| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Particle sliding track with friction sections |
| Difficulty | Standard +0.8 This is a multi-stage energy problem requiring careful application of work-energy principles across three different sections with varying friction conditions. Part (a) requires tracking energy through smooth and rough sections to find distance, while part (b) involves working backwards from rest to find μ and then analyzing return motion. The 'no change in speed' constraint at transitions adds conceptual complexity beyond standard inclined plane problems. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(CD\): \(R = mg\cos 30\) | B1 | May be seen in later working without. |
| Use of \(F = 0.1R\) for either \(BC\) or \(CD\); \(F_{BC} = 0.1mg\ [=m]\) OR \(F_{CD} = 0.1mg\cos 30\ \left[=\frac{\sqrt{3}}{2}m\right]\) | M1 | Note: The first two marks are often gained in the work-energy equation. |
| \(0.1mg\cos 30d + mgd\sin 30\ \left[\left(\frac{\sqrt{3}}{2}+5\right)md\right]\) | A1 | For sum of work done by friction and the change in PE. Note: Allow terms on different sides of a work energy equation as long as they have different signs. |
| \(mg \times 2\sin 30 = 0.1mg \times 2 + 0.1mg\cos 30d + mgd\sin 30 + \frac{1}{2}m \times 1^2\) \(\left[10m = 2m + m\cos 30d + 5md + \frac{1}{2}m\right]\) | M1 | Attempt at work energy equation with five relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. Note: Initial PE \(= mg\). |
| \(d = 1.28\) m or \(\dfrac{15(10-\sqrt{3})}{97}\ [1.27854\ldots]\) | A1 | ISW if go on to find total distance \(= 2 + 2 + 1.28\) having already found 1.28. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(CD\): \(R = mg\cos 30\) | (B1) | May be seen in later working without \(m\). If not seen in working check diagram but must be a reaction force, not a downward component of the weight. |
| Use of \(F = 0.1R\) for either \(BC\) or \(CD\); \(F_{BC} = 0.1mg\ [=m]\) or \(F_{CD} = 0.1mg\cos 30\left[=\frac{\sqrt{3}}{2}m\right]\) | (M1) | |
| For \(a_{CD}\): \(-mg\sin 30 - 0.1mg\cos 30 = ma\) \(\left[a = -g\sin 30 - 0.1g\cos 30 = -5.866\ldots = -\left(5 + \frac{\sqrt{3}}{2}\right)\right]\) | (A1) | For correct equation for \(a\) or \(ma\) in section \(CD\). Note: Allow if acceleration in the opposite sense and both signs positive. |
| For \(a_{AB}\): \(mg\sin 30 = ma \Rightarrow a = 5 \Rightarrow v_B^2 = 0 + 2\times5\times2\ [=20]\) For \(a_{BC}\): \(-0.1mg = ma\), \(a = -1\) so \(v_C^2 = 20 - 2\times1\times2\ [=16]\) \(1^2 = 16 - 2(g\sin 30 + 0.1g\cos 30)d\ \left[1 = 16 - 2\left(5+\frac{\sqrt{3}}{2}\right)d\right]\) | (M1) | Attempt to find \(d\). Allow sign errors in Newton's second law. Allow sin/cos errors but must be consistent. Should include a valid attempt at \(v_C^2\) to get M1. Must get to final line of working. Note: this mark can be earned even if A0 above. Must have 2 term acceleration though could have sign error. |
| \(d = 1.28\) m or \(\dfrac{15(10-\sqrt{3})}{97}\ [1.27854\ldots]\) | (A1) | ISW if go on to find total distance \(= 2 + 2 + 1.28\) having already found 1.28. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(a_{AB}\): \(mg\sin 30 = ma \Rightarrow a=5 \Rightarrow v_B^2 = 0+2\times5\times2\ [=20]\) For \(a_{BC}\): \(-0.1mg = ma\), \(a=-1\) so \(v_C^2 = 20-2\times1\times2\ [=16]\) \(\frac{1}{2}m(1^2-4^2) = -mgd\sin 30 - 0.1mg\cos 30 \times d\) | (M1) | Attempt at work energy equation for the third phase with four relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. Must get to final line. |
