| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between vertical fixed points |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring energy methods and force analysis with Hooke's law. While it involves several steps (energy conservation, finding equilibrium, calculating maximum acceleration), the techniques are standard for M2 level: applying EPE = λx²/(2l), using energy conservation, and finding where resultant force is zero. The geometry is straightforward and the problem guides students through the solution structure. More challenging than basic Hooke's law recall but less demanding than problems requiring novel geometric insight or complex optimization. |
| 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 |
|---|---|---|
| Energy conservation, no KE, 2 EE terms | M1 | |
| \(45 \times 1^2/(2 \times 1.5) + 0.6 \, gh = 45 \, h^2/(2 \times 1.5)\) | A1 | |
| \(5h^2 - 2h - 5 = 0\) | M1 | Simplifies, tries to solve a 3 term quadratic equation |
| \(h = 1.22 \, \text{m}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Finds equilibrium position (e = 0.6) | M1 | |
| \(45e/1.5 = 45(1 - e)/1.5 + 6\) | M1 | |
| \(AP = (1.5 + 0.6) = 2.1\) | AG A1 | |
| \(0.6 \, v^2/2 = 0.6 \, g \times 0.6 + 45(1)^2/(2 \times 1.5) - 4.5(0.6)^2/(2 \times 1.5) - 45(0.4)^2/(2 \times 1.5)\) | M1 A1 | Energy conservation with KE/PE/EE terms |
| \(v = 6 \, \text{ms}^{-1}\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.6 \, a = \pm(0.6g + 45 \times 1/1.5)\) | M1* | Top \(a = \pm 60 \, \text{ms}^{-2}\) |
| \(0.6 \, a = \pm(0.6g - 45 \times 1.22/1.5)\) | M1* | Bottom \(a = \pm 51 \, \text{ms}^{-2}\) |
| \( | a | = 60 \, \text{ms}^{-2}\) |
## (i)
Energy conservation, no KE, 2 EE terms | M1 |
$45 \times 1^2/(2 \times 1.5) + 0.6 \, gh = 45 \, h^2/(2 \times 1.5)$ | A1 |
$5h^2 - 2h - 5 = 0$ | M1 | Simplifies, tries to solve a 3 term quadratic equation
$h = 1.22 \, \text{m}$ | A1 [4] |
## (ii)
Finds equilibrium position (e = 0.6) | M1 |
$45e/1.5 = 45(1 - e)/1.5 + 6$ | M1 |
$AP = (1.5 + 0.6) = 2.1$ | AG A1 |
$0.6 \, v^2/2 = 0.6 \, g \times 0.6 + 45(1)^2/(2 \times 1.5) - 4.5(0.6)^2/(2 \times 1.5) - 45(0.4)^2/(2 \times 1.5)$ | M1 A1 | Energy conservation with KE/PE/EE terms
$v = 6 \, \text{ms}^{-1}$ | A1 [5] |
## (iii)
$0.6 \, a = \pm(0.6g + 45 \times 1/1.5)$ | M1* | Top $a = \pm 60 \, \text{ms}^{-2}$
$0.6 \, a = \pm(0.6g - 45 \times 1.22/1.5)$ | M1* | Bottom $a = \pm 51 \, \text{ms}^{-2}$
$|a| = 60 \, \text{ms}^{-2}$ | A**1 [3] | Needs acceleration at both extreme positions consideredA light elastic string has natural length $3$ m and modulus of elasticity $45$ N. A particle $P$ of weight $6$ N is attached to the mid-point of the string. The ends of the string are attached to fixed points $A$ and $B$ which lie in the same vertical line with $A$ above $B$ and $AB = 4$ m. The particle $P$ is released from rest at the point $1.5$ m vertically below $A$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the distance $P$ moves after its release before first coming to instantaneous rest at a point vertically above $B$. (You may assume that at this point the part of the string joining $P$ to $B$ is slack.) [4]
\item Show that the greatest speed of $P$ occurs when it is $2.1$ m below $A$, and calculate this greatest speed. [5]
\item Calculate the greatest magnitude of the acceleration of $P$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2012 Q7 [12]}}