| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Pulley at edge of table, specific geometry |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with friction requiring Newton's second law, connected particles equations, and SUVAT kinematics. While it has multiple parts and involves friction calculations, it follows a completely routine template with no novel problem-solving required—slightly easier than average due to its predictable structure. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For applying Newton's 2nd law to A or to B | |
| \(T - \frac{2}{7}(1.26)g = 1.26a\) or \(0.9g - T = 0.9a\) | A1 | |
| \(0.9g - T = 0.9a\) or \(T - \frac{2}{7}(1.26)g = 1.26a\) or \(0.9g - \frac{2}{7}(1.26)g = (0.9 + 1.26)a\) | B1 | |
| Acceleration is \(2.5 \text{ ms}^{-2}\) | B1 | AG |
| Tension is \(6.75 \text{ N}\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v^2 = 2 \times (2.5) \times 0.45]\) | M1 | For using \(v^2 = 2ah\) |
| Speed is \(1.5 \text{ ms}^{-1}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[-\frac{2}{7}(1.26)g = 1.26a\right]\) | M1 | For applying Newton's 2nd law to A |
| \(a = -\dfrac{20}{7}\) | A1 | |
| \(\left[v^2 = 2.25 + 2\left(-\dfrac{20}{7}\right)(0.03)\right]\) | M1 | For using \(v^2 = v_B^2 + 2as\) |
| Speed is \(1.44 \text{ ms}^{-1}\) | A1 | [4] |
## Question 7:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's 2nd law to A or to B |
| $T - \frac{2}{7}(1.26)g = 1.26a$ or $0.9g - T = 0.9a$ | A1 | |
| $0.9g - T = 0.9a$ or $T - \frac{2}{7}(1.26)g = 1.26a$ or $0.9g - \frac{2}{7}(1.26)g = (0.9 + 1.26)a$ | B1 | |
| Acceleration is $2.5 \text{ ms}^{-2}$ | B1 | AG |
| Tension is $6.75 \text{ N}$ | A1 | [5] |
**Part (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v^2 = 2 \times (2.5) \times 0.45]$ | M1 | For using $v^2 = 2ah$ |
| Speed is $1.5 \text{ ms}^{-1}$ | A1 | [2] |
**Part (iii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[-\frac{2}{7}(1.26)g = 1.26a\right]$ | M1 | For applying Newton's 2nd law to A |
| $a = -\dfrac{20}{7}$ | A1 | |
| $\left[v^2 = 2.25 + 2\left(-\dfrac{20}{7}\right)(0.03)\right]$ | M1 | For using $v^2 = v_B^2 + 2as$ |
| Speed is $1.44 \text{ ms}^{-1}$ | A1 | [4] |7\\
\includegraphics[max width=\textwidth, alt={}, center]{ceb367ee-4e12-4cb2-9020-078ea5724d6e-3_430_860_1585_641}
Particle $A$ of mass 1.26 kg and particle $B$ of mass 0.9 kg are attached to the ends of a light inextensible string. The string passes over a small smooth pulley $P$ which is fixed at the edge of a rough horizontal table. $A$ is held at rest at a point 0.48 m from $P$, and $B$ hangs vertically below $P$, at a height of 0.45 m above the floor (see diagram). The coefficient of friction between $A$ and the table is $\frac { 2 } { 7 } . A$ is released and the particles start to move.\\
(i) Show that the magnitude of the acceleration of the particles is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and find the tension in the string.\\
(ii) Find the speed with which $B$ reaches the floor.\\
(iii) Find the speed with which $A$ reaches the pulley.
\hfill \mbox{\textit{CAIE M1 2013 Q7 [11]}}