| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Standard +0.3 This is a standard M3 elastic string problem requiring application of work-energy principle with friction on an inclined plane. The question guides students through parts (a) and (b) before asking them to combine results in part (c). While it involves multiple energy components (elastic PE, gravitational PE, friction work), the structure is typical and the calculations are straightforward once the correct approach is identified. Slightly easier than average due to the scaffolded nature. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02g 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 |
|---|---|---|
| (a) resolve \(\curvearrowleft\): \(R - 2g\cos 30° = 0 \therefore R = \sqrt{3}g\) | M1 A1 | |
| friction = \(\mu R = \frac{1}{6}\sqrt{3} \times \sqrt{3}g = \frac{1}{2}g\) | A1 | |
| work done = \(Fs = \frac{1}{2} \times 9.8 \times 2.2 = 10.78 \text{ J}\) | M1 A1 | |
| (b) change in GPE = \(mgh = 2 \times 9.8 \times 2.2\sin 30° = 21.56 \text{ J (loss)}\) | M1 A1 | |
| (c) work done = loss of GPE - gain of EPE | M1 | |
| \(\therefore 10.78 = 21.56 - \frac{\lambda x^2}{2x}\) | M1 A1 | |
| giving \(\frac{\lambda60 \times 2}{2 \times 15} = 10.78\) so \(\lambda = 66 \text{ N}\) | M1 A1 | (12) |
**(a)** resolve $\curvearrowleft$: $R - 2g\cos 30° = 0 \therefore R = \sqrt{3}g$ | M1 A1 |
friction = $\mu R = \frac{1}{6}\sqrt{3} \times \sqrt{3}g = \frac{1}{2}g$ | A1 |
work done = $Fs = \frac{1}{2} \times 9.8 \times 2.2 = 10.78 \text{ J}$ | M1 A1 |
**(b)** change in GPE = $mgh = 2 \times 9.8 \times 2.2\sin 30° = 21.56 \text{ J (loss)}$ | M1 A1 |
**(c)** work done = loss of GPE - gain of EPE | M1 |
$\therefore 10.78 = 21.56 - \frac{\lambda x^2}{2x}$ | M1 A1 |
giving $\frac{\lambda60 \times 2}{2 \times 15} = 10.78$ so $\lambda = 66 \text{ N}$ | M1 A1 | **(12)**
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e0668f31-4b72-4dfd-9cf7-470acef0bfdb-3_316_536_1087_639}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A particle $P$ of mass 2 kg is attached to one end of a light elastic string of natural length 1.5 m and modulus of elasticity $\lambda$. The other end of the string is fixed to a point $A$ on a rough plane inclined at an angle of $30 ^ { \circ }$ to the horizontal as shown in Figure 2. The coefficient of friction between $P$ and the plane is $\frac { 1 } { 6 } \sqrt { 3 }$.\\
$P$ is held at rest at $A$ and then released. It first comes to instantaneous rest at the point $B , 2.2 \mathrm {~m}$ from $A$. For the motion of $P$ from $A$ to $B$,
\begin{enumerate}[label=(\alph*)]
\item show that the work done against friction is 10.78 J ,
\item find the change in the gravitational potential energy of $P$.
By using the work-energy principle, or otherwise,
\item find $\lambda$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4 [12]}}