| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Maximum/minimum speed in elastic motion |
| Difficulty | Standard +0.3 This is a standard A-level mechanics problem involving elastic strings, requiring Hooke's law for tensions, Newton's second law for acceleration, and energy/force balance for maximum speed. While it has multiple parts and requires careful setup, the techniques are routine for M2 students with no novel insight needed—slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02d Mechanical energy: KE and PE concepts6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(24 \times 0.65/0.6\) or \(18 \times 0.35/0.9\) | M1 | For using \(T = \lambda\, x/L\) |
| Tension in \(AP\) is 26N | A1 | |
| Tension in \(BP\) is 7N | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(26 - 7 = 0.95a\) | M1 | For using Newton's second law (3 terms) |
| Acceleration is \(20 \text{ ms}^{-2}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(T_{AP} = T_{BP}\) | |
| \(24x/0.6 = 18(1-x)/0.9\) | A1 | |
| \(x = 1/3\) | DM1 | For attempting to solve for \(x\) |
| Distance is 0.933 m | A1 | 4 |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $24 \times 0.65/0.6$ or $18 \times 0.35/0.9$ | M1 | For using $T = \lambda\, x/L$ |
| Tension in $AP$ is 26N | A1 | |
| Tension in $BP$ is 7N | A1 | 3 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $26 - 7 = 0.95a$ | M1 | For using Newton's second law (3 terms) |
| Acceleration is $20 \text{ ms}^{-2}$ | A1 | 2 | ft $|T_{AP} - T_{BP}| = 0.95a$ |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $T_{AP} = T_{BP}$ |
| $24x/0.6 = 18(1-x)/0.9$ | A1 | |
| $x = 1/3$ | DM1 | For attempting to solve for $x$ |
| Distance is 0.933 m | A1 | 4 | |
**Total: 9 marks**
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{57f7ca89-f028-447a-9ac9-55f931201e6b-3_83_771_1978_689}\\
$A$ and $B$ are fixed points on a smooth horizontal table. The distance $A B$ is 2.5 m . An elastic string of natural length 0.6 m and modulus of elasticity 24 N has one end attached to the table at $A$, and the other end attached to a particle $P$ of mass 0.95 kg . Another elastic string of natural length 0.9 m and modulus of elasticity 18 N has one end attached to the table at $B$, and the other end attached to $P$. The particle $P$ is held at rest at the mid-point of $A B$ (see diagram).\\
(i) Find the tensions in the strings.
The particle is released from rest.\\
(ii) Find the acceleration of $P$ immediately after its release.\\
(iii) $P$ reaches its maximum speed at the point $C$. Find the distance $A C$.
\hfill \mbox{\textit{CAIE M2 2007 Q6 [9]}}