| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Multi-stage motion: changing surface conditions or external intervention |
| Difficulty | Challenging +1.2 This is a standard connected particles problem requiring systematic application of Newton's second law to two stages of motion (rough then smooth surface). While it involves multiple steps and careful bookkeeping of forces across different sections, the techniques are routine M1 content with no novel insights required. The multi-stage nature and need to track velocities between sections elevates it slightly above average difficulty. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Particle \(P\): \(2g\sin 30 - F - T = 2a\ [10 - F - T = 2a]\); Particle \(Q\): \(T - 0.25g = 0.25a\); System: \(2g\sin 30 - F - 0.25g = (2 + 0.25)a\) | M1 | Newton's second law on either particle or for the system with correct masses; correct number of terms, allow sin/cos mix, allow sign errors. Allow with *their* \(F\). |
| Both particle equations correct (with the same \(T\)) or system equation correct. Allow with *their* \(F\). If *their a* direction is different to ours, allow if *their a* is consistently used e.g. \(-2g\sin 30 + F + T = 2a'\) and \(0.25g - T = 0.25a'\). | A1 | |
| \(F = 0.3R = 0.3\times 2g\cos 30\left[= 3\sqrt{3} = 5.1961\ldots\right]\) | M1 | Use of \(F = 0.3R\), where \(R\) is a component of weight. |
| Acceleration from \(A\) to \(B = 1.02\ \text{ms}^{-2}\) | A1 | Solving for the acceleration from \(A\) to \(B\). Allow \(\dfrac{10 - 4\sqrt{3}}{3}\); AWRT 1.02. May see \(T = \dfrac{10 - \sqrt{3}}{3} = 2.7559\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of suvat from \(A\) to \(B\) to get an equation in \(v^2\) or \(v\). | \*M1 | Using \(u = 0\) and *their* \( |
| \(v^2 = \left[\dfrac{80 - 32\sqrt{3}}{15}\right] = 1.64\) OR \(v = 1.28\) | A1 | Not \(v = 1.29\). Allow 2sf or better without wrong work, i.e. \(v^2 = 1.6\) or \(v = 1.3\). |
| Find the acceleration from \(B\) to \(C\): \(2g\sin 30 - 0.25g = (2 + 0.25)a\) | \*M1 | Resolving on both particles and eliminate \(T\) (if *their a* direction is different to ours, allow if *their a* is consistently used) OR for the system to get an equation in \(a\) only. Correct number of relevant terms, allow sin/cos mix, allow sign errors. For reference \(a = \dfrac{10}{3}\) or \(3.33\). May see \(T = \dfrac{10}{3}\). |
| \(v^2 = (\text{their } 1.28)^2 + 2\times\left(\text{their } \dfrac{10}{3}\right)\times 1.2\) | DM1 | Use of suvat from \(B\) to \(C\), allow *their* positive \(a \neq g\) from (a)(ii) not *their a* from (a)(i) and *their* 1.28. Dependent on previous two marks. |
| Velocity \(= 3.1(0)\ \text{ms}^{-1}\) | A1 | AWRT 3.1(0) to 3sf. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of suvat from \(A\) to \(B\) to get an equation in \(v^2\) or \(v\) | \*M1 | Using their positive \(a\) from (a)(i). E.g. \(v^2 = 2 \times 0.8 \times (their\ 1.02)\) |
| \(v^2 = \left[\frac{80-32\sqrt{3}}{15}\right] = 1.64\) OR \(v = 1.28\) | A1 | Allow 2sf or better, i.e. \(v^2 = 1.6\) or \(v = 1.3\) |
| Change in PE \(= \pm(2g \times 1.2\sin 30 - 0.25g \times 1.2)\) OR Change in KE \(= \pm\left[\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(their\ 1.28)^2\right]\) | B1 | |
| \(\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(their\ 1.28)^2 = 2g \times 1.2\sin 30 - 0.25g \times 1.2\) | DM1 | Use of work-energy 6 terms; dimensionally correct. Allow sign errors. Allow sin/cos mix on PE. Dependent on previous M. |
