| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Standard +0.3 This is a standard vertical elastic string energy problem requiring application of conservation of energy with elastic potential energy, gravitational potential energy, and kinetic energy. The setup is straightforward with given values, and both parts follow routine procedures: (i) equating energies at two positions to find speed, (ii) finding equilibrium where KE=0. While it requires careful bookkeeping of energy terms and understanding that the string only has tension when extended, these are well-practiced techniques in M2 with no novel problem-solving required. |
| Spec | 6.02d 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 |
| Gain in Elastic \(PE = 140(0.1)^2/(2 \times 0.5)\) | B1 | |
| Loss in \(GPE = 0.8\,g(0.5 + 0.1)\) | B1 | |
| \(\frac{1}{2}(0.8)v^2 + 140 \times 0.1^2 = 0.8\,g(0.5+0.1)\) | M1 | For using Gain in \(KE\) + Gain in \(EPE\) = Loss in \(GPE\) |
| Speed is \(2.92 \text{ ms}^{-1}\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using Gain in \(EPE\) = Loss in \(GPE\) | |
| \(140x^2 = 0.8\,g(0.5 + x)\) | A1 | |
| \((5x-1)(28x+4) = 0\) | M1 | For solving the resulting 3 term quadratic equation |
| Extension is 0.2 m | A1 | 4 |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gain in Elastic $PE = 140(0.1)^2/(2 \times 0.5)$ | B1 | |
| Loss in $GPE = 0.8\,g(0.5 + 0.1)$ | B1 | |
| $\frac{1}{2}(0.8)v^2 + 140 \times 0.1^2 = 0.8\,g(0.5+0.1)$ | M1 | For using Gain in $KE$ + Gain in $EPE$ = Loss in $GPE$ |
| Speed is $2.92 \text{ ms}^{-1}$ | A1 | 4 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using Gain in $EPE$ = Loss in $GPE$ |
| $140x^2 = 0.8\,g(0.5 + x)$ | A1 | |
| $(5x-1)(28x+4) = 0$ | M1 | For solving the resulting 3 term quadratic equation |
| Extension is 0.2 m | A1 | 4 | |
**Total: 8 marks**
---5 One end of a light elastic string, of natural length 0.5 m and modulus of elasticity 140 N , is attached to a fixed point $O$. A particle of mass 0.8 kg is attached to the other end of the string. The particle is released from rest at $O$. By considering the energy of the system, find\\
(i) the speed of the particle when the extension of the string is 0.1 m ,\\
(ii) the extension of the string when the particle is at its lowest point.
\hfill \mbox{\textit{CAIE M2 2007 Q5 [8]}}