha = kmg\cos\alpha\); Triangle of forces: \(\tan\alpha = \frac{kmg}{mg} = \frac{4}{3}\) |
| \(\frac{20mg}{3} \times \frac{1}{4} \times \frac{4}{5} = kmg\) | M1 | Complete method to produce equation in \(k\) only (replace \(T\) and trig) |
| \(k = \frac{4}{3}\) | A1 | (4) Any equivalent fraction. Accept 1.3 or better |
## Question 1:
### Part (a):
$$T = \frac{\lambda a}{4a}$$ | B1 | Award for correct expression for tension in terms of $\lambda$ and $a$
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*Note: Only one question/part is visible in the provided pages of the mark scheme. The document appears to be cut off after Question 1(a). Further questions would need additional pages to be shared.*
## Question 1:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T\cos\alpha = mg$ | M1A1 | Resolve vertically. Must be dimensionally correct with correct number of terms, condone sign errors and sin/cos confusion. Eg: Parallel to string: $T = kmg\sin\alpha + mg\cos\alpha$; Triangle of forces: $T = \sqrt{(mg)^2 + (kmg)^2}$ |
| $\frac{\lambda a}{4a} \times \frac{3}{5} = mg \Rightarrow \lambda = \frac{20mg}{3}$ | A1* | (4) Given answer from complete correct working. Must include a line of working before reaching given answer. |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T\sin\alpha = kmg$ | M1A1 | Resolve horizontally. Must be dimensionally correct, condone sign errors and sin/cos confusion. Eg: Perp to string: $mg\sin\alpha = kmg\cos\alpha$; Triangle of forces: $\tan\alpha = \frac{kmg}{mg} = \frac{4}{3}$ |
| $\frac{20mg}{3} \times \frac{1}{4} \times \frac{4}{5} = kmg$ | M1 | Complete method to produce equation in $k$ only (replace $T$ and trig) |
| $k = \frac{4}{3}$ | A1 | (4) Any equivalent fraction. Accept 1.3 or better |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{176ae8f8-7de9-4993-825a-6067614526ae-02_314_677_296_696}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A light elastic string $A B$ has natural length $4 a$ and modulus of elasticity $\lambda$. The end $A$ is attached to a fixed point and the end $B$ is attached to a particle of mass $m$. The particle is held in equilibrium, with the string stretched, by a horizontal force of magnitude $k m g$.\\
The line of action of the horizontal force lies in the vertical plane containing the elastic string.\\
The string $A B$ makes an angle $\alpha$ with the vertical, where $\tan \alpha = \frac { 4 } { 3 }$\\
With the particle in this position, $A B = 5 a$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = \frac { 20 m g } { 3 }$
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2024 Q1 [8]}}