| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Smooth ring on string |
| Difficulty | Standard +0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions and solving simultaneous equations. The setup is straightforward with clearly defined geometry (90° angle), and the solution method is explicitly guided through parts (i) and (ii). While it requires understanding of tension forces and trigonometry, it's a routine textbook exercise with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolving forces horizontally or vertically | M1 | For resolving forces horizontally or vertically |
| \(T\cos\theta + T\sin\theta = 11.2\) (or \(-T\cos\theta + T\sin\theta = 0.16g\)) | A1 | |
| \(-T\cos\theta + T\sin\theta = 0.16g\) (or \(T\cos\theta + T\sin\theta = 11.2\)) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T\cos\theta = 4.8\) and \(T\sin\theta = 6.4\) and \(T^2 = 4.8^2 + 6.4^2\) or \(\tan\theta = 6.4/4.8]\) \([4T^2(\cos^2\theta + \sin^2\theta) = (11.2-1.6)^2 + (11.2+1.6)^2]\) or \(2T\sin\theta \div 2T\cos\theta =\) \((11.2+1.6) \div (11.2-1.6)\) or \((T\cos\theta + T\sin\theta) \div (-T\cos\theta + T\sin\theta) = 11.2 \div 1.6\) | M1 | For finding \(T\cos\theta\) and \(T\sin\theta\) and hence finding \(T\) or \(\theta\), OR for finding the value of \(4T^2(\cos^2\theta + \sin^2\theta)\) or of \(2T\sin\theta \div 2T\cos\theta\) or of \((T\cos\theta + T\sin\theta) \div (-T\cos\theta + T\sin\theta)\) |
| \(T = 8\) (or \(\theta = 53.1\)) | A1 | |
| \(\theta = 53.1\) or \(T = 8\) | A1 [3] |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving forces horizontally or vertically | M1 | For resolving forces horizontally or vertically |
| $T\cos\theta + T\sin\theta = 11.2$ (or $-T\cos\theta + T\sin\theta = 0.16g$) | A1 | |
| $-T\cos\theta + T\sin\theta = 0.16g$ (or $T\cos\theta + T\sin\theta = 11.2$) | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T\cos\theta = 4.8$ and $T\sin\theta = 6.4$ and $T^2 = 4.8^2 + 6.4^2$ or $\tan\theta = 6.4/4.8]$ $[4T^2(\cos^2\theta + \sin^2\theta) = (11.2-1.6)^2 + (11.2+1.6)^2]$ or $2T\sin\theta \div 2T\cos\theta =$ $(11.2+1.6) \div (11.2-1.6)$ or $(T\cos\theta + T\sin\theta) \div (-T\cos\theta + T\sin\theta) = 11.2 \div 1.6$ | M1 | For finding $T\cos\theta$ and $T\sin\theta$ and hence finding $T$ or $\theta$, OR for finding the value of $4T^2(\cos^2\theta + \sin^2\theta)$ or of $2T\sin\theta \div 2T\cos\theta$ or of $(T\cos\theta + T\sin\theta) \div (-T\cos\theta + T\sin\theta)$ |
| $T = 8$ (or $\theta = 53.1$) | A1 | |
| $\theta = 53.1$ or $T = 8$ | A1 [3] | |
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\includegraphics[max width=\textwidth, alt={}, center]{918b65cc-617d-4942-8d96-b02eef21e417-2_471_621_870_762}
A smooth ring $R$ of mass 0.16 kg is threaded on a light inextensible string. The ends of the string are attached to fixed points $A$ and $B$. A horizontal force of magnitude 11.2 N acts on $R$, in the same vertical plane as $A$ and $B$. The ring is in equilibrium. The string is taut with angle $A R B = 90 ^ { \circ }$, and the part $A R$ of the string makes an angle of $\theta ^ { \circ }$ with the horizontal (see diagram). The tension in the string is $T \mathrm {~N}$.\\
(i) Find two simultaneous equations involving $T \sin \theta$ and $T \cos \theta$.\\
(ii) Hence find $T$ and $\theta$.
\hfill \mbox{\textit{CAIE M1 2012 Q2 [6]}}