| \(d = 1.28\) m or \(\dfrac{15(10-\sqrt{3})}{97}\ [1.27854\ldots]\) ignore units | (A1) | ISW if go on to find total distance \(= 2 + 2 + 1.28\) having already found 1.28. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(mg \times 2\sin 30 = 2\mu mg + 1\times\mu mg\cos 30 + mg\times1\sin 30\) \(\left[10m = 20m\mu + 10m\cos 30\mu + 5m\ \text{ OR }\ 10m = 20m\mu + m5\sqrt{3}\mu + 5m\right]\) | M1 | Attempt at work energy equation with four relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. |
| \(\mu = 0.174\) or \(\dfrac{4-\sqrt{3}}{13}\ [0.174457\ldots]\) | A1 | |
| \(mg\times1\sin 30 - 1\times\mu mg\cos 30\) \(\left[5m - 5\sqrt{3}m\mu\right]\) | M1 | For difference between the change in PE and the work done by friction. Note: Allow terms on different sides of a work energy equation as long as both have the same sign. Allow sin/cos errors but must be consistent. Using \(\mu\), their \(\mu\) or the correct value of \(\mu\) to at least 2 sf. Must be as part of an attempt to find speed, not \(\mu\), although this could be the first step. |
| \(mg\times1\sin 30 - 1\times\mu mg\cos 30 = \frac{1}{2}mv^2\ \left[5m - 5\sqrt{3}m\mu = \frac{1}{2}mv^2\right]\) Speed \(= 2.64\ \text{ms}^{-1}\ [2.64164\ldots]\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(a_{AB}\): \(mg\sin 30 = ma \Rightarrow a=5 \Rightarrow v_B^2 = 0+2\times5\times2\ [=20]\) For \(a_{BC}\): \(-\mu mg = ma\), \(a=-\mu g\) so \(v_C^2 = 20-2\mu g\times2\) For \(a_{CD}\): \(-mg\sin 30 - \mu mg\cos 30 = ma\ \left[a = -5.866\ldots = -\left(5+\frac{\sqrt{3}}{2}\right)\right]\) \(0 = 20 - 2\mu g\times2 - 2(g\sin 30 + \mu g\cos 30)\times1\) \(\left[20 - 40\mu - 10 - 10\sqrt{3}\mu = 0\right]\) | (M1) | For attempt at equation for \(\mu\). Allow sign errors. Allow sin/cos errors but must be consistent. Must get to fourth line for M1. |
| \(\mu = 0.174\) or \(\dfrac{4-\sqrt{3}}{13}\ [0.174457\ldots]\) | (A1) | |
| \(a = g\sin 30 - \mu g\cos 30\ \left[= 5 - 5\sqrt{3}\mu\right]\) | (M1) | For correct equation for \(a\) or \(ma\) in section \(CD\) down plane (weight component \(-\) friction). Allow sin/cos errors but must be consistent. Using \(\mu\), their \(\mu\) or the correct value of \(\mu\) to at least 2sf. Must be as part of an attempt to find speed, not \(\mu\), although this could be the first step. |
| \(v^2 = 0 + 2(g\sin 30 - \mu g\cos 30)\times1 \Rightarrow\) Speed \(= 2.64\ \text{ms}^{-1}\ [2.64164\ldots]\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg \times 2\sin 30 - (\mu mg \times 2 + \mu mg \cos 30 \times 2) = \frac{1}{2}mv^2\) | (M1) | For \(PE_A\) − total work done against friction \([= KE_C]\). Allow sin/cos errors but must be consistent. Using \(\mu\), *their* \(\mu\) or the correct value of \(\mu\) to at least 2sf. |
| \(\left[10m - (20m\mu + 20m\cos 30\,\mu) = \frac{1}{2}mv^2\right]\) | (A1) | |
| \(\left[10m - (20m\mu + 10m\sqrt{3}\,\mu) = \frac{1}{2}mv^2\right]\) | ||
| \(\text{Speed} = 2.64\ \text{ms}^{-1}\ [2.64164\ldots]\) | ||
| 4 |
## Question 7(a):
### Method 1 (Work-Energy):
| Answer | Mark | Guidance |
|--------|------|----------|
| For $CD$: $R = mg\cos 30$ | B1 | May be seen in later working without. |
| Use of $F = 0.1R$ for either $BC$ or $CD$; $F_{BC} = 0.1mg\ [=m]$ OR $F_{CD} = 0.1mg\cos 30\ \left[=\frac{\sqrt{3}}{2}m\right]$ | M1 | Note: The first two marks are often gained in the work-energy equation. |
| $0.1mg\cos 30d + mgd\sin 30\ \left[\left(\frac{\sqrt{3}}{2}+5\right)md\right]$ | A1 | For sum of work done by friction and the change in PE. Note: Allow terms on different sides of a work energy equation as long as they have different signs. |
