| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: vertical motion |
| Difficulty | Standard +0.8 This M3 question requires multiple techniques: resolving forces at equilibrium with geometry (finding string lengths and angles), applying Hooke's law to find the modulus, then using energy conservation to find elastic PE at maximum speed. The geometric setup with the bead at midpoint and the energy method for part (c) require careful reasoning beyond routine mechanics, though the individual steps are standard M3 content. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\cos\theta = \frac{4}{5}\) | B1 | Seen or implied |
| \(2T\cos\theta = 12\) | M1A1 | Resolving vertically with 2 equal tensions (implied) and weight; correct equation |
| \(2T\left(\frac{4}{5}\right) = 12 \Rightarrow T = \frac{60}{8} = 7.5\text{ (N)}\) | A1 | \(T = \frac{15}{2} = 7.5\) N; accept any equivalent fraction since \(g\) not used |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{Ext} = 2(2.5) - 3 = 2\text{ m}\) | — | — |
| \(7.5 = \frac{\lambda \times 2}{3} \Rightarrow \lambda = 11.25\text{ (N)}\) | M1A1* | Use Hooke's Law with their tension; must use natural length 3 (or 1.5 for half string); condone incorrect extension for M mark; 11.25 or any equivalent fraction |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{EPE} = \frac{11.25 \times 2^2}{2 \times 3} = 7.5\text{ (J)}\) | M1A1 | Attempt EPE in equilibrium position; same extension as (b); condone missing half in formula; 7.5 J, must be clear answer (not embedded) |
## Question 2(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\cos\theta = \frac{4}{5}$ | B1 | Seen or implied |
| $2T\cos\theta = 12$ | M1A1 | Resolving vertically with 2 equal tensions (implied) and weight; correct equation |
| $2T\left(\frac{4}{5}\right) = 12 \Rightarrow T = \frac{60}{8} = 7.5\text{ (N)}$ | A1 | $T = \frac{15}{2} = 7.5$ N; accept any equivalent fraction since $g$ not used |
**(4)**
## Question 2(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Ext} = 2(2.5) - 3 = 2\text{ m}$ | — | — |
| $7.5 = \frac{\lambda \times 2}{3} \Rightarrow \lambda = 11.25\text{ (N)}$ | M1A1* | Use Hooke's Law with their tension; must use natural length 3 (or 1.5 for half string); condone incorrect extension for M mark; 11.25 or any equivalent fraction |
**(2)**
## Question 2(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{EPE} = \frac{11.25 \times 2^2}{2 \times 3} = 7.5\text{ (J)}$ | M1A1 | Attempt EPE in equilibrium position; same extension as (b); condone missing half in formula; 7.5 J, must be clear answer (not embedded) |
**(2) [8]**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ace84823-db30-463e-b24b-f0cd7df73746-04_542_831_301_552}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A smooth bead of weight 12 N is threaded onto a light elastic string of natural length 3 m . The points $A$ and $B$ are on a horizontal ceiling, with $A B = 3 \mathrm {~m}$. One end of the string is attached to $A$ and the other end of the string is attached to $B$.
The bead hangs freely in equilibrium, 2 m below the ceiling, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string.
\item Show that the modulus of elasticity of the string is 11.25 N .
The bead is now pulled down to a point vertically below its equilibrium position and released from rest.
\item Find the elastic energy stored in the string at the instant when the bead is moving at its maximum speed.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2020 Q2 [8]}}