| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with two parts: (a) uses work-energy theorem with given work against resistance, requiring one equation; (b) uses energy methods to find friction coefficient on an inclined plane. Both parts are standard M1 applications with clear given values and no conceptual surprises, making it slightly easier than average. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PE lost \(=mgh=25g\times1.8\ [=450]\) | B1 | |
| For work energy equation | M1 | Must have correct number of terms. Allow sign errors. Dimensionally correct. Must use 25, not \(m\). Candidates who try to use constant acceleration can only score B1. |
| \(25g\times1.8-50=\frac{1}{2}\times25v^2\) | A1 | OE. Must be correct. |
| \(v=4\sqrt{2}\ [\text{ms}^{-1}]\) or \(5.66\ [5.6568\ldots]\) | A1 | Allow \(\sqrt{32}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PE gained/lost \(=\pm25g\times2\times0.28\ [=\pm140]\) or KE gained/lost \(=\pm\frac{1}{2}\times25\left(their\ 4\sqrt{2}\right)^2\ [\text{KE}=\pm400]\) | B1FT | For either. FT from their \(v\) for KE. Must have \(\alpha\) substituted for PE. Allow \(25g\times2\sin16.26°\) or \(25g\times2\sin16.3°\). |
| For work energy equation | *M1 | Must have correct number of terms. Allow sign errors. Dimensionally correct. Allow sin/cos mix. Do not allow with WD instead of \(F\times2\). Must have substituted \(\alpha\) and \(v\). |
| \(F\times2=25g\times2\times0.28+\frac{1}{2}\times25\left(4\sqrt{2}\right)^2\ [\Rightarrow F=270]\) | A1FT | FT their \(v^2\) or \(v\). |
| \(R=25g\times0.96\ [=240]\) | B1 | Allow \(25g\cos16.26°\) or \(25g\cos16.3°\). |
| Use of \(F=\mu R\) to form an equation in \(\mu\) only | DM1 | Must be from 3 term \(F\), dimensionally correct and single term \(R\). Allow sin/cos mix but must be different components of weight. \(F\) and \(R\) must be numerical expressions. |
| \(\mu=\frac{9}{8}\) | A1 | CAO. Allow \(1\frac{1}{8}\), but no other answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| PE lost \(=\pm25g\times(1.8+2\times0.28)\ [=\pm590]\) | B1 | Allow \(25g\times(1.8+2\sin16.26°)\) or \(25g\times(1.8+2\sin16.3°)\). |
| For work energy equation | *M1 | Must have correct number of terms. Allow sign errors. Dimensionally correct. Allow sin/cos mix. Do not allow with WD instead of \(F\times2\). Must have substituted \(\alpha\). |
| \(F\times2=25g\times(1.8+2\times0.28)-50\ [\Rightarrow F=270]\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = (4\sqrt{2})^2 + 2(\pm a) \times 2\) | Use of \(v^2 = u^2 + 2as\) | |
| \(a = \pm 8\) | SC B1FT | FT *their* \(v^2\) or \(v\). Note: 8.01 or 8.0089 from use of 5.66 |
| \(R = 25g \times 0.96\) | SC B1 | Allow \(25g\cos16.26°\) or \(25g\cos16.3°\) |
| Use of \(F = \mu R\) and attempt at N2L | SC M1 | To form an equation in \(\mu\) only. Using *their a*. Allow sign errors. Allow sin/cos mix but must be different components of weight. \(F\) and \(R\) must be numerical expressions. Must have substituted \(a\). |
| If correct should get \(25g\sin16.3° - \mu \times 25g\cos16.3° = 25 \times (-8)\) \([\Rightarrow 70 - 240\mu = -200]\) | ||
| \(\mu = \dfrac{9}{8}\) | SC A1 | CAO. Allow \(1\dfrac{1}{8}\), but no other answer. |
| 6 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| PE lost $=mgh=25g\times1.8\ [=450]$ | **B1** | |
| For work energy equation | **M1** | Must have correct number of terms. Allow sign errors. Dimensionally correct. Must use 25, not $m$. Candidates who try to use constant acceleration can only score B1. |
