| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Particle on rough horizontal, string over pulley |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring force resolution, limiting equilibrium (μR = F), and then Newton's second law when the system moves. It involves multiple steps but uses routine mechanics techniques without requiring novel insight—slightly above average due to the two-part nature and intermediate connection point. |
| Spec | 3.03l Newton's third law: extend to situations requiring force resolution3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (Resolving forces horizontally on \(B\)) | M1 | For resolving forces horizontally on \(B\), including frictional force and using tensions in \(PB\) and \(BQ\) equal to weights of \(P\) and \(Q\) respectively |
| Frictional force \(= \mu \times 0.25g\) | B1 | |
| \(0.3g = 0.2g + \mu \times 0.25g \rightarrow\) Coefficient of friction is \(0.4\) | A1 | [Total: 3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (Applying Newton's 2nd law to \(P\) or \(B\)) | M1 | For applying Newton's 2nd law to \(P\) or to \(B\) |
| \(0.2g - T = 0.2a\) or \(T - 0.4 \times 0.25g = 0.25a\) | A1 | |
| \(T - 0.4 \times 0.25g = 0.25a\) or \(0.2g - T = 0.2a\) or \(0.2g - \mu \times 0.25g = (0.2 + 0.25)a\) | B1 | |
| (Solving for \(a\) and for \(T\)) | M1 | For solving for \(a\) and for \(T\) |
| Acceleration is \(2.22\) ms\(^{-2}\) | B1 | |
| Tension is \(1.56\) N | A1 | [Total: 6] |
## Question 5 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (Resolving forces horizontally on $B$) | M1 | For resolving forces horizontally on $B$, including frictional force and using tensions in $PB$ and $BQ$ equal to weights of $P$ and $Q$ respectively |
| Frictional force $= \mu \times 0.25g$ | B1 | |
| $0.3g = 0.2g + \mu \times 0.25g \rightarrow$ Coefficient of friction is $0.4$ | A1 | **[Total: 3]** |
## Question 5 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (Applying Newton's 2nd law to $P$ or $B$) | M1 | For applying Newton's 2nd law to $P$ or to $B$ |
| $0.2g - T = 0.2a$ **or** $T - 0.4 \times 0.25g = 0.25a$ | A1 | |
| $T - 0.4 \times 0.25g = 0.25a$ **or** $0.2g - T = 0.2a$ **or** $0.2g - \mu \times 0.25g = (0.2 + 0.25)a$ | B1 | |
| (Solving for $a$ and for $T$) | M1 | For solving for $a$ and for $T$ |
| Acceleration is $2.22$ ms$^{-2}$ | B1 | |
| Tension is $1.56$ N | A1 | **[Total: 6]** |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{ffefbc81-402f-4048-8741-23c8bae30d5a-3_250_846_260_648}
A small block $B$ of mass 0.25 kg is attached to the mid-point of a light inextensible string. Particles $P$ and $Q$, of masses 0.2 kg and 0.3 kg respectively, are attached to the ends of the string. The string passes over two smooth pulleys fixed at opposite sides of a rough table, with $B$ resting in limiting equilibrium on the table between the pulleys and particles $P$ and $Q$ and block $B$ are in the same vertical plane (see diagram).\\
(i) Find the coefficient of friction between $B$ and the table.\\
$Q$ is now removed so that $P$ and $B$ begin to move.\\
(ii) Find the acceleration of $P$ and the tension in the part $P B$ of the string.
\hfill \mbox{\textit{CAIE M1 2014 Q5 [9]}}