| Velocity \(= 3.1(0)\) m s\(^{-1}\) | A1 | AWRT 3.1(0) to 3sf. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Change in PE \(= \pm(2g \times 2\sin 30 - 0.25g \times 2)\) | B1 | |
| Work done against friction \(= 0.3 \times 2g\cos 30 \times 0.8\) | B1 | |
| Change in KE \(= \frac{1}{2}(2+0.25)v^2\) | B1 | |
| \(\frac{1}{2}(2+0.25)v^2 + 0.3 \times 2g\cos 30 \times 0.8 = (2g \times 2\sin 30 - 0.25g \times 2)\) | M1 | Use of work-energy 5 terms; dimensionally correct. Must be considering both particles. Allow sign errors. Allow sin/cos mix on PE and/or WD against friction. |
| Velocity \(= 3.1(0)\) m s\(^{-1}\) | A1 | AWRT 3.1(0) to 3sf. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}(2+0.25)v^2 + 0.3 \times 2g\cos 30 \times 0.8 = 2g \times 0.8\sin 30 - 0.25g \times 0.8\) OR \(\frac{1}{2}\times 2 \times v^2 + (their\ 2.7559\ldots)\times 0.8 + 0.3\times 2g\cos 30\times 0.8 = 2g\times 0.8\sin 30\) OR \(\frac{1}{2}\times 0.25\times v^2 + 0.25g\times 0.8 = (their\ 2.7559\ldots)\times 0.8\) | \*M1 | Use of work-energy 5 terms; dimensionally correct. Allow sign errors. Allow sin/cos mix on PE. OR 4 terms; OR 3 terms; dimensionally correct. Allow sign errors. |
| \(v^2 = \left[\frac{80-32\sqrt{3}}{15}\right] = 1.64\) OR \(v = 1.28\) | A1 | Allow 2sf or better, i.e. \(v^2 = 1.6\) or \(v = 1.3\) |
| Change in PE \(= \pm(2g \times 1.2\sin 30 - 0.25g \times 1.2)\) OR Change in KE \(= \pm\left[\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(1.28)^2\right]\) | B1 | |
| \(\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(1.28)^2 = 2g \times 1.2\sin 30 - 0.25g \times 1.2\) | DM1 | Use of work-energy 6 terms; dimensionally correct. Allow sign errors. Allow sin/cos mix on PE. Dependent on previous 2 marks. |
| Velocity \(= 3.1(0)\) m s\(^{-1}\) | A1 | AWRT 3.1(0) to 3sf. |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.8 = 0 + \frac{1}{2}\times(their\ positive\ a\ from\ \textbf{(a)(i)})\times t_1^2\) and solve for \(t_1\); OR \(0.8 = \frac{1}{2}(0 + their\ positive\ 1.28)t_1\); OR \((their\ positive\ 1.28) = (their\ positive\ a)\cdot t_1\) and solve for \(t_1\) | M1 | Use of suvat from \(A\) to \(B\) to find \(t_1\), using \(s=0.8\) and their positive \(a \neq g\) from (a)(i). Must get to \(t_1 = \ldots\) OR using their positive 1.28; OR using their positive \(a \neq g\) from (a)(i) and positive 1.28. |
| \(1.2 = (their\ 1.28)t_2 + \frac{1}{2}\times\left(their\ \frac{10}{3}\right)\times t_2^2\) and solve for \(t_2\); OR \(1.2 = \frac{1}{2}((their\ 1.28)+(their\ answer\ to\ \textbf{(a)(ii)}))t_2\); OR \((their\ answer\ to\ \textbf{(a)(ii)}) = (their\ 1.28) + \left(their\ \frac{10}{3}\right)t_2\) | M1 | Use of suvat from \(B\) to \(C\) to find \(t_2\), using \(s=1.2\) and their positive \(a \neq g\) from (a)(ii) and their 1.28 leading to positive \(t_2\). Must get to \(t_2 = \ldots\) |
| \(t_1 = 1.25\) or \(t_2 = 0.547\) | A1 | Allow as expression. Allow \(t_1 = \frac{0.8}{0.64}\), \(t_2 = \frac{1.2}{2.19}\) OE. Allow 2sf or better. |
| Total time \(= 1.8(0)\) s | A1 | WWW. AWRT 1.8(0) to 3sf. |
| 4 |
## Question 7(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Particle $P$: $2g\sin 30 - F - T = 2a\ [10 - F - T = 2a]$; Particle $Q$: $T - 0.25g = 0.25a$; System: $2g\sin 30 - F - 0.25g = (2 + 0.25)a$ | M1 | Newton's second law on either particle or for the system with correct masses; correct number of terms, allow sin/cos mix, allow sign errors. Allow with *their* $F$. |
| Both particle equations correct (with the same $T$) or system equation correct. Allow with *their* $F$. If *their a* direction is different to ours, allow if *their a* is consistently used e.g. $-2g\sin 30 + F + T = 2a'$ and $0.25g - T = 0.25a'$. | A1 | |
