| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough incline, particle hanging |
| Difficulty | Standard +0.8 This is a two-part pulley system question requiring equilibrium analysis with friction (part a) and motion in two phases with energy considerations (part b). While the individual techniques are standard M1 content, part (b) requires careful tracking of motion after B hits the ground and solving for when A comes to rest, involving multiple equations and phases of motion. This is more demanding than typical single-phase pulley problems. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolving for both particles or for the system to form equation(s) | M1* | Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow \(g\) missing. M0 if acceleration included unless subsequently equated to zero. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. |
| Either \(T - F - 2.4g\sin 30 = 0\) AND \(3.3g - T = 0\), Or \(3.3g - F - 2.4g\sin 30 = 0\) | A1 | Both correct or system correct. May get \(F = 21\). Can be with a wrong non-zero \(F\). |
| \(R = 2.4g\cos 30 \left[= 12\sqrt{3} = 20.7846...\right]\) | B1 | |
| Use of \(F = \mu R\) to get an equation in \(\mu\) only; \(\left[3.3g - 2.4g\mu\cos 30 - 2.4g\sin 30 = 0\right]\) | DM1 | Must be from \(F\) dimensionally correct and single term \(R\) which is equal to a component the 2.4 kg weight. Allow consistent sin/cos mix but must be different components of weight. \(F\) and \(R\) must be numerical expressions. |
| \(\mu = 1.01\) [sight of \(1.01036...\) or \(1.0104\)] | A1 | AG perhaps from one of \(\mu = \dfrac{3.3g - 24\sin 30}{2.4g\cos 30} = \dfrac{33-12}{12\sqrt{3}} = \dfrac{21}{12\sqrt{3}} = \dfrac{7\sqrt{3}}{12} = \dfrac{21}{20.7846...} = \dfrac{21}{20.8}\). Do not allow unless evidence of 30 substituted for \(\theta\). E.g.: sight of \(1.01036...\) or \(1.0104\). |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using Newton's second law for both particles or the system | M1* | Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow \(g\) missing. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. |
| Either \(3.3g - T = 3.3a\) and \(T - F - 2.4g\sin 20 = 2.4a\); \(\left[T - 22.778 - 8.208 = 2.4a\right]\) or \(\left[T - 30.986 = 2.4a\right]\); or \(3.3g - F - 2.4g\sin 20 = (2.4 + 3.3)a\ \left[2.013367... = 5.7a\right]\) | A1 | Both correct or system equation correct. Can be with a wrong non-zero \(F\). |
| \(F = 1.01 \times 2.4g\cos 20 \left[= 22.778\right]\) | B1 | For correct expression for \(F\). |
| Attempt to solve for \(a\); \(a = 0.353\ \left[0.353222...\right]\) | DM1 | Using their \(F\). Must get to '\(a =\)'. If sin/cos mix must be consistent. |
| \(v^2 = 2 \times 0.353 \times 1 \left[= 0.706444...\right]\) or \(\left[v = 0.841\right]\); Or \(1 = 0 + \frac{1}{2} \times 0.353t^2 \Rightarrow t = 2.3795 \Rightarrow v = 0.353 \times 2.38\) | A1FT | FT their value of \(a \neq \pm g\) to get an expression for \(v^2\) or \(v\). Can be implied by awrt 0.84 for \(v\) or awrt 0.71 for \(v^2\). This mark does not depend on previous A or B mark, but both Ms must have been awarded. |
| Using Newton's second law on \(A\) after \(B\) reaches the ground: \(-F - 2.4g\sin 20 = 2.4a\); \(\left[-1.01 \times 2.4g\cos 20 - 2.4g\sin 20 = 2.4a\right]\); \(\left[-22.78814... - 8.20848... = 2.4a\right]\) | M1* | Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow \(g\) missing. \(\left[\Rightarrow a = -12.911...\right]\) |
| Use of suvat to find \(s\): \(\left[0 = \text{their } 0.841^2 + 2 \times \text{their } {-12.911...} \times s \Rightarrow s = 0.027358...\right]\) | DM1 | Using their \(a \neq \pm g\). Must get to '\(s =\)'. May find and use \(t = 0.0651\). |
| Total distance \(= 1.03\) m | A1 | |
