| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| 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 equilibrium analysis (resolving forces, using Hooke's law) and energy conservation. Part (i) involves routine resolution of forces and algebraic manipulation. Part (ii) applies conservation of energy with elastic potential energy, gravitational potential energy, and kinetic energy—a standard M2 technique. The geometry is straightforward and the multi-step nature is typical for this level, making it slightly easier than average. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \([0.35g = 2T/0.7 / (2(4^2 + 0.7^2)^{1/2})]\) Tension is 6.25N \([6.25 = 2 \times \frac{1}{4}]\) Modulus is 25N | M1 A1 M1 A1 [4] | For resolving forces on P vertically. For using \(T = 2v/L\) AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(25 \times 2^2/(2 \times 4) = 0.35g \times 1.8 + 25 \times 0.8^2/(2 \times 4) + \frac{1}{2} 0.35v^2\) Speed is 4.90ms\(^{-1}\) | M1 A1 A1 M1 A1 A1 [6] | For using \(EE = 2x^2/2L\). For using \(EE\) on release \(= mgh + EE\) when P is at M \(+ \frac{1}{2}mv^2\) |
## (i)
$[0.35g = 2T/0.7 / (2(4^2 + 0.7^2)^{1/2})]$ Tension is 6.25N $[6.25 = 2 \times \frac{1}{4}]$ Modulus is 25N | M1 A1 M1 A1 [4] | For resolving forces on P vertically. For using $T = 2v/L$ AG
## (ii)
EE on release $= 25 \times 2^2/(2 \times 4)$ EE when P is at M $= 25 \times 0.8^2/(2 \times 4)$
$25 \times 2^2/(2 \times 4) = 0.35g \times 1.8 + 25 \times 0.8^2/(2 \times 4) + \frac{1}{2} 0.35v^2$ Speed is 4.90ms$^{-1}$ | M1 A1 A1 M1 A1 A1 [6] | For using $EE = 2x^2/2L$. For using $EE$ on release $= mgh + EE$ when P is at M $+ \frac{1}{2}mv^2$
---\includegraphics{figure_6}
A particle $P$ of mass 0.35 kg is attached to the mid-point of a light elastic string of natural length 4 m. The ends of the string are attached to fixed points $A$ and $B$ which are 4.8 m apart at the same horizontal level. $P$ hangs in equilibrium at a point 0.7 m vertically below the mid-point $M$ of $AB$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string and hence show that the modulus of elasticity of the string is 25 N. [4]
\end{enumerate}
$P$ is now held at rest at a point 1.8 m vertically below $M$, and is then released.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the speed with which $P$ passes through $M$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q6 [10]}}