| $mg \times 2\sin 30 = 0.1mg \times 2 + 0.1mg\cos 30d + mgd\sin 30 + \frac{1}{2}m \times 1^2$ $\left[10m = 2m + m\cos 30d + 5md + \frac{1}{2}m\right]$ | M1 | Attempt at work energy equation with five relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. Note: Initial PE $= mg$. |
| $d = 1.28$ m or $\dfrac{15(10-\sqrt{3})}{97}\ [1.27854\ldots]$ | A1 | ISW if go on to find total distance $= 2 + 2 + 1.28$ having already found 1.28. |
### Method 2 (Newton's Second Law):
| Answer | Mark | Guidance |
|--------|------|----------|
| For $CD$: $R = mg\cos 30$ | (B1) | May be seen in later working without $m$. If not seen in working check diagram but must be a reaction force, not a downward component of the weight. |
| Use of $F = 0.1R$ for either $BC$ or $CD$; $F_{BC} = 0.1mg\ [=m]$ or $F_{CD} = 0.1mg\cos 30\left[=\frac{\sqrt{3}}{2}m\right]$ | (M1) | |
| For $a_{CD}$: $-mg\sin 30 - 0.1mg\cos 30 = ma$ $\left[a = -g\sin 30 - 0.1g\cos 30 = -5.866\ldots = -\left(5 + \frac{\sqrt{3}}{2}\right)\right]$ | (A1) | For correct equation for $a$ or $ma$ in section $CD$. Note: Allow if acceleration in the opposite sense and both signs positive. |
| For $a_{AB}$: $mg\sin 30 = ma \Rightarrow a = 5 \Rightarrow v_B^2 = 0 + 2\times5\times2\ [=20]$ For $a_{BC}$: $-0.1mg = ma$, $a = -1$ so $v_C^2 = 20 - 2\times1\times2\ [=16]$ $1^2 = 16 - 2(g\sin 30 + 0.1g\cos 30)d\ \left[1 = 16 - 2\left(5+\frac{\sqrt{3}}{2}\right)d\right]$ | (M1) | Attempt to find $d$. Allow sign errors in Newton's second law. Allow sin/cos errors but must be consistent. Should include a valid attempt at $v_C^2$ to get M1. Must get to final line of working. Note: this mark can be earned even if A0 above. Must have 2 term acceleration though could have sign error. |
| $d = 1.28$ m or $\dfrac{15(10-\sqrt{3})}{97}\ [1.27854\ldots]$ | (A1) | ISW if go on to find total distance $= 2 + 2 + 1.28$ having already found 1.28. |
### Alternative Method for last 2 marks (Energy method for third phase):
| Answer | Mark | Guidance |
|--------|------|----------|
| For $a_{AB}$: $mg\sin 30 = ma \Rightarrow a=5 \Rightarrow v_B^2 = 0+2\times5\times2\ [=20]$ For $a_{BC}$: $-0.1mg = ma$, $a=-1$ so $v_C^2 = 20-2\times1\times2\ [=16]$ $\frac{1}{2}m(1^2-4^2) = -mgd\sin 30 - 0.1mg\cos 30 \times d$ | (M1) | Attempt at work energy equation for the third phase with four relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. Must get to final line. |
| $d = 1.28$ m or $\dfrac{15(10-\sqrt{3})}{97}\ [1.27854\ldots]$ ignore units | (A1) | ISW if go on to find total distance $= 2 + 2 + 1.28$ having already found 1.28. |
**Total: 5 marks**
---
## Question 7(b):
### Method 1 (Work-Energy):
| Answer | Mark | Guidance |
|--------|------|----------|
| $mg \times 2\sin 30 = 2\mu mg + 1\times\mu mg\cos 30 + mg\times1\sin 30$ $\left[10m = 20m\mu + 10m\cos 30\mu + 5m\ \text{ OR }\ 10m = 20m\mu + m5\sqrt{3}\mu + 5m\right]$ | M1 | Attempt at work energy equation with four relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. |
| $\mu = 0.174$ or $\dfrac{4-\sqrt{3}}{13}\ [0.174457\ldots]$ | A1 | |
| $mg\times1\sin 30 - 1\times\mu mg\cos 30$ $\left[5m - 5\sqrt{3}m\mu\right]$ | M1 | For difference between the change in PE and the work done by friction. Note: Allow terms on different sides of a work energy equation as long as both have the same sign. Allow sin/cos errors but must be consistent. Using $\mu$, their $\mu$ or the correct value of $\mu$ to at least 2 sf. Must be as part of an attempt to find speed, not $\mu$, although this could be the first step. |
