| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle attached to two separate elastic strings |
| Difficulty | Challenging +1.2 This is a standard energy conservation problem with elastic strings requiring systematic application of EPE, GPE, and KE formulas. While it involves two strings and requires careful bookkeeping of extensions and energy terms, the method is straightforward once set up. The algebraic manipulation is routine, and part (ii) follows directly from differentiating the given result. More challenging than basic mechanics but less demanding than problems requiring novel geometric insight or complex multi-stage reasoning. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using \(\text{EPE} = \dfrac{\lambda x^2}{2L}\) | M1 | |
| EPE gain \(= 2\!\left(\dfrac{200x^2}{2\times 4}\right)\) \((= 50x^2)\) | A1 | |
| GPE loss \(= 10g(4 + x)\) | B1 | |
| For using the principle of conservation of energy to form an equation containing EPE, GPE and KE terms | M1 | |
| \(\left[\tfrac{1}{2}(10)v^2 + 50x^2 = 10g(4+x)\right]\) | ||
| Given answer obtained correctly | A1 | 5 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = \dfrac{200x}{4}\) | B1 | |
| \(100 - 2\!\left(\dfrac{200x}{4}\right) = 10v\dfrac{dv}{dx}\) | M1 | |
| \(\tfrac{1}{2}v^2 = 10x - 5x^2\) \((+C)\) | A1 | |
| Use \(x = 0\), \(v^2 = 8g\) | M1 | |
| \(v^2 = 10(8 + 2x - x^2)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using \(v = 0\) and factorising or using formula method for solving | M1 | |
| \(x = 4\) (only) | A1 | 2 marks total |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using $\text{EPE} = \dfrac{\lambda x^2}{2L}$ | M1 | |
| EPE gain $= 2\!\left(\dfrac{200x^2}{2\times 4}\right)$ $(= 50x^2)$ | A1 | |
| GPE loss $= 10g(4 + x)$ | B1 | |
| For using the principle of conservation of energy to form an equation containing EPE, GPE and KE terms | M1 | |
| $\left[\tfrac{1}{2}(10)v^2 + 50x^2 = 10g(4+x)\right]$ | | |
| Given answer obtained correctly | A1 | **5 marks total** |
**Alternative Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \dfrac{200x}{4}$ | B1 | |
| $100 - 2\!\left(\dfrac{200x}{4}\right) = 10v\dfrac{dv}{dx}$ | M1 | |
| $\tfrac{1}{2}v^2 = 10x - 5x^2$ $(+C)$ | A1 | |
| Use $x = 0$, $v^2 = 8g$ | M1 | |
| $v^2 = 10(8 + 2x - x^2)$ | A1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using $v = 0$ and factorising or using formula method for solving | M1 | |
| $x = 4$ (only) | A1 | **2 marks total** |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{835616aa-0b2b-4e8c-bbbf-60b72dc5ea3e-3_321_698_1692_726}
One end of a light elastic string of natural length 4 m and modulus of elasticity 200 N is attached to a fixed point $A$. The other end is attached to the end $C$ of a uniform rod $C D$ of mass 10 kg . One end of another light elastic string, which is identical to the first, is attached to a fixed point $B$ and the other end is attached to $D$, as shown in the diagram. The distance $A B$ is equal to the length of the rod, and $A B$ is horizontal. The rod is released from rest with $C$ at $A$ and $D$ at $B$. While the strings are taut, the speed of the rod is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when the rod is at a distance of $( 4 + x ) \mathrm { m }$ below $A B$.\\
(i) Show that $v ^ { 2 } = 10 \left( 8 + 2 x - x ^ { 2 } \right)$.\\
(ii) Hence find the value of $x$ when the rod is at its lowest point.
\hfill \mbox{\textit{CAIE M2 2004 Q5 [7]}}