| Total: 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[\text{KE gained} =\right] \frac{1}{2} \times (2.4 + 3.3)v^2 \left[= 2.85v^2\right]\) | B1 | |
| \(\left[\text{PE lost} =\right] 3.3g \times 1 - 2.4g \times 1\sin 20 \left[= 24.791...\right]\) | B1 | Allow omission of 1 in either or both terms. |
| \(\left[\text{Friction} =\right] 1.01 \times 2.4g\cos 20 \left[= 22.778...\right]\) | B1 | For correct expression for \(F\). |
| \(\frac{1}{2} \times (2.4+3.3)v^2 = 3.3g \times 1 - 2.4g \times 1\sin 20 - 1.01 \times 2.4g\cos 20 \times 1\); Or \(2.85v^2 = 24.791... - 22.778...\) | M1 | For attempt at energy equation. Allow sign errors, allow sin/cos mix but must have sin/cos where needed. Correct number of terms, dimensionally correct. Allow omission of 1 in any or all the three relevant terms. Must have \(\cos 20\) and \(\sin 20\). |
| To get a correct expression for \(v^2\): \(\left[v^2 = 0.706444...\text{ or } v = 0.841...\right]\) | A1 | Can be implied by awrt 0.84 for \(v\) or awrt 0.71 for \(v^2\) if expression not seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{KE} = \frac{1}{2} \times 2.4 \times 0.841^2\) | M1 | Using their \(v^2\). |
| \(1.01 \times 2.4g\cos 20 \times s + 2.4g\sin 20 \times s = \frac{1}{2} \times 2.4 \times 0.841^2\); \(\left[\Rightarrow s = 0.027358..\right]\) | M1 | For attempt at 3 term energy equation and solved to get '\(s =\)'. Allow sign errors, allow consistent sin/cos mix but must have sin/cos where needed. Correct number of terms, dimensionally correct. |
| Total distance \(= 1.03\) m | A1 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolving for both particles or for the system to form equation(s) | M1* | Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow $g$ missing. M0 if acceleration included unless subsequently equated to zero. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. |
| Either $T - F - 2.4g\sin 30 = 0$ AND $3.3g - T = 0$, Or $3.3g - F - 2.4g\sin 30 = 0$ | A1 | Both correct or system correct. May get $F = 21$. Can be with a wrong non-zero $F$. |
| $R = 2.4g\cos 30 \left[= 12\sqrt{3} = 20.7846...\right]$ | B1 | |
| Use of $F = \mu R$ to get an equation in $\mu$ only; $\left[3.3g - 2.4g\mu\cos 30 - 2.4g\sin 30 = 0\right]$ | DM1 | Must be from $F$ dimensionally correct and single term $R$ which is equal to a component the 2.4 kg weight. Allow consistent sin/cos mix but must be different components of weight. $F$ and $R$ must be numerical expressions. |
| $\mu = 1.01$ [sight of $1.01036...$ or $1.0104$] | A1 | AG perhaps from one of $\mu = \dfrac{3.3g - 24\sin 30}{2.4g\cos 30} = \dfrac{33-12}{12\sqrt{3}} = \dfrac{21}{12\sqrt{3}} = \dfrac{7\sqrt{3}}{12} = \dfrac{21}{20.7846...} = \dfrac{21}{20.8}$. Do not allow unless evidence of 30 substituted for $\theta$. E.g.: sight of $1.01036...$ or $1.0104$. |
| **Total: 5** | | |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Using Newton's second law for both particles or the system | M1* | Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow $g$ missing. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. |
| Either $3.3g - T = 3.3a$ and $T - F - 2.4g\sin 20 = 2.4a$; $\left[T - 22.778 - 8.208 = 2.4a\right]$ or $\left[T - 30.986 = 2.4a\right]$; or $3.3g - F - 2.4g\sin 20 = (2.4 + 3.3)a\ \left[2.013367... = 5.7a\right]$ | A1 | Both correct or system equation correct. Can be with a wrong non-zero $F$. |
| $F = 1.01 \times 2.4g\cos 20 \left[= 22.778\right]$ | B1 | For correct expression for $F$. |
| Attempt to solve for $a$; $a = 0.353\ \left[0.353222...\right]$ | DM1 | Using their $F$. Must get to '$a =$'. If sin/cos mix must be consistent. |
