| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Challenging +1.2 This is a multi-part M3 elastic string problem requiring energy methods and force analysis. Part (a) uses work-energy principle with friction and elastic PE (standard technique), part (b) requires finding where acceleration=0 to locate maximum speed (requires insight but follows from M3 syllabus), and part (c) needs force analysis at rest position. More demanding than typical M1/M2 questions due to elastic strings and multi-step reasoning, but follows standard M3 patterns without requiring novel approaches. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02h Elastic PE: 1/2 k x^26.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| EPE Gain \(= \frac{2mgx^2}{2a}\) | B1 | B1 correct expression |
| PE loss \(= mgx\sin\alpha\) | B1 | B1 correct expression |
| WD against friction \(= \mu mg\cos\alpha \times x\) | B1 | B1 correct expression |
| \(\mu mg\cos\alpha \times x = mgx\sin\alpha - \frac{2mgx^2}{2a}\) | M1 | M1 for energy equation dim correct with correct terms, condone sign errors |
| \(x = AB = a(\sin\alpha - \mu\cos\alpha)\) * | A1* | A1* for given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\mu mg\cos\alpha \times y = mgy\sin\alpha - \frac{1}{2}mv^2 - \frac{2mgy^2}{2a}\) | M1 A2,1,0 | M1 for energy equation dim correct with correct terms, condone sign errors; A2 for correct unsimplified equation, A1 for correct unsimplified equation with at most one error |
| At max speed: \(\mu mg\cos\alpha = mg\sin\alpha - \frac{2mgy}{a}\) | M1 | M1 for finding resultant force parallel to plane and equating to 0, or differentiating energy equation wrt \(y\) and equating \(\frac{dv}{dy}\) to 0 |
| \(y = \frac{1}{10}a\) | A1 | A1 for correct value of \(y\) |
| Use their \(y\) value to find the max speed | M1 | M1 for using their \(y\) value to find max speed |
| \(v = \sqrt{\frac{ag}{50}}\) oe | A1 | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| At \(B\), nett force down plane \(= \frac{3}{5}mg - \frac{2mgx}{a} = \frac{1}{5}mg\) | M1 | M1 for finding nett force up or down plane, correct terms, condone sign errors |
| Max friction available \(= \frac{1}{2} \times mg \times \frac{4}{5} = \frac{2}{5}mg\) | B1 | B1 for max friction |
| Hence, friction \(= \frac{1}{5}mg\) up and \(P\) remains at \(B\) | A1 | A1 correct conclusion and justification |
## Question 7:
### Part 7(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| EPE Gain $= \frac{2mgx^2}{2a}$ | B1 | B1 correct expression |
| PE loss $= mgx\sin\alpha$ | B1 | B1 correct expression |
| WD against friction $= \mu mg\cos\alpha \times x$ | B1 | B1 correct expression |
| $\mu mg\cos\alpha \times x = mgx\sin\alpha - \frac{2mgx^2}{2a}$ | M1 | M1 for energy equation dim correct with correct terms, condone sign errors |
| $x = AB = a(\sin\alpha - \mu\cos\alpha)$ * | A1* | A1* for given answer correctly obtained |
**(5 marks)**
### Part 7(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\mu mg\cos\alpha \times y = mgy\sin\alpha - \frac{1}{2}mv^2 - \frac{2mgy^2}{2a}$ | M1 A2,1,0 | M1 for energy equation dim correct with correct terms, condone sign errors; A2 for correct unsimplified equation, A1 for correct unsimplified equation with at most one error |
| At max speed: $\mu mg\cos\alpha = mg\sin\alpha - \frac{2mgy}{a}$ | M1 | M1 for finding resultant force parallel to plane and equating to 0, or differentiating energy equation wrt $y$ and equating $\frac{dv}{dy}$ to 0 |
| $y = \frac{1}{10}a$ | A1 | A1 for correct value of $y$ |
| Use their $y$ value to find the max speed | M1 | M1 for using their $y$ value to find max speed |
| $v = \sqrt{\frac{ag}{50}}$ oe | A1 | A1 cao |
**(7 marks)**
### Part 7(c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| At $B$, nett force down plane $= \frac{3}{5}mg - \frac{2mgx}{a} = \frac{1}{5}mg$ | M1 | M1 for finding nett force up or down plane, correct terms, condone sign errors |
| Max friction available $= \frac{1}{2} \times mg \times \frac{4}{5} = \frac{2}{5}mg$ | B1 | B1 for max friction |
| Hence, friction $= \frac{1}{5}mg$ up and $P$ remains at $B$ | A1 | A1 correct conclusion and justification |
**(3 marks)**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a1365c54-4910-449b-b270-c56c1bc5a751-24_396_992_246_539}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length $a$ and modulus of elasticity 2 mg . The other end of the string is attached to a fixed point $O$ on a rough plane which is inclined to the horizontal at an angle $\alpha$
The string lies along a line of greatest slope of the plane.\\
The particle $P$ is held at rest on the plane at the point $A$, where $O A = a$, as shown in Figure 5.
The particle $P$ is released from $A$ and slides down the plane, coming to rest at the point $B$. The coefficient of friction between $P$ and the plane is $\mu$, where $\mu < \tan \alpha$ Air resistance is modelled as being negligible.
\begin{enumerate}[label=(\alph*)]
\item Show that $A B = a ( \sin \alpha - \mu \cos \alpha )$.
Given that $\tan \alpha = \frac { 3 } { 4 }$ and $\mu = \frac { 1 } { 2 }$
\item find, in terms of $a$ and $g$, the maximum speed of $P$ as it moves from $A$ to $B$
\item Describe the motion of $P$ after it reaches $B$, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2022 Q7 [15]}}