| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on smooth inclined plane |
| Difficulty | Standard +0.3 This is a standard M3 elastic string energy problem with straightforward application of conservation of energy. Part (a) requires setting up energy equation between two rest positions (routine for this topic), and part (b) requires finding maximum speed by differentiating or recognizing it occurs at equilibrium position. The inclined plane at 30° and the given extension values make calculations clean, and the problem follows a very typical textbook pattern for this module. |
| 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 | Marks | Guidance |
| \(mg\frac{l}{4}\sin 30° = \frac{\lambda}{2l}\left(\frac{l}{4}\right)^2\) | M1A1A1 | Attempt energy equation from C to D. Must use vertical height for PE. EPE must have form \(kx^2\). Must have 1 PE term and 1 EPE term. A1 correct loss of PE, A1 correct final EPE |
| \(\lambda = 4mg\) | A1* (4) | Correct answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Greatest speed when acceleration of \(B\) is zero | ||
| \((\nearrow)\) \(T = mg\sin 30° = \frac{4mge}{l}\) | M1 | Resolve along plane using HL to find \(T\) |
| \(e = \frac{l}{8}\) | A1 | Correct value for extension |
| \(\frac{1}{2}mv^2 + \frac{4mg}{2l}\left(\frac{l}{8}\right)^2 = mg\frac{l}{8}\sin 30°\) | M1A1A1 | Form energy equation. M0 if \(l/4\) used for extension. Must have 1 PE term, 1 KE term, 1 EPE term. A1 two correct terms, A1 completely correct equation |
| \(v = \sqrt{\left(\frac{gl}{16}\right)} = \frac{\sqrt{gl}}{4}\) | DM1A1 (7) | Solve for \(v\), dependent on previous M. Correct expression for \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Correct value for \(e\) | M1A1 | May be embedded in complete method |
| \(\int g\sin 30 - \frac{4gx}{l}\,dx = \int v\,dv\) leading to \(\frac{gx}{2} - \frac{2gx^2}{l} = \frac{v^2}{2} + c\) | M1 | Uses F=ma, correct number of terms, weight resolved |
| Correct integration with at most one slip/error; completely correct but \(c\) may be missing | A1, A1 | |
| Find \(c\) (when \(x = \frac{l}{4}\), \(v=0\) gives \(c=0\)) and sub in \(e\) to find expression for \(v\) | DM1, A1 | Correct expression for \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Correct value for \(e\) | M1A1 | May be embedded in complete method |
| Correctly uses F=ma to show motion is SHM; correct proof | M1A1, A1 | |
| Uses \(v = a\omega\) to find expression for \(v\); correct expression | M1, A1 |
# Question 4:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $mg\frac{l}{4}\sin 30° = \frac{\lambda}{2l}\left(\frac{l}{4}\right)^2$ | M1A1A1 | Attempt energy equation from C to D. Must use vertical height for PE. EPE must have form $kx^2$. Must have 1 PE term and 1 EPE term. A1 correct loss of PE, A1 correct final EPE |
| $\lambda = 4mg$ | A1* (4) | Correct answer correctly obtained |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Greatest speed when acceleration of $B$ is zero | | |
| $(\nearrow)$ $T = mg\sin 30° = \frac{4mge}{l}$ | M1 | Resolve along plane using HL to find $T$ |
| $e = \frac{l}{8}$ | A1 | Correct value for extension |
| $\frac{1}{2}mv^2 + \frac{4mg}{2l}\left(\frac{l}{8}\right)^2 = mg\frac{l}{8}\sin 30°$ | M1A1A1 | Form energy equation. M0 if $l/4$ used for extension. Must have 1 PE term, 1 KE term, 1 EPE term. A1 two correct terms, A1 completely correct equation |
| $v = \sqrt{\left(\frac{gl}{16}\right)} = \frac{\sqrt{gl}}{4}$ | DM1A1 (7) | Solve for $v$, dependent on previous M. Correct expression for $v$ |
**ALT 1 (Integration):**
| Working | Marks | Guidance |
|---------|-------|----------|
| Correct value for $e$ | M1A1 | May be embedded in complete method |
| $\int g\sin 30 - \frac{4gx}{l}\,dx = \int v\,dv$ leading to $\frac{gx}{2} - \frac{2gx^2}{l} = \frac{v^2}{2} + c$ | M1 | Uses F=ma, correct number of terms, weight resolved |
| Correct integration with at most one slip/error; completely correct but $c$ may be missing | A1, A1 | |
| Find $c$ (when $x = \frac{l}{4}$, $v=0$ gives $c=0$) and sub in $e$ to find expression for $v$ | DM1, A1 | Correct expression for $v$ |
**ALT 2 (SHM):**
| Working | Marks | Guidance |
|---------|-------|----------|
| Correct value for $e$ | M1A1 | May be embedded in complete method |
| Correctly uses F=ma to show motion is SHM; correct proof | M1A1, A1 | |
| Uses $v = a\omega$ to find expression for $v$; correct expression | M1, A1 | |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e837bb9-4ada-4f0f-8b21-2730611335f2-12_357_737_260_664}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
One end of a light elastic string, of natural length $l$ and modulus of elasticity $\lambda$, is fixed to a point $A$ on a smooth plane. The plane is inclined at $30 ^ { \circ }$ to the horizontal.
A small ball $B$ of mass $m$ is attached to the other end of the elastic string. Initially, $B$ is held at rest at the point $C$ on the plane with the elastic string lying along a line of greatest slope of the plane.
The point $C$ is below $A$ and $A C = l$, as shown in Figure 2 .
The ball is released and comes to instantaneous rest at a point $D$ on the plane.\\
The points $A , C$ and $D$ all lie along a line of greatest slope of the plane and $A D = \frac { 5 l } { 4 }$\\
The ball is modelled as a particle and air resistance is modelled as being negligible.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item show that $\lambda = 4 \mathrm { mg }$
\item find, in terms of $g$ and $l$, the greatest speed of $B$ as it moves from $C$ to $D$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2022 Q4 [11]}}