| $v^2 = 2 \times 0.353 \times 1 \left[= 0.706444...\right]$ or $\left[v = 0.841\right]$; Or $1 = 0 + \frac{1}{2} \times 0.353t^2 \Rightarrow t = 2.3795 \Rightarrow v = 0.353 \times 2.38$ | A1FT | FT their value of $a \neq \pm g$ to get an expression for $v^2$ or $v$. Can be implied by awrt 0.84 for $v$ or awrt 0.71 for $v^2$. This mark does not depend on previous A or B mark, but both Ms must have been awarded. |
| Using Newton's second law on $A$ after $B$ reaches the ground: $-F - 2.4g\sin 20 = 2.4a$; $\left[-1.01 \times 2.4g\cos 20 - 2.4g\sin 20 = 2.4a\right]$; $\left[-22.78814... - 8.20848... = 2.4a\right]$ | M1* | Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow $g$ missing. $\left[\Rightarrow a = -12.911...\right]$ |
| Use of suvat to find $s$: $\left[0 = \text{their } 0.841^2 + 2 \times \text{their } {-12.911...} \times s \Rightarrow s = 0.027358...\right]$ | DM1 | Using their $a \neq \pm g$. Must get to '$s =$'. May find and use $t = 0.0651$. |
| Total distance $= 1.03$ m | A1 | |
| **Total: 8** | | |
**Alternative method using energy for first 5 marks:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\text{KE gained} =\right] \frac{1}{2} \times (2.4 + 3.3)v^2 \left[= 2.85v^2\right]$ | B1 | |
| $\left[\text{PE lost} =\right] 3.3g \times 1 - 2.4g \times 1\sin 20 \left[= 24.791...\right]$ | B1 | Allow omission of 1 in either or both terms. |
| $\left[\text{Friction} =\right] 1.01 \times 2.4g\cos 20 \left[= 22.778...\right]$ | B1 | For correct expression for $F$. |
| $\frac{1}{2} \times (2.4+3.3)v^2 = 3.3g \times 1 - 2.4g \times 1\sin 20 - 1.01 \times 2.4g\cos 20 \times 1$; Or $2.85v^2 = 24.791... - 22.778...$ | M1 | For attempt at energy equation. Allow sign errors, allow sin/cos mix but must have sin/cos where needed. Correct number of terms, dimensionally correct. Allow omission of 1 in any or all the three relevant terms. Must have $\cos 20$ and $\sin 20$. |
| To get a correct expression for $v^2$: $\left[v^2 = 0.706444...\text{ or } v = 0.841...\right]$ | A1 | Can be implied by awrt 0.84 for $v$ or awrt 0.71 for $v^2$ if expression not seen. |
**Alternative method using energy for final 3 marks:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{KE} = \frac{1}{2} \times 2.4 \times 0.841^2$ | M1 | Using their $v^2$. |
| $1.01 \times 2.4g\cos 20 \times s + 2.4g\sin 20 \times s = \frac{1}{2} \times 2.4 \times 0.841^2$; $\left[\Rightarrow s = 0.027358..\right]$ | M1 | For attempt at 3 term energy equation and solved to get '$s =$'. Allow sign errors, allow consistent sin/cos mix but must have sin/cos where needed. Correct number of terms, dimensionally correct. |
| Total distance $= 1.03$ m | A1 | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{99f20949-471d-4da3-a680-ec24abf6baa5-10_335_937_255_605}
Particles $A$ and $B$, of masses 2.4 kg and 3.3 kg respectively, are connected by a light inextensible string that passes over a smooth pulley which is fixed to the top of a rough plane. The plane makes an angle of $\theta ^ { \circ }$ with horizontal ground. Particle $A$ is on the plane and the section of the string between $A$ and the pulley is parallel to a line of greatest slope of the plane. Particle $B$ hangs vertically below the pulley and is 1 m above the ground (see diagram). The coefficient of friction between the plane and $A$ is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item It is given that $\theta = 30$ and the system is in equilibrium with $A$ on the point of moving directly up the plane.
Show that $\mu = 1.01$ correct to 3 significant figures.
\item It is given instead that $\theta = 20$ and $\mu = 1.01$. The system is released from rest with the string taut.
Find the total distance travelled by $A$ before coming to instantaneous rest. You may assume that $A$ does not reach the pulley and that $B$ remains at rest after it hits the ground.\\
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 2023 Q7 [13]}}