| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string vertical motion |
| Difficulty | Standard +0.8 This is a multi-stage elastic string problem requiring work-energy principle with air resistance, involving careful tracking of elastic PE changes and finding equilibrium position for maximum speed. Part (a) requires setting up energy equation with extension calculations; part (b) needs recognizing that maximum speed occurs where net force is zero. More demanding than standard M3 questions due to the resistance modeling and the two-part structure requiring different approaches. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| WD against air resistance \(= kmga\); PE Gain \(= \frac{1}{2}mga\); KE Gain \(= \frac{1}{2} \times \frac{1}{2}m \times 3ag\) | B2,1,0 | B2 for all 3 unsimplified terms; B1B0 for 2 out of 3 correct |
| Initial EPE \(= \frac{2mg}{4a}(2a)^2\); Final EPE \(= \frac{2mg}{4a}a^2\) | B1; B1 | B1 for initial EPE; B1 for final EPE |
| \(kmga = \frac{2mg}{4a}\left((2a)^2 - a^2\right) - \frac{1}{2}mga - \frac{1}{2} \times \frac{1}{2}m \times 3ag\) | M1A1 | M1 for work-energy equation with all necessary terms; condone sign errors |
| \(k = \frac{1}{4}\) | A1* (7) | Given answer correctly obtained; at least one step of working to be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{2}mg - \frac{1}{4}mg - T = 0\) | M1 | Vertical resolution with correct terms; must have acceleration \(= 0\); \(T\) does not need to be substituted |
| \(\frac{1}{2}mg - \frac{1}{4}mg - \frac{2mg}{2a}x = 0\) | A1 | Correct equation with \(T\) replaced |
| \(x = \frac{1}{4}a\) | A1 | cao |
| \(OB = \frac{9a}{4}\) | A1ft (4) | ft \(2a +\) their \(x\) |
# Question 4:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| WD against air resistance $= kmga$; PE Gain $= \frac{1}{2}mga$; KE Gain $= \frac{1}{2} \times \frac{1}{2}m \times 3ag$ | B2,1,0 | B2 for all 3 unsimplified terms; B1B0 for 2 out of 3 correct |
| Initial EPE $= \frac{2mg}{4a}(2a)^2$; Final EPE $= \frac{2mg}{4a}a^2$ | B1; B1 | B1 for initial EPE; B1 for final EPE |
| $kmga = \frac{2mg}{4a}\left((2a)^2 - a^2\right) - \frac{1}{2}mga - \frac{1}{2} \times \frac{1}{2}m \times 3ag$ | M1A1 | M1 for work-energy equation with all necessary terms; condone sign errors |
| $k = \frac{1}{4}$ | A1* (7) | Given answer correctly obtained; at least one step of working to be seen |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mg - \frac{1}{4}mg - T = 0$ | M1 | Vertical resolution with correct terms; must have acceleration $= 0$; $T$ does not need to be substituted |
| $\frac{1}{2}mg - \frac{1}{4}mg - \frac{2mg}{2a}x = 0$ | A1 | Correct equation with $T$ replaced |
| $x = \frac{1}{4}a$ | A1 | cao |
| $OB = \frac{9a}{4}$ | A1ft (4) | ft $2a +$ their $x$ |
---\begin{enumerate}
\item A light elastic string has natural length $2 a$ and modulus of elasticity $2 m g$.
\end{enumerate}
One end of the elastic string is attached to a fixed point $O$. A particle $P$ of mass $\frac { 1 } { 2 } m$ is attached to the other end of the elastic string.
The point $A$ is vertically below $O$ with $O A = 4 a$.
Particle $P$ is held at $A$ and released from rest. The speed of $P$ at the instant when it has moved a distance $a$ upwards is $\sqrt { 3 a g }$
Air resistance to the motion of $P$ is modelled as having magnitude $k m g$, where $k$ is a constant.
Using the model and the work-energy principle,\\
(a) show that $k = \frac { 1 } { 4 }$
Particle $P$ is now held at $O$ and released from rest. As $P$ moves downwards, it reaches its maximum speed as it passes through the point $B$.\\
(b) Find the distance $O B$.
\hfill \mbox{\textit{Edexcel M3 2021 Q4 [11]}}