| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Elastic string with variable force |
| Difficulty | Challenging +1.8 This is a challenging multi-part mechanics problem requiring energy methods, equilibrium analysis, and calculus to find maximum speed. It involves elastic strings with variable tension, energy conservation across multiple phases (slack/taut), and finding extrema through differentiation. The conceptual setup is non-trivial (particle at midpoint, vertical configuration), requiring careful geometric reasoning and systematic application of Hooke's law and energy principles across three distinct parts with increasing sophistication. |
| 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 |
|---|---|---|
| \(45 \times 1^2/(2 \times 1.5) + 0.6 gh = 45 h^2/(2 \times 1.5)\) | M1 | Energy conservation, no KE, 2 EE terms |
| \(5h^2 - 2h - 5 = 0\) | A1 | Simplifies, tries to solve a 3 term quadratic equation |
| \(h = 1.22 \text{ m}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(45e/1.5 = 45(1-e)/1.5 + 6\) | M1 | Finds equilibrium position (\(e = 0.6\)) |
| \(AP = (1.5 + 0.6) = 2.1\) | 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)**
$45 \times 1^2/(2 \times 1.5) + 0.6 gh = 45 h^2/(2 \times 1.5)$ | M1 | Energy conservation, no KE, 2 EE terms
$5h^2 - 2h - 5 = 0$ | A1 | Simplifies, tries to solve a 3 term quadratic equation
$h = 1.22 \text{ m}$ | A1 [4] |
**(ii)**
$45e/1.5 = 45(1-e)/1.5 + 6$ | M1 | Finds equilibrium position ($e = 0.6$)
$AP = (1.5 + 0.6) = 2.1$ | 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 considered7 A 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 $A B = 4 \mathrm {~m}$. The particle $P$ is released from rest at the point 1.5 m vertically below $A$.\\
(i) 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.)\\
(ii) Show that the greatest speed of $P$ occurs when it is 2.1 m below $A$, and calculate this greatest speed.\\
(iii) Calculate the greatest magnitude of the acceleration of $P$.
\hfill \mbox{\textit{CAIE M2 2012 Q7 [12]}}