| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ring on wire with string |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem requiring resolution of forces and basic trigonometry. Students must find tension using the ring's equilibrium, then resolve forces at the bead. The geometry is straightforward (3-4-5 triangle), and all steps follow routine procedures with no novel insight required. Slightly above average due to the multi-part nature and need to coordinate geometry with force resolution. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For resolving forces on R vertically |
| \(2T\cos\alpha = 0.6g\) | A1 | Where \(\alpha = \frac{1}{2}\) angle ARB |
| Tension is 5N | A1 | [3] |
| (ii) | M1 | For resolving forces on B horizontally |
| \([F = T\sin\alpha]\) | ||
| Frictional component is 4N | A1 | |
| M1 | For resolving forces on B vertically | |
| \([N = 0.4g + T\cos\alpha]\) | ||
| Normal component is 7 N | A1 | [4] |
| (iii) | M1 | For using \(\mu = F/N\) |
| Coefficient is 4/7 or 0.571 | A1, B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| (i) For finding the relevant angles and using Lami's theorem | M1 | |
| \(6/\sin 106.26° = T/\sin 126.87°\) | A1 | |
| Tension is 5N | A1 | [3] |
| (ii) \(F/\sin 126.87° = 5/\sin 90°\) | B1 | |
| Frictional component is 4N | B1 | |
| \((R - 4)/\sin 143.13° = 5/\sin 90°\) | B1 | |
| Normal component is 7 N | B1 | [4] |
**(i)** | M1 | For resolving forces on R vertically
$2T\cos\alpha = 0.6g$ | A1 | Where $\alpha = \frac{1}{2}$ angle ARB
Tension is 5N | A1 | [3]
**(ii)** | M1 | For resolving forces on B horizontally
$[F = T\sin\alpha]$ |
Frictional component is 4N | A1 |
| M1 | For resolving forces on B vertically
$[N = 0.4g + T\cos\alpha]$ |
Normal component is 7 N | A1 | [4]
**(iii)** | M1 | For using $\mu = F/N$
Coefficient is 4/7 or 0.571 | A1, B1 | [2] | $ft$ conditional on both M1 marks scored in (ii); If F and/or N
**Alternative for Q6(i)/(ii)**
**(i)** For finding the relevant angles and using Lami's theorem | M1 |
$6/\sin 106.26° = T/\sin 126.87°$ | A1 |
Tension is 5N | A1 | [3]
**(ii)** $F/\sin 126.87° = 5/\sin 90°$ | B1 |
Frictional component is 4N | B1 |
$(R - 4)/\sin 143.13° = 5/\sin 90°$ | B1 |
Normal component is 7 N | B1 | [4]6\\
\includegraphics[max width=\textwidth, alt={}, center]{d3bb6702-231d-42a0-830e-9f844dca78d7-3_387_1095_1724_525}
A small smooth ring $R$, of mass 0.6 kg , is threaded on a light inextensible string of length 100 cm . One end of the string is attached to a fixed point $A$. A small bead $B$ of mass 0.4 kg is attached to the other end of the string, and is threaded on a fixed rough horizontal rod which passes through $A$. The system is in equilibrium with $B$ at a distance of 80 cm from $A$ (see diagram).\\
(i) Find the tension in the string.\\
(ii) Find the frictional and normal components of the contact force acting on $B$.\\
(iii) Given that the equilibrium is limiting, find the coefficient of friction between the bead and the rod.
\hfill \mbox{\textit{CAIE M1 2011 Q6 [9]}}