| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Ring or bead on wire/rod, equilibrium |
| Difficulty | Standard +0.3 This is a standard statics problem requiring resolution of forces at point B (using perpendicular components of the string tensions to balance the 8N force) and then applying equilibrium conditions at ring A with limiting friction. The right-angle geometry simplifies calculations, and the method is routine for M1 level—slightly easier than average due to the straightforward setup and clear geometric relationships. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| M1 | For resolving forces vertically and horizontally at B | |
| \(T_C \times (2/2.5) - T_A \times (1.5/2.5) = 0\) | A1 | |
| \(T_C \times (1.5/2.5) + T_A \times (2/2.5) = 8\) | A1 | |
| \([0.6\,T_C + 0.8\,(4T_C/3) = 8 \Rightarrow (5/3)\,T_C = 8\) or \(0.6(0.75T_A) + 0.8T_A = 8 \Rightarrow 1.25T_A = 8]\) | M1 | For eliminating \(T_A\) or \(T_C\) and attempting to find \(T_C\) or \(T_A\) |
| Tension in AB is 6.4 N; tension in BC is 4.8 N | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| M1 | For resolving forces vertically | |
| \(F + 0.2g = T_A \times (1.5/2.5)\) | A1 | |
| \(N = T_A \times (2/2.5)\) | B1 | |
| \([\mu = (3.84 - 2)/5.12]\) | M1 | For using \(\mu = F/N\) with \(F\) vertical and \(N\) horizontal |
| Coefficient is 0.359 | A1 [5] | Accept 0.36 |
## Question 7(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces vertically and horizontally at B |
| $T_C \times (2/2.5) - T_A \times (1.5/2.5) = 0$ | A1 | |
| $T_C \times (1.5/2.5) + T_A \times (2/2.5) = 8$ | A1 | |
| $[0.6\,T_C + 0.8\,(4T_C/3) = 8 \Rightarrow (5/3)\,T_C = 8$ or $0.6(0.75T_A) + 0.8T_A = 8 \Rightarrow 1.25T_A = 8]$ | M1 | For eliminating $T_A$ or $T_C$ and attempting to find $T_C$ or $T_A$ |
| Tension in AB is 6.4 N; tension in BC is 4.8 N | A1 [5] | |
## Question 7(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces vertically |
| $F + 0.2g = T_A \times (1.5/2.5)$ | A1 | |
| $N = T_A \times (2/2.5)$ | B1 | |
| $[\mu = (3.84 - 2)/5.12]$ | M1 | For using $\mu = F/N$ with $F$ vertical and $N$ horizontal |
| Coefficient is 0.359 | A1 [5] | Accept 0.36 |7\\
\includegraphics[max width=\textwidth, alt={}, center]{01e73486-5a95-4e65-bf18-518d1adc7cfb-4_529_481_255_831}
A small ring of mass 0.2 kg is threaded on a fixed vertical rod. The end $A$ of a light inextensible string is attached to the ring. The other end $C$ of the string is attached to a fixed point of the rod above $A$. A horizontal force of magnitude 8 N is applied to the point $B$ of the string, where $A B = 1.5 \mathrm {~m}$ and $B C = 2 \mathrm {~m}$. The system is in equilibrium with the string taut and $A B$ at right angles to $B C$ (see diagram).\\
(i) Find the tension in the part $A B$ of the string and the tension in the part $B C$ of the string.
The equilibrium is limiting with the ring on the point of sliding up the rod.\\
(ii) Find the coefficient of friction between the ring and the rod.
\hfill \mbox{\textit{CAIE M1 2012 Q7 [10]}}