| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Limiting equilibrium on incline |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system on inclines with straightforward application of Newton's second law and kinematics. Part (a) requires resolving forces and applying limiting equilibrium (a routine M1 technique), while part (b) involves finding acceleration and using SUVAT equations. The angles and masses are simple, requiring no geometric insight beyond standard resolution of forces parallel and perpendicular to inclines. This is slightly easier than average because it's a textbook application of well-practiced methods with no conceptual surprises. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([T = 2g\sin 10]\) or \([3g\sin 20 = F + T]\) | M1 | Resolve forces parallel to plane \(P\) for particle \(A\) or parallel to plane \(Q\) for particle \(B\) |
| \(T = 2g\sin 10\) and \(3g\sin 20 = F + T\) | A1 | |
| \(R = 30\cos 20\ (= 28.19...)\) | B1 | Resolving forces perpendicular to plane \(Q\) for particle \(B\) |
| \(\mu = \dfrac{3g\sin 20 - 2g\sin 10}{30\cos 20}\) | M1 | Using \(\mu = F/R\) |
| \(\mu = 0.241\ (=0.2407...)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3g\sin 20 - T = 3a\) or \(T - 2g\sin 10 = 2a\) or System: \(3g\sin 20 - 2g\sin 10 = 5a\) | M1 | For applying Newton's second law to either \(A\) or to \(B\) or to the system |
| \(a = \dfrac{(3g\sin 20 - 2g\sin 10)}{5}\) | M1 | For applying Newton's second law to the second particle and/or solving for \(a\) |
| \(a = 1.3575...\) | A1 | |
| \(h_1 = x\sin 20\); \(h_2 = x\sin 10\); \(x\sin 20 + x\sin 10 = 1\) | B1 | Using expressions for height change of each particle after each moves a distance \(x\) along the plane, to obtain equation in \(x\) |
| \(\dfrac{1}{\sin 10 + \sin 20} = 0 + \dfrac{1}{2} \times 1.3575 \times t^2\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) for either particle with \(s = x\), \(u = 0\) and using their \(a\ (= 1.3575)\) |
| \(t = 1.69\) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[T = 2g\sin 10]$ or $[3g\sin 20 = F + T]$ | M1 | Resolve forces parallel to plane $P$ for particle $A$ or parallel to plane $Q$ for particle $B$ |
| $T = 2g\sin 10$ **and** $3g\sin 20 = F + T$ | A1 | |
| $R = 30\cos 20\ (= 28.19...)$ | B1 | Resolving forces perpendicular to plane $Q$ for particle $B$ |
| $\mu = \dfrac{3g\sin 20 - 2g\sin 10}{30\cos 20}$ | M1 | Using $\mu = F/R$ |
| $\mu = 0.241\ (=0.2407...)$ | A1 | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3g\sin 20 - T = 3a$ or $T - 2g\sin 10 = 2a$ or System: $3g\sin 20 - 2g\sin 10 = 5a$ | M1 | For applying Newton's second law to either $A$ or to $B$ or to the system |
| $a = \dfrac{(3g\sin 20 - 2g\sin 10)}{5}$ | M1 | For applying Newton's second law to the second particle and/or solving for $a$ |
| $a = 1.3575...$ | A1 | |
| $h_1 = x\sin 20$; $h_2 = x\sin 10$; $x\sin 20 + x\sin 10 = 1$ | B1 | Using expressions for height change of each particle after each moves a distance $x$ along the plane, to obtain equation in $x$ |
| $\dfrac{1}{\sin 10 + \sin 20} = 0 + \dfrac{1}{2} \times 1.3575 \times t^2$ | M1 | For using $s = ut + \frac{1}{2}at^2$ for either particle with $s = x$, $u = 0$ and using their $a\ (= 1.3575)$ |
| $t = 1.69$ | A1 | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{ac4bb5a0-c7c0-4e1d-9e76-64f92ae28066-10_214_1461_255_342}
As shown in the diagram, particles $A$ and $B$ of masses 2 kg and 3 kg respectively are attached to the ends of a light inextensible string. The string passes over a small fixed smooth pulley which is attached to the top of two inclined planes. Particle $A$ is on plane $P$, which is inclined at an angle of $10 ^ { \circ }$ to the horizontal. Particle $B$ is on plane $Q$, which is inclined at an angle of $20 ^ { \circ }$ to the horizontal. The string is taut, and the two parts of the string are parallel to lines of greatest slope of their respective planes.
\begin{enumerate}[label=(\alph*)]
\item It is given that plane $P$ is smooth, plane $Q$ is rough, and the particles are in limiting equilibrium.
Find the coefficient of friction between particle $B$ and plane $Q$.
\item It is given instead that both planes are smooth and that the particles are released from rest at the same horizontal level.
Find the time taken until the difference in the vertical height of the particles is 1 m . [You should assume that this occurs before $A$ reaches the pulley or $B$ reaches the bottom of plane $Q$.] [6]\\
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 2020 Q7 [11]}}