| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: vertical motion |
| Difficulty | Standard +0.3 This is a standard elastic string equilibrium and energy problem requiring Hooke's law application, Pythagoras for geometry, force resolution at equilibrium, and energy conservation. The steps are methodical and follow typical M2 patterns, though the geometry setup and energy calculation require care. Slightly above average difficulty due to the multi-step nature and need to combine mechanics principles, but well within standard M2 scope. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) 2Tcos θ = 0.28g | M1 |
| 2T x 0.7/2.5 = 2.8, T = 5 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 = x 0.5/2 | M1 | Hookes Law |
| Answer | Marks |
|---|---|
| = 20 N | A1 |
| Answer | Marks |
|---|---|
| 0.28gx0.7 +2x20x0.4 /(2x2) | M1 |
| A1 | PE/EE/KE conservation with 4 terms |
| Answer | Marks |
|---|---|
| v = 2.75 ms | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme: Teachers’ version | Syllabus |
| GCE A LEVEL – October/November 2010 | 9709 | 52 |
Question 5:
5 | (i) 2Tcos θ = 0.28g | M1 | Tension component = weight
2T x 0.7/2.5 = 2.8, T = 5 | A1
λ
5 = x 0.5/2 | M1 | Hookes Law
λ
= 20 N | A1
[4]
(ii) 0.28v 2 /2 + 2x20x0.5 2 /(2x2) =
2
0.28gx0.7 +2x20x0.4 /(2x2) | M1
A1 | PE/EE/KE conservation with 4 terms
–1
v = 2.75 ms | A1
[3]
Page 6 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2010 | 9709 | 52A particle $P$ of mass $0.28 \text{ kg}$ is attached to the mid-point of a light elastic string of natural length $4 \text{ m}$. The ends of the string are attached to fixed points $A$ and $B$ which are at the same horizontal level and $4.8 \text{ m}$ apart. $P$ is released from rest at the mid-point of $AB$. In the subsequent motion, the acceleration of $P$ is zero when $P$ is at a distance $0.7 \text{ m}$ below $AB$.
\begin{enumerate}[label=(\roman*)]
\item Show that the modulus of elasticity of the string is $20 \text{ N}$. [4]
\item Calculate the maximum speed of $P$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q5 [7]}}