| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| 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 two-part Hooke's law problem requiring (i) resolving forces at equilibrium using geometry and tension formula, and (ii) applying energy conservation with elastic potential energy. While it involves multiple steps and careful geometry (Pythagoras to find extensions), the techniques are routine for M2 students with no novel problem-solving required. Slightly easier than average due to the 'show that' structure in part (i) providing a target value. |
| 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 |
|---|---|---|
| 5 | (i) T = λ ( 1.22 +0.52 – 1)/1 | B1 |
| Answer | Marks |
|---|---|
| 2xTx0.5/1.3 = 6 | B1 |
| Answer | Marks |
|---|---|
| T = 0.3 = 7.8 | M1 |
| Answer | Marks |
|---|---|
| = 26 AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) EE1 = 2x26x0.3 2 /2x1 | M1 | (= 2.34) Use of EPE formula, either |
| Answer | Marks | Guidance |
|---|---|---|
| EE2 = 2x26( – 1) /2x1 | A1 | (= 6.5) Both expressions correct |
| M1 | Conservation of energy (including |
| Answer | Marks |
|---|---|
| 0.6v /2 + 0.6x10x(0.9 – 0.5) = 6.5 – 2.34 | A1 |
| Answer | Marks |
|---|---|
| V = 2.42ms | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme: Teachers’ version | Syllabus |
| GCE A LEVEL – October/November 2010 | 9709 | 53 |
Question 5:
5 | (i) T = λ ( 1.22 +0.52 – 1)/1 | B1 | λ
T = 0.3 or T = 0.3x26
2xTx0.5/1.3 = 6 | B1
λ
T = 0.3 = 7.8 | M1
λ
= 26 AG | A1
[4]
(ii) EE1 = 2x26x0.3 2 /2x1 | M1 | (= 2.34) Use of EPE formula, either
1.22 +0.92 2
EE2 = 2x26( – 1) /2x1 | A1 | (= 6.5) Both expressions correct
M1 | Conservation of energy (including
KE/GPE/EPE)
2
0.6v /2 + 0.6x10x(0.9 – 0.5) = 6.5 – 2.34 | A1
–1
V = 2.42ms | A1
[5]
Page 6 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2010 | 9709 | 53
\includegraphics{figure_5}
A light elastic string has natural length $2$ m and modulus of elasticity $\lambda$ N. The ends of the string are attached to fixed points $A$ and $B$ which are at the same horizontal level and $2.4$ m apart. A particle $P$ of mass $0.6$ kg is attached to the mid-point of the string and hangs in equilibrium at a point $0.5$ m below $AB$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that $\lambda = 26$. [4]
\end{enumerate}
$P$ is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point $0.9$ m below $AB$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the speed of projection of $P$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q5 [9]}}