| $F = 0.3R = 0.3\times 2g\cos 30\left[= 3\sqrt{3} = 5.1961\ldots\right]$ | M1 | Use of $F = 0.3R$, where $R$ is a component of weight. |
| Acceleration from $A$ to $B = 1.02\ \text{ms}^{-2}$ | A1 | Solving for the acceleration from $A$ to $B$. Allow $\dfrac{10 - 4\sqrt{3}}{3}$; AWRT 1.02. May see $T = \dfrac{10 - \sqrt{3}}{3} = 2.7559\ldots$ |
---
## Question 7(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of suvat from $A$ to $B$ to get an equation in $v^2$ or $v$. | \*M1 | Using $u = 0$ and *their* $|a|$ from **(a)(i)** to get a positive $v^2$. E.g. $v^2 = 2\times 0.8\times(\text{their } 1.02)$ or $v^2 = 2\times(-0.8)\times(-\text{their } 1.02)$. |
| $v^2 = \left[\dfrac{80 - 32\sqrt{3}}{15}\right] = 1.64$ OR $v = 1.28$ | A1 | Not $v = 1.29$. Allow 2sf or better without wrong work, i.e. $v^2 = 1.6$ or $v = 1.3$. |
| Find the acceleration from $B$ to $C$: $2g\sin 30 - 0.25g = (2 + 0.25)a$ | \*M1 | Resolving on both particles and eliminate $T$ (if *their a* direction is different to ours, allow if *their a* is consistently used) OR for the system to get an equation in $a$ only. Correct number of relevant terms, allow sin/cos mix, allow sign errors. For reference $a = \dfrac{10}{3}$ or $3.33$. May see $T = \dfrac{10}{3}$. |
| $v^2 = (\text{their } 1.28)^2 + 2\times\left(\text{their } \dfrac{10}{3}\right)\times 1.2$ | DM1 | Use of suvat from $B$ to $C$, allow *their* positive $a \neq g$ from **(a)(ii)** not *their a* from **(a)(i)** and *their* 1.28. Dependent on previous two marks. |
| Velocity $= 3.1(0)\ \text{ms}^{-1}$ | A1 | AWRT 3.1(0) to 3sf. |
## Question 7(a)(ii) – Alternative Method 1: suvat in first stage, energy in second stage
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of suvat from $A$ to $B$ to get an equation in $v^2$ or $v$ | \*M1 | Using their positive $a$ from (a)(i). E.g. $v^2 = 2 \times 0.8 \times (their\ 1.02)$ |
| $v^2 = \left[\frac{80-32\sqrt{3}}{15}\right] = 1.64$ OR $v = 1.28$ | A1 | Allow 2sf or better, i.e. $v^2 = 1.6$ or $v = 1.3$ |
| Change in PE $= \pm(2g \times 1.2\sin 30 - 0.25g \times 1.2)$ OR Change in KE $= \pm\left[\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(their\ 1.28)^2\right]$ | B1 | |
| $\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(their\ 1.28)^2 = 2g \times 1.2\sin 30 - 0.25g \times 1.2$ | DM1 | Use of work-energy 6 terms; dimensionally correct. Allow sign errors. Allow sin/cos mix on PE. Dependent on previous M. |
| Velocity $= 3.1(0)$ m s$^{-1}$ | A1 | AWRT 3.1(0) to 3sf. |
---
## Question 7(a)(ii) – Alternative Method 2: energy for complete motion
| Answer | Mark | Guidance |
|--------|------|----------|
| Change in PE $= \pm(2g \times 2\sin 30 - 0.25g \times 2)$ | B1 | |
| Work done against friction $= 0.3 \times 2g\cos 30 \times 0.8$ | B1 | |
| Change in KE $= \frac{1}{2}(2+0.25)v^2$ | B1 | |
| $\frac{1}{2}(2+0.25)v^2 + 0.3 \times 2g\cos 30 \times 0.8 = (2g \times 2\sin 30 - 0.25g \times 2)$ | M1 | Use of work-energy 5 terms; dimensionally correct. Must be considering both particles. Allow sign errors. Allow sin/cos mix on PE and/or WD against friction. |
| Velocity $= 3.1(0)$ m s$^{-1}$ | A1 | AWRT 3.1(0) to 3sf. |
---
