| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| 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 elastic string energy problem requiring conservation of energy to find maximum extension, then applying Newton's second law at the lowest point. While it involves multiple steps (energy equation setup, solving a quadratic, then force analysis), the approach is entirely routine for M3 students with no novel insight required. Slightly above average difficulty due to the algebraic manipulation needed. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EE gain \(= 44.1x^2 \div (2 \times 0.75)\) | B1 | Allow use of \((e + x)\) for \(x\) |
| PE loss \(= 1.8g(0.75 + x)\) | B1 | Ignore signs |
| \([x^2 - 0.6x - 0.45 = 0]\) | M1 | For using EE gain \(=\) PE loss; \(44.1x^2 - 26.46x - 19.845 = 0\) allow sign errors |
| Extension is \(1.03\) m | A1 | \(1.0348469\ldots\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{44.1 \times 1.03}{0.75} - 1.8 \times 9.8 = -1.8\ddot{x}\) | M1 | For using \(T = \lambda x/L\) |
| M1 | For using Newton's 2nd law; allow missed \(g\), \(m\), sign error | |
| A1ft | ft their '\(1.03\)' from (i); allow sign error | |
| Acceleration is \(-24.0\text{ ms}^{-2}\) | A1 | Direction must be clear; \(1.03 \to -23.84666\); \(1.035 \to -24.01\) |
| [4] |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| EE gain $= 44.1x^2 \div (2 \times 0.75)$ | B1 | Allow use of $(e + x)$ for $x$ |
| PE loss $= 1.8g(0.75 + x)$ | B1 | Ignore signs |
| $[x^2 - 0.6x - 0.45 = 0]$ | M1 | For using EE gain $=$ PE loss; $44.1x^2 - 26.46x - 19.845 = 0$ allow sign errors |
| Extension is $1.03$ m | A1 | $1.0348469\ldots$ |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{44.1 \times 1.03}{0.75} - 1.8 \times 9.8 = -1.8\ddot{x}$ | M1 | For using $T = \lambda x/L$ |
| | M1 | For using Newton's 2nd law; allow missed $g$, $m$, sign error |
| | A1ft | ft their '$1.03$' from (i); allow sign error |
| Acceleration is $-24.0\text{ ms}^{-2}$ | A1 | Direction must be clear; $1.03 \to -23.84666$; $1.035 \to -24.01$ |
| **[4]** | | |
---4 One end of a light elastic string, of natural length 0.75 m and modulus of elasticity 44.1 N , is attached to a fixed point $O$. A particle $P$ of mass 1.8 kg is attached to the other end of the string. $P$ is released from rest at $O$ and falls vertically. Assuming there is no air resistance, find\\
(i) the extension of the string when $P$ is at its lowest position,\\
(ii) the acceleration of $P$ at its lowest position.
\hfill \mbox{\textit{OCR M3 2012 Q4 [8]}}