| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Particle on slope with pulley |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system on inclined planes requiring resolution of forces, Newton's second law, and friction calculations. Part (i) involves routine application of F=ma with smooth planes (4-5 steps), while part (ii) requires setting up limiting equilibrium with friction forces. The symmetry and standard setup make this slightly easier than average, though the two-part structure and friction component keep it near typical difficulty. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T - 0.8g\sin 30 = 0.8a\) | M1 | Resolve along the plane for either \(A\) or \(B\) or for the system |
| \(1.2g\sin 60 - T = 1.2a\) | ||
| \([1.2g\sin 60 - 0.8g\sin 30 = 2a]\) | ||
| For \(A\): \(T - 4 = 0.8a\) | A1 | |
| For \(B\): \(6\sqrt{3} - T = 10.4 - T = 1.2a\) | A1 | System equation is \(6\sqrt{3} - 4 = 6.4 = 2a\) |
| M1 | Solve for \(a\) or \(T\) | |
| \(a = 3\sqrt{3} - 2 = 3.20\ \text{ms}^{-2}\) | A1 | |
| \(T = \frac{12}{5}(1 + \sqrt{3}) = 6.56\) N | A1 | |
| Total: | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R_A = 0.8g\cos 30 = 4\sqrt{3}\) | B1 | For either \(R_A\) or \(R_B\) |
| \(R_B = 1.2g\cos 60 = 6\) | ||
| \(F_A = 4\sqrt{3}\,\mu\) and \(F_B = 6\mu\) | M1 | Either \(F_A\) or \(F_B\) used |
| M1 | Resolve parallel to the plane for both particles \(A\) and \(B\) or system | |
| \(12\sin 60 - 6\mu - T = 0\) or \(T - 8\sin 30 - 4\sqrt{3}\,\mu = 0\) | A1 | System equation is \(12\sin 60 - 8\sin 30 - 6\mu - 4\sqrt{3}\,\mu = 0\) |
| M1 | Eliminate \(T\) and/or find \(\mu\) | |
| \(\mu = (6\sqrt{3} - 4)/(6 + 4\sqrt{3}) = 0.494\) | A1 | |
| Total: | 6 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T - 0.8g\sin 30 = 0.8a$ | M1 | Resolve along the plane for either $A$ or $B$ or for the system |
| $1.2g\sin 60 - T = 1.2a$ | | |
| $[1.2g\sin 60 - 0.8g\sin 30 = 2a]$ | | |
| For $A$: $T - 4 = 0.8a$ | A1 | |
| For $B$: $6\sqrt{3} - T = 10.4 - T = 1.2a$ | A1 | System equation is $6\sqrt{3} - 4 = 6.4 = 2a$ |
| | M1 | Solve for $a$ or $T$ |
| $a = 3\sqrt{3} - 2 = 3.20\ \text{ms}^{-2}$ | A1 | |
| $T = \frac{12}{5}(1 + \sqrt{3}) = 6.56$ N | A1 | |
| **Total:** | **6** | |
---
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R_A = 0.8g\cos 30 = 4\sqrt{3}$ | B1 | For either $R_A$ or $R_B$ |
| $R_B = 1.2g\cos 60 = 6$ | | |
| $F_A = 4\sqrt{3}\,\mu$ and $F_B = 6\mu$ | M1 | Either $F_A$ or $F_B$ used |
| | M1 | Resolve parallel to the plane for both particles $A$ and $B$ or system |
| $12\sin 60 - 6\mu - T = 0$ or $T - 8\sin 30 - 4\sqrt{3}\,\mu = 0$ | A1 | System equation is $12\sin 60 - 8\sin 30 - 6\mu - 4\sqrt{3}\,\mu = 0$ |
| | M1 | Eliminate $T$ and/or find $\mu$ |
| $\mu = (6\sqrt{3} - 4)/(6 + 4\sqrt{3}) = 0.494$ | A1 | |
| **Total:** | **6** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{4941e074-2f93-4a0c-80ba-0ca96a48389e-10_374_762_259_688}
As shown in the diagram, a particle $A$ of mass 0.8 kg lies on a plane inclined at an angle of $30 ^ { \circ }$ to the horizontal and a particle $B$ of mass 1.2 kg lies on a plane inclined at an angle of $60 ^ { \circ }$ to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley $P$ fixed at the top of the planes. The parts $A P$ and $B P$ of the string are parallel to lines of greatest slope of the respective planes. The particles are released from rest with both parts of the string taut.\\
(i) Given that both planes are smooth, find the acceleration of $A$ and the tension in the string.\\
(ii) It is given instead that both planes are rough, with the same coefficient of friction, $\mu$, for both particles. Find the value of $\mu$ for which the system is in limiting equilibrium.\\
\hfill \mbox{\textit{CAIE M1 2017 Q7 [12]}}