## Question 7(a)(ii) – Alternative Method 3: energy in two stages
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}(2+0.25)v^2 + 0.3 \times 2g\cos 30 \times 0.8 = 2g \times 0.8\sin 30 - 0.25g \times 0.8$ OR $\frac{1}{2}\times 2 \times v^2 + (their\ 2.7559\ldots)\times 0.8 + 0.3\times 2g\cos 30\times 0.8 = 2g\times 0.8\sin 30$ OR $\frac{1}{2}\times 0.25\times v^2 + 0.25g\times 0.8 = (their\ 2.7559\ldots)\times 0.8$ | \*M1 | Use of work-energy 5 terms; dimensionally correct. Allow sign errors. Allow sin/cos mix on PE. OR 4 terms; OR 3 terms; dimensionally correct. Allow sign errors. |
| $v^2 = \left[\frac{80-32\sqrt{3}}{15}\right] = 1.64$ OR $v = 1.28$ | A1 | Allow 2sf or better, i.e. $v^2 = 1.6$ or $v = 1.3$ |
| Change in PE $= \pm(2g \times 1.2\sin 30 - 0.25g \times 1.2)$ OR Change in KE $= \pm\left[\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(1.28)^2\right]$ | B1 | |
| $\frac{1}{2}(2+0.25)v^2 - \frac{1}{2}(2+0.25)(1.28)^2 = 2g \times 1.2\sin 30 - 0.25g \times 1.2$ | DM1 | Use of work-energy 6 terms; dimensionally correct. Allow sign errors. Allow sin/cos mix on PE. Dependent on previous 2 marks. |
| Velocity $= 3.1(0)$ m s$^{-1}$ | A1 | AWRT 3.1(0) to 3sf. |
| | **5** | |
---
## Question 7(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.8 = 0 + \frac{1}{2}\times(their\ positive\ a\ from\ \textbf{(a)(i)})\times t_1^2$ and solve for $t_1$; OR $0.8 = \frac{1}{2}(0 + their\ positive\ 1.28)t_1$; OR $(their\ positive\ 1.28) = (their\ positive\ a)\cdot t_1$ and solve for $t_1$ | M1 | Use of suvat from $A$ to $B$ to find $t_1$, using $s=0.8$ and their positive $a \neq g$ from (a)(i). Must get to $t_1 = \ldots$ OR using their positive 1.28; OR using their positive $a \neq g$ from (a)(i) and positive 1.28. |
| $1.2 = (their\ 1.28)t_2 + \frac{1}{2}\times\left(their\ \frac{10}{3}\right)\times t_2^2$ and solve for $t_2$; OR $1.2 = \frac{1}{2}((their\ 1.28)+(their\ answer\ to\ \textbf{(a)(ii)}))t_2$; OR $(their\ answer\ to\ \textbf{(a)(ii)}) = (their\ 1.28) + \left(their\ \frac{10}{3}\right)t_2$ | M1 | Use of suvat from $B$ to $C$ to find $t_2$, using $s=1.2$ and their positive $a \neq g$ from (a)(ii) and their 1.28 leading to positive $t_2$. Must get to $t_2 = \ldots$ |
| $t_1 = 1.25$ or $t_2 = 0.547$ | A1 | Allow as expression. Allow $t_1 = \frac{0.8}{0.64}$, $t_2 = \frac{1.2}{2.19}$ OE. Allow 2sf or better. |
| Total time $= 1.8(0)$ s | A1 | WWW. AWRT 1.8(0) to 3sf. |
| | **4** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{e5ee28f2-5876-4149-9a77-18c5792c1bd8-10_551_776_260_689}
Two particles $P$ and $Q$, of masses 2 kg and 0.25 kg respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. Particle $P$ is on an inclined plane at an angle of $30 ^ { \circ }$ to the horizontal. Particle $Q$ hangs below the pulley. Three points $A , B$ and $C$ lie on a line of greatest slope of the plane with $A B = 0.8 \mathrm {~m}$ and $B C = 1.2 \mathrm {~m}$ (see diagram).
Particle $P$ is released from rest at $A$ with the string taut and slides down the plane. During the motion of $P$ from $A$ to $C , Q$ does not reach the pulley. The part of the plane from $A$ to $B$ is rough, with coefficient of friction 0.3 between the plane and $P$. The part of the plane from $B$ to $C$ is smooth.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the acceleration of $P$ between $A$ and $B$.
\item Hence, find the speed of $P$ at $C$.
\end{enumerate}\item Find the time taken for $P$ to travel from $A$ to $C$.\\
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]}}