| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string vertical motion |
| Difficulty | Challenging +1.2 This is a multi-stage energy problem involving elastic strings, requiring students to apply conservation of energy across different phases (string slack vs taut) and solve a quadratic equation. While it requires careful bookkeeping of energy terms (gravitational PE, elastic PE, kinetic energy) and understanding when the string becomes taut/slack, the problem structure is fairly standard for M2 elastic string questions with clear signposting of what to find and show. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EE at \(A = (5 \times 0.3^2) \div (2 \times 0.6)\) | M1 | For using \(EE = \dfrac{\lambda x^2}{2L}\) |
| PE loss at \(B = 0.2 \times 10 \times 0.3\) | B1 | |
| \(0.375 = \frac{1}{2} \times 0.2v^2 - 0.6\) | M1 | For using EE at \(A\) = KE gain \(-\) PE loss (allow sign errors) |
| Velocity is \(3.12\ \text{ms}^{-1}\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using EE at \(A\) = EE at \(C\) \(-\) PE loss | |
| \(0.375 = 5x^2 \div (2 \times 0.6) - 0.2 \times 10(1.5 + x)\) | A1 | |
| \(x^2 - 0.48x - 0.81 = 0\) | A1 (3) | From correct working (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (M1) | For using KE at \(B\) + PE at \(B\) = EE at \(C\) | |
| \(0.975 + 0.2 \times 10(1.2 + x) = 5x^2/1.2\) | (A1) | |
| \(x^2 - 0.48x - 0.81 = 0\) | (A1) (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For solving and adding \(1.5\) to \(+\)ve root | |
| Distance is \(2.67\)m | A1 (2) | From \(1.5 + 1.17\) |
## Question 8:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| EE at $A = (5 \times 0.3^2) \div (2 \times 0.6)$ | M1 | For using $EE = \dfrac{\lambda x^2}{2L}$ |
| PE loss at $B = 0.2 \times 10 \times 0.3$ | B1 | |
| $0.375 = \frac{1}{2} \times 0.2v^2 - 0.6$ | M1 | For using EE at $A$ = KE gain $-$ PE loss (allow sign errors) |
| Velocity is $3.12\ \text{ms}^{-1}$ | A1 (4) | |
### Part (ii)(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using EE at $A$ = EE at $C$ $-$ PE loss |
| $0.375 = 5x^2 \div (2 \times 0.6) - 0.2 \times 10(1.5 + x)$ | A1 | |
| $x^2 - 0.48x - 0.81 = 0$ | A1 (3) | From correct working **(AG)** |
### OR Part (ii)(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | (M1) | For using KE at $B$ + PE at $B$ = EE at $C$ |
| $0.975 + 0.2 \times 10(1.2 + x) = 5x^2/1.2$ | (A1) | |
| $x^2 - 0.48x - 0.81 = 0$ | (A1) (3) | |
### Part (ii)(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For solving and adding $1.5$ to $+$ve root |
| Distance is $2.67$m | A1 (2) | From $1.5 + 1.17$ |8\\
\includegraphics[max width=\textwidth, alt={}, center]{0cb05368-9ddf-4564-8428-725c77193a1e-4_933_275_689_934}
The diagram shows a light elastic string of natural length 0.6 m and modulus of elasticity 5 N with one end attached to a fixed point $O$. A particle $P$ of mass 0.2 kg is attached to the other end of the string. $P$ is held at the point $A$, which is 0.9 m vertically above $O$. The particle is released from rest and travels vertically downwards through $O$ to the point $C$, where it starts to move upwards. $B$ is the point of the line $A C$ where the string first becomes slack.\\
(i) Find the speed of $P$ at $B$.\\
(ii) The extension of the string when $P$ is at $C$ is $x \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x ^ { 2 } - 0.48 x - 0.81 = 0$.
\item Hence find the distance $A C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2006 Q8 [9]}}