| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| 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 energy conservation problem with elastic strings requiring application of EPE = λx²/(2l) and conservation of energy principles. Part (i) involves straightforward energy equation setup with given boundary conditions, while part (ii) requires solving a quadratic from energy conservation. The multi-step nature and need to handle elastic potential energy correctly places it slightly above average, but it follows a well-practiced template for M3 vertical motion questions. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(EE = \lambda \times (5-3)^2/(2\times3)\) | M1, A1 | For use of EE formula |
| \(2\lambda/3 = 1.6\times9.8\times5\) | M1 | For equating EE and PE |
| \(\lambda = 117.6\) N | A1 | [4] AG |
| (ii) \(0.5\times1.6v^2 = 1.6\times9.8\times4.5 - 117.6\times1.5^2/(2\times3)\) | M1, A2,1,0 | For use of conservation of energy; \(-1\) each error |
| \(v = 5.75\) ms\(^{-1}\) | A1 | [4] |
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $EE = \lambda \times (5-3)^2/(2\times3)$ | M1, A1 | For use of EE formula |
| $2\lambda/3 = 1.6\times9.8\times5$ | M1 | For equating EE and PE |
| $\lambda = 117.6$ N | A1 | [4] AG |
| **(ii)** $0.5\times1.6v^2 = 1.6\times9.8\times4.5 - 117.6\times1.5^2/(2\times3)$ | M1, A2,1,0 | For use of conservation of energy; $-1$ each error |
| $v = 5.75$ ms$^{-1}$ | A1 | [4] |
---3 A light elastic string has natural length 3 m . One end is attached to a fixed point $O$ and the other end is attached to a particle of mass 1.6 kg . The particle is released from rest in a position 5 m vertically below $O$. Air resistance may be neglected.\\
(i) Given that in the subsequent motion the particle just reaches $O$, show that the modulus of elasticity of the string is 117.6 N .\\
(ii) Calculate the speed of the particle when it is 4.5 m below $O$.
\hfill \mbox{\textit{OCR M3 2006 Q3 [8]}}