| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Standard +0.3 This is a straightforward three-part elastic spring question requiring standard applications of Hooke's law, elastic potential energy formula, and energy conservation. Part (i) uses F=ma with tension from compressed spring, (ii) applies EPE = λx²/2L directly, and (iii) uses energy conservation. All are routine M2 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T = \frac{88 \times 0.1}{0.4}\) | B1 | |
| For using Newton's second law \((22 - 0.2 \times 10 = 0.2a)\) | M1 | 3 term equation needed |
| Initial acceleration is 100 ms\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For using \(\text{EPE} = \frac{\lambda x^2}{2L}\) \(\left(\frac{88 \times 0.1^2}{2 \times 0.4}\right)\) | M1 | |
| Initial elastic energy is 1.1 J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Change in GPE \(= 0.2 \times 10 \times 0.1\) | B1 | |
| For using the principle of conservation of energy (KE, EPE and GPE must all be represented) | M1 | |
| \(\left[\frac{1}{2}(0.2)v^2 = 1.1 - 0.2\right]\) | ||
| Speed is 3 ms\(^{-1}\) | A1 |
# Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T = \frac{88 \times 0.1}{0.4}$ | B1 | |
| For using Newton's second law $(22 - 0.2 \times 10 = 0.2a)$ | M1 | 3 term equation needed |
| Initial acceleration is 100 ms$^{-2}$ | A1 | |
# Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| For using $\text{EPE} = \frac{\lambda x^2}{2L}$ $\left(\frac{88 \times 0.1^2}{2 \times 0.4}\right)$ | M1 | |
| Initial elastic energy is 1.1 J | A1 | |
# Question 3(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Change in GPE $= 0.2 \times 10 \times 0.1$ | B1 | |
| For using the principle of conservation of energy (KE, EPE and GPE must all be represented) | M1 | |
| $\left[\frac{1}{2}(0.2)v^2 = 1.1 - 0.2\right]$ | | |
| Speed is 3 ms$^{-1}$ | A1 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{be83d46f-bf5b-4382-b424-bb5067626adc-2_433_446_1635_854}
One end of a light elastic spring, of natural length 0.4 m and modulus of elasticity 88 N , is attached to a fixed point $O$. A particle $P$ of mass 0.2 kg is attached to the other end of the spring and is held, with the spring compressed, at a point 0.3 m vertically above $O$, as shown in the diagram. $P$ is now released from rest and moves vertically upwards.\\
(i) Find the initial acceleration of $P$.\\
(ii) Find the initial elastic potential energy of the spring.\\
(iii) Find the speed of $P$ when the distance $O P$ is 0.4 m .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be83d46f-bf5b-4382-b424-bb5067626adc-3_362_657_269_744}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Fig. 1 shows a uniform lamina $A B C D$ with dimensions $A B = 15.5 \mathrm {~cm} , B C = 8 \mathrm {~cm}$ and $C D = 9.5 \mathrm {~cm}$. Angles $A B C$ and $B C D$ are right angles.\\
\hfill \mbox{\textit{CAIE M2 2003 Q3 [8]}}