| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 10 |
| 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 +0.3 This is a straightforward elastic string equilibrium problem requiring standard application of Hooke's law, Pythagoras, and resolving forces vertically. The symmetry argument in part (a) is routine, and all calculations follow directly from standard M3 techniques with no novel problem-solving required. Slightly easier than average due to the clear setup and step-by-step structure. |
| Spec | 3.03h Newton's third law: action-reaction pairs3.03m Equilibrium: sum of resolved forces = 06.02g 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 |
|---|---|
| (a) by symmetry | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| by Pythag. distance of object from A \(= 1.25\) | M1 | |
| \(T = \frac{F}{r} = \frac{280 \times 0.25}{?} = 70\) | M1 A1 | |
| resolve \(\uparrow\): \(2T \sin \alpha - mg = 0\) | M1 A1 | |
| \(\sin \alpha = \frac{?}{?} \therefore 2 \times 70 \times \frac{7}{25} = m \times 9.8\) | M1 | |
| giving \(m = 4 \text{ kg}\) | A1 | |
| (c) EPE \(= 2 \times \frac{\Delta l^2}{l} = 2 \times \frac{280 \times 0.25^2}{2 \times l} = 17.5 \text{ J}\) | M1 A1 | (10) |
(a) by symmetry | B1 |
(b) [Diagram showing configuration with distances 1.2, 1.2, 0.35 from A to B through T, with mg below]
by Pythag. distance of object from A $= 1.25$ | M1 |
$T = \frac{F}{r} = \frac{280 \times 0.25}{?} = 70$ | M1 A1 |
resolve $\uparrow$: $2T \sin \alpha - mg = 0$ | M1 A1 |
$\sin \alpha = \frac{?}{?} \therefore 2 \times 70 \times \frac{7}{25} = m \times 9.8$ | M1 |
giving $m = 4 \text{ kg}$ | A1 |
(c) EPE $= 2 \times \frac{\Delta l^2}{l} = 2 \times \frac{280 \times 0.25^2}{2 \times l} = 17.5 \text{ J}$ | M1 A1 | (10)5. A physics student is set the task of finding the mass of an object without using a set of scales. She decides to use a light elastic string of natural length 2 m and modulus of elasticity 280 N attached to two points $A$ and $B$ which are on the same horizontal level and 2.4 m apart.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{00776cc0-0214-4029-8ef1-c1cba89f4b87-3_307_1072_993_438}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
She attaches the object to the midpoint of the string so that it hangs in equilibrium 0.35 m below $A B$ as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Explain why it is reasonable to assume that the tensions in each half of the string are equal.
\item Find the mass of the object.
\item Find the elastic potential energy of the string when the object is suspended from it.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [10]}}