| $mg\times1\sin 30 - 1\times\mu mg\cos 30 = \frac{1}{2}mv^2\ \left[5m - 5\sqrt{3}m\mu = \frac{1}{2}mv^2\right]$ Speed $= 2.64\ \text{ms}^{-1}\ [2.64164\ldots]$ | A1 | |
### Method 2 (Newton's Second Law):
| Answer | Mark | Guidance |
|--------|------|----------|
| For $a_{AB}$: $mg\sin 30 = ma \Rightarrow a=5 \Rightarrow v_B^2 = 0+2\times5\times2\ [=20]$ For $a_{BC}$: $-\mu mg = ma$, $a=-\mu g$ so $v_C^2 = 20-2\mu g\times2$ For $a_{CD}$: $-mg\sin 30 - \mu mg\cos 30 = ma\ \left[a = -5.866\ldots = -\left(5+\frac{\sqrt{3}}{2}\right)\right]$ $0 = 20 - 2\mu g\times2 - 2(g\sin 30 + \mu g\cos 30)\times1$ $\left[20 - 40\mu - 10 - 10\sqrt{3}\mu = 0\right]$ | (M1) | For attempt at equation for $\mu$. Allow sign errors. Allow sin/cos errors but must be consistent. Must get to fourth line for M1. |
| $\mu = 0.174$ or $\dfrac{4-\sqrt{3}}{13}\ [0.174457\ldots]$ | (A1) | |
| $a = g\sin 30 - \mu g\cos 30\ \left[= 5 - 5\sqrt{3}\mu\right]$ | (M1) | For correct equation for $a$ or $ma$ in section $CD$ down plane (weight component $-$ friction). Allow sin/cos errors but must be consistent. Using $\mu$, their $\mu$ or the correct value of $\mu$ to at least 2sf. Must be as part of an attempt to find speed, not $\mu$, although this could be the first step. |
| $v^2 = 0 + 2(g\sin 30 - \mu g\cos 30)\times1 \Rightarrow$ Speed $= 2.64\ \text{ms}^{-1}\ [2.64164\ldots]$ | (A1) | |
## Question 7(b) — Alternative Method:
**Using energy at the start – total work done against friction**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg \times 2\sin 30 - (\mu mg \times 2 + \mu mg \cos 30 \times 2) = \frac{1}{2}mv^2$ | **(M1)** | For $PE_A$ − total work done against friction $[= KE_C]$. Allow sin/cos errors but must be consistent. Using $\mu$, *their* $\mu$ or the correct value of $\mu$ to at least 2sf. |
| $\left[10m - (20m\mu + 20m\cos 30\,\mu) = \frac{1}{2}mv^2\right]$ | **(A1)** | |
| $\left[10m - (20m\mu + 10m\sqrt{3}\,\mu) = \frac{1}{2}mv^2\right]$ | | |
| $\text{Speed} = 2.64\ \text{ms}^{-1}\ [2.64164\ldots]$ | | |
| | **4** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{c3246fbe-6f77-48f7-98eb-19e9166008bc-10_323_1308_292_376}
The diagram shows a track $A B C D$ which lies in a vertical plane. The section $A B$ is a straight line inclined at an angle of $30 ^ { \circ }$ to the horizontal and is smooth. The section $B C$ is a horizontal straight line and is rough. The section CD is a straight line inclined at an angle of $30 ^ { \circ }$ to the horizontal and is rough. The lengths $A B , B C$ and $C D$ are each 2 m .
A particle is released from rest at $A$. The coefficient of friction between the particle and both $B C$ and $C D$ is $\mu$. There is no change in the speed of the particle when it passes through either of the points $B$ or $C$.
\begin{enumerate}[label=(\alph*)]
\item It is given that $\mu = 0.1$.
Find the distance which the particle has moved up the section $C D$ when its speed is $1 \mathrm {~ms} ^ { - 1 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{c3246fbe-6f77-48f7-98eb-19e9166008bc-10_2716_33_143_2014}
\item It is given instead that with a different value of $\mu$ the particle travels 1 m up the track from $C$ before it comes instantaneously to rest.
Find the value of $\mu$ and the speed of the particle at the instant that it passes $C$ for the second time.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q7 [9]}}