| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Multi-part pulley system, subsequent motion |
| Difficulty | Standard +0.3 This is a standard connected particles problem with friction requiring resolution of forces, Newton's second law, and kinematics across three parts. Part (a) is routine equilibrium with friction; part (b) applies F=ma to the system; part (c) requires careful consideration of motion after impact but uses standard SUVAT equations. While multi-step, it follows predictable mechanics patterns without requiring novel insight, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 5g,\ F = 6g - 4g\) | M1 | For resolving forces to find \(F\) and \(R\) |
| \(\mu = \frac{2g}{5g} = 0.4\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_1 - 4g = 4a\) or \(8g - T_2 = 8a\) | M1 | For applying Newton's 2nd law to the 4 kg particle or the 8 kg particle |
| \(T_1 - 4g = 4a\) and \(8g - T_2 = 8a\) | A1 | Both equations correct |
| \(T_2 - T_1 - F = 5a\) and \(F = 0.4 \times 5g\) | B1 | |
| Adding gives \(8g - 4g - 2g = 17a\) leading to \(a = \ldots\) | M1 | Attempt to solve for \(a\), \(T_1\) or \(T_2\) |
| \(a = 1.18\ \text{ms}^{-2},\ T_1 = 44.7\) N, \(T_2 = 70.6\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 4g = 4a,\ {-T} - F = 5a,\ F = 2g\) or \(-4g - 2g = 9a\) | M1 | Applying Newton's 2nd law to both active particles |
| \(a = -\frac{60}{9}\) | A1 | |
| \(v^2 = 2 \times \frac{20}{17} \times 0.5 = \frac{20}{17}\) leading to \(v = \ldots\ [v = 1.0846...]\) | M1 | Use of \(v^2 = u^2 + 2as\) or equivalent to find \(v\) or \(v^2\) when the 8 kg particle reaches the ground |
| \(0 = \sqrt{\frac{20}{17}} - \frac{60}{9}t\) | M1 | Use of \(v = u + at\) or equivalent to find \(t\) |
| \(t = 0.163\) s | A1 | From \(t = 0.1626978\ldots\) |
## Question 6:
### Part (a):
| $R = 5g,\ F = 6g - 4g$ | M1 | For resolving forces to find $F$ and $R$ |
| $\mu = \frac{2g}{5g} = 0.4$ | A1 | AG |
### Part (b):
| $T_1 - 4g = 4a$ or $8g - T_2 = 8a$ | M1 | For applying Newton's 2nd law to the 4 kg particle or the 8 kg particle |
| $T_1 - 4g = 4a$ and $8g - T_2 = 8a$ | A1 | Both equations correct |
| $T_2 - T_1 - F = 5a$ and $F = 0.4 \times 5g$ | B1 | |
| Adding gives $8g - 4g - 2g = 17a$ leading to $a = \ldots$ | M1 | Attempt to solve for $a$, $T_1$ or $T_2$ |
| $a = 1.18\ \text{ms}^{-2},\ T_1 = 44.7$ N, $T_2 = 70.6$ N | A1 | |
### Part (c):
| $T - 4g = 4a,\ {-T} - F = 5a,\ F = 2g$ or $-4g - 2g = 9a$ | M1 | Applying Newton's 2nd law to both active particles |
| $a = -\frac{60}{9}$ | A1 | |
| $v^2 = 2 \times \frac{20}{17} \times 0.5 = \frac{20}{17}$ leading to $v = \ldots\ [v = 1.0846...]$ | M1 | Use of $v^2 = u^2 + 2as$ or equivalent to find $v$ or $v^2$ when the 8 kg particle reaches the ground |
| $0 = \sqrt{\frac{20}{17}} - \frac{60}{9}t$ | M1 | Use of $v = u + at$ or equivalent to find $t$ |
| $t = 0.163$ s | A1 | From $t = 0.1626978\ldots$ |6\\
\includegraphics[max width=\textwidth, alt={}, center]{cb2cec83-6f8d-4c13-90a1-03bbf4e4452f-10_451_1315_258_415}
The diagram shows a particle of mass 5 kg on a rough horizontal table, and two light inextensible strings attached to it passing over smooth pulleys fixed at the edges of the table. Particles of masses 4 kg and 6 kg hang freely at the ends of the strings. The particle of mass 6 kg is 0.5 m above the ground. The system is in limiting equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Show that the coefficient of friction between the 5 kg particle and the table is 0.4 .\\
The 6 kg particle is now replaced by a particle of mass 8 kg and the system is released from rest.
\item Find the acceleration of the 4 kg particle and the tensions in the strings.
\item In the subsequent motion the 8 kg particle hits the ground and does not rebound.
Find the time that elapses after the 8 kg particle hits the ground before the other two particles come to instantaneous rest. (You may assume this occurs before either particle reaches a pulley.)\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q6 [12]}}