| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| 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 M2 elastic string energy problem requiring application of the work-energy principle with three routine parts: deriving a given energy equation, finding maximum speed by differentiation, and calculating tension using Hooke's law. While it involves multiple steps and careful bookkeeping of energy terms (KE, GPE, EPE), the method is entirely standard with no novel insight required, making it slightly easier than average. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{0.4v^2}{2} + \frac{24x^2}{(2 \times 3)}\) | M1 | PE, EE, KE terms |
| \(0.4g(3+x) + \frac{0.4 \times 2^2}{2}\) | A2 | \(-1\) each error to zero |
| \(v^2 = 64 + 20x - 20x^2\) | A1 AG [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2v\frac{dv}{dx} = 20 - 40x = 0\) | M1 | \(0.4g = \frac{24x}{3}\) |
| \(x = 0.5\) | A1ft | |
| \(v = 8.31\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(20x^2 - 20x - 64 = 0\) | M1 | And attempts to solve |
| \(x = 2.357\) | A1 | |
| \(T = \frac{24 \times 2.357}{3}\) | M1 | |
| \(T = 18.9\) | A1 [4] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{0.4v^2}{2} + \frac{24x^2}{(2 \times 3)}$ | M1 | PE, EE, KE terms |
| $0.4g(3+x) + \frac{0.4 \times 2^2}{2}$ | A2 | $-1$ each error to zero |
| $v^2 = 64 + 20x - 20x^2$ | A1 AG **[4]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2v\frac{dv}{dx} = 20 - 40x = 0$ | M1 | $0.4g = \frac{24x}{3}$ |
| $x = 0.5$ | A1ft | |
| $v = 8.31$ | A1 **[3]** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $20x^2 - 20x - 64 = 0$ | M1 | And attempts to solve |
| $x = 2.357$ | A1 | |
| $T = \frac{24 \times 2.357}{3}$ | M1 | |
| $T = 18.9$ | A1 **[4]** | |7 One end of a light elastic string of natural length 3 m and modulus of elasticity 24 N is attached to a fixed point $O$. A particle $P$ of mass 0.4 kg is attached to the other end of the string. $P$ is projected vertically downwards from $O$ with initial speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When the extension of the string is $x \mathrm {~m}$ the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $v ^ { 2 } = 64 + 20 x - 20 x ^ { 2 }$.\\
(ii) Find the greatest speed of the particle.\\
(iii) Calculate the greatest tension in the string.
\hfill \mbox{\textit{CAIE M2 2010 Q7 [11]}}