| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ring on wire with string |
| Difficulty | Standard +0.3 This is a standard A-level mechanics equilibrium problem involving tension, friction, and resolving forces. Part (i) is straightforward force resolution at point M, part (ii) requires applying friction laws at limiting equilibrium, and part (iii) reverses the friction direction. While it requires careful diagram interpretation and systematic force resolution, the techniques are all standard M1 content with no novel insight required. |
| Spec | 3.03n Equilibrium in 2D: particle under forces3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_{BM} = T_{AM}\) or \(T_{BM}\cos 30° = T_{AM}\cos 30°\) | B1 | |
| For resolving forces at M horizontally \((27\sin 30° = 5)\) or for using the sine rule in the triangle of forces \((T + \sin 60° = 5 \div \sin 60°)\) or for using Lami's theorem \((T + \sin 120° = 5 \div \sin 120°)\) | M1 | |
| Tension is 5 N | A1 | 3 marks |
| For resolving forces at B horizontally \((N = T\sin 30°)\) or from symmetry \((N = 5/2)\) or for using Lami's theorem \((N + \sin 150° = 5 \div \sin 90°)\) | M1 | |
| For resolving forces at B vertically (3 terms needed) or for using Lami's theorem | M1 | |
| \(0.2 \times 10 + F = T\cos 30°\) or \((0.2g + F) \div \sin 120° = T \div \sin 90°\) | A1 | |
| For using \(F = \mu R\) with \((2.33 = 2.5\mu)\) | M1 | |
| Coefficient is 0.932 | A1 | 5 marks |
| \((0.2 + m)g - 2.33 = 5\cos 30°\) or \(mg = 2(2.33)\) | B1 √ | |
| \(m = 0.466\) | B1 | 2 marks |
$T_{BM} = T_{AM}$ or $T_{BM}\cos 30° = T_{AM}\cos 30°$ | B1 |
For resolving forces at M horizontally $(27\sin 30° = 5)$ or for using the sine rule in the triangle of forces $(T + \sin 60° = 5 \div \sin 60°)$ or for using Lami's theorem $(T + \sin 120° = 5 \div \sin 120°)$ | M1 |
Tension is 5 N | A1 | 3 marks
For resolving forces at B horizontally $(N = T\sin 30°)$ or from symmetry $(N = 5/2)$ or for using Lami's theorem $(N + \sin 150° = 5 \div \sin 90°)$ | M1 |
For resolving forces at B vertically (3 terms needed) or for using Lami's theorem | M1 |
$0.2 \times 10 + F = T\cos 30°$ or $(0.2g + F) \div \sin 120° = T \div \sin 90°$ | A1 |
For using $F = \mu R$ with $(2.33 = 2.5\mu)$ | M1 |
Coefficient is 0.932 | A1 | 5 marks
$(0.2 + m)g - 2.33 = 5\cos 30°$ or $mg = 2(2.33)$ | B1 √
$m = 0.466$ | B1 | 2 marks
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\includegraphics[max width=\textwidth, alt={}, center]{5cba3e17-3979-4c22-a415-2cdd60f09289-3_579_469_1142_840}
One end of a light inextensible string is attached to a fixed point $A$ of a fixed vertical wire. The other end of the string is attached to a small ring $B$, of mass 0.2 kg , through which the wire passes. A horizontal force of magnitude 5 N is applied to the mid-point $M$ of the string. The system is in equilibrium with the string taut, with $B$ below $A$, and with angles $A B M$ and $B A M$ equal to $30 ^ { \circ }$ (see diagram).\\
(i) Show that the tension in $B M$ is 5 N .\\
(ii) The ring is on the point of sliding up the wire. Find the coefficient of friction between the ring and the wire.\\
(iii) A particle of mass $m \mathrm {~kg}$ is attached to the ring. The ring is now on the point of sliding down the wire. Given that the coefficient of friction between the ring and the wire is unchanged, find the value of $m$.
\hfill \mbox{\textit{CAIE M1 2003 Q6 [10]}}