| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | November |
| Marks | 8 |
| 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 requiring resolution of forces and friction calculations. The geometry forms a 3-4-5 right triangle (making angle calculations straightforward), and the solution involves routine application of resolving forces, tension calculations, and the limiting friction formula μR=F. While it requires multiple steps across three parts, each step follows standard mechanics procedures without requiring novel insight or complex problem-solving strategies. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) For angle between AP and vertical = \(36.9°\) (or \(\sin^{-1}0.6\)) or for angle between PS and vertical = \(53.1°\) (or \(\sin^{-1}0.8\)) | B1 | May be implied |
| \([T_{PS} + (T_{PA}\cos 90°) = 5\sin 36.9°]\) | M1 | For resolving forces on P in the direction of PS (2 non-zero terms required) |
| Answer | Marks | Guidance |
|---|---|---|
| For the angle between PA and the horizontal through P is \(53.1°\) and the angle between PS and the horizontal through P is \(36.9°\) | B1 | May be implied |
| \([0.6T_{PA} = 0.8T_{PS}\) and \(0.8T_{PA} + 0.6T_{PS} = 5 \Rightarrow \{0.8(0.8/0.6) + 0.6\}T_{PS} = 5]\) | M1 | For resolving forces on P vertically and horizontally and eliminating \(T_{PA}\) |
| Answer | Marks | Guidance |
|---|---|---|
| For \(\Delta\) of forces with sides \(T_{PA}\), \(T_{PS}\) and 5, with angles opposite \(T_{PS}\) and 5 shown as \(36.9°\) and \(90°\) | B1 | May be implied |
| \([T_{PS} = 5\sin 36.9°]\) | M1 | For using trig. in \(\Delta\) |
| Answer | Marks | Guidance |
|---|---|---|
| For force diag. showing \(T_{PA}\), \(T_{PS}\) and 5, with angles between \(T_{PS}\) and \(T_{PA}\), and between 5 and \(T_{PA}\) being shown as \(90°\) and \(143.1°\) | B1 | May be implied |
| \([T_{PS}/\sin 143.1° = 5/\sin 90°]\) | M1 | For using Lami's rule |
| Tension is 3N | A1 | 3 |
| (ii) \([F = T\cos(\sin^{-1}0.6)]\) | M1 | For resolving forces on S horizontally |
| Frictional force is 2.4N | A1 | 2 |
| (iii) \(R = 2.4/0.75\) | B1ft | |
| \([W + T\sin(\sin^{-1}0.6) = R]\) | M1 | For resolving forces on S vertically |
| \(W = 1.4\) | A1ft | 3 |
(i) For angle between AP and vertical = $36.9°$ (or $\sin^{-1}0.6$) or for angle between PS and vertical = $53.1°$ (or $\sin^{-1}0.8$) | B1 | May be implied
$[T_{PS} + (T_{PA}\cos 90°) = 5\sin 36.9°]$ | M1 | For resolving forces on P in the direction of PS (2 non-zero terms required)
**(First alternative)**
For the angle between PA and the horizontal through P is $53.1°$ and the angle between PS and the horizontal through P is $36.9°$ | B1 | May be implied
$[0.6T_{PA} = 0.8T_{PS}$ and $0.8T_{PA} + 0.6T_{PS} = 5 \Rightarrow \{0.8(0.8/0.6) + 0.6\}T_{PS} = 5]$ | M1 | For resolving forces on P vertically and horizontally and eliminating $T_{PA}$
**(Second alternative)**
For $\Delta$ of forces with sides $T_{PA}$, $T_{PS}$ and 5, with angles opposite $T_{PS}$ and 5 shown as $36.9°$ and $90°$ | B1 | May be implied
$[T_{PS} = 5\sin 36.9°]$ | M1 | For using trig. in $\Delta$
**(Third alternative)**
For force diag. showing $T_{PA}$, $T_{PS}$ and 5, with angles between $T_{PS}$ and $T_{PA}$, and between 5 and $T_{PA}$ being shown as $90°$ and $143.1°$ | B1 | May be implied
$[T_{PS}/\sin 143.1° = 5/\sin 90°]$ | M1 | For using Lami's rule
Tension is 3N | A1 | 3 | Accept 3.00
(ii) $[F = T\cos(\sin^{-1}0.6)]$ | M1 | For resolving forces on S horizontally
Frictional force is 2.4N | A1 | 2 | Accept 2.40
(iii) $R = 2.4/0.75$ | B1ft |
$[W + T\sin(\sin^{-1}0.6) = R]$ | M1 | For resolving forces on S vertically
$W = 1.4$ | A1ft | 3 | ft $W = 7T/15$ or $W = 4F/3 - 1.8$4\\
\includegraphics[max width=\textwidth, alt={}, center]{a9f3480e-7a8a-497d-a26a-b2aba9b05512-3_335_751_264_696}
A particle $P$ of weight 5 N is attached to one end of each of two light inextensible strings of lengths 30 cm and 40 cm . The other end of the shorter string is attached to a fixed point $A$ of a rough rod which is fixed horizontally. A small ring $S$ of weight $W \mathrm {~N}$ is attached to the other end of the longer string and is threaded on to the rod. The system is in equilibrium with the strings taut and $A S = 50 \mathrm {~cm}$ (see diagram).\\
(i) By resolving the forces acting on $P$ in the direction of $P S$, or otherwise, find the tension in the longer string.\\
(ii) Find the magnitude of the frictional force acting on $S$.\\
(iii) Given that the coefficient of friction between $S$ and the rod is 0.75 , and that $S$ is in limiting equilibrium, find the value of $W$.
\hfill \mbox{\textit{CAIE M1 2009 Q4 [8]}}