| $25g\times1.8-50=\frac{1}{2}\times25v^2$ | **A1** | OE. Must be correct. |
| $v=4\sqrt{2}\ [\text{ms}^{-1}]$ or $5.66\ [5.6568\ldots]$ | **A1** | Allow $\sqrt{32}$. |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| PE gained/lost $=\pm25g\times2\times0.28\ [=\pm140]$ or KE gained/lost $=\pm\frac{1}{2}\times25\left(their\ 4\sqrt{2}\right)^2\ [\text{KE}=\pm400]$ | **B1FT** | For either. FT from their $v$ for KE. Must have $\alpha$ substituted for PE. Allow $25g\times2\sin16.26°$ or $25g\times2\sin16.3°$. |
| For work energy equation | ***M1** | Must have correct number of terms. Allow sign errors. Dimensionally correct. Allow sin/cos mix. Do not allow with WD instead of $F\times2$. Must have substituted $\alpha$ and $v$. |
| $F\times2=25g\times2\times0.28+\frac{1}{2}\times25\left(4\sqrt{2}\right)^2\ [\Rightarrow F=270]$ | **A1FT** | FT their $v^2$ or $v$. |
| $R=25g\times0.96\ [=240]$ | **B1** | Allow $25g\cos16.26°$ or $25g\cos16.3°$. |
| Use of $F=\mu R$ to form an equation in $\mu$ only | **DM1** | Must be from 3 term $F$, dimensionally correct and single term $R$. Allow sin/cos mix but must be different components of weight. $F$ and $R$ must be numerical expressions. |
| $\mu=\frac{9}{8}$ | **A1** | CAO. Allow $1\frac{1}{8}$, but no other answer. |
**Alternative method 1 (using energy from initial position):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| PE lost $=\pm25g\times(1.8+2\times0.28)\ [=\pm590]$ | **B1** | Allow $25g\times(1.8+2\sin16.26°)$ or $25g\times(1.8+2\sin16.3°)$. |
| For work energy equation | ***M1** | Must have correct number of terms. Allow sign errors. Dimensionally correct. Allow sin/cos mix. Do not allow with WD instead of $F\times2$. Must have substituted $\alpha$. |
| $F\times2=25g\times(1.8+2\times0.28)-50\ [\Rightarrow F=270]$ | **A1** | |
## Question 7(b):
**Special Case: Use of constant acceleration. Award max 4/6**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = (4\sqrt{2})^2 + 2(\pm a) \times 2$ | | Use of $v^2 = u^2 + 2as$ |
| $a = \pm 8$ | **SC B1FT** | FT *their* $v^2$ or $v$. Note: 8.01 or 8.0089 from use of 5.66 |
| $R = 25g \times 0.96$ | **SC B1** | Allow $25g\cos16.26°$ or $25g\cos16.3°$ |
| Use of $F = \mu R$ and attempt at N2L | **SC M1** | To form an equation in $\mu$ only. Using *their a*. Allow sign errors. Allow sin/cos mix but must be different components of weight. $F$ and $R$ must be numerical expressions. Must have substituted $a$. |
| If correct should get $25g\sin16.3° - \mu \times 25g\cos16.3° = 25 \times (-8)$ $[\Rightarrow 70 - 240\mu = -200]$ | | |
| $\mu = \dfrac{9}{8}$ | **SC A1** | CAO. Allow $1\dfrac{1}{8}$, but no other answer. |
| | **6** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{2a680bda-4ba2-44eb-8592-95b4e1aed263-10_525_885_264_625}
The diagram shows the vertical cross-section $X Y Z$ of a rough slide. The section $Y Z$ is a straight line of length 2 m inclined at an angle of $\alpha$ to the horizontal, where $\sin \alpha = 0.28$. The section $Y Z$ is tangential to the curved section $X Y$ at $Y$, and $X$ is 1.8 m above the level of $Y$. A child of mass 25 kg slides down the slide, starting from rest at $X$. The work done by the child against the resistance force in moving from $X$ to $Y$ is 50 J .
\begin{enumerate}[label=(\alph*)]
\item Find the speed of the child at $Y$.\\
It is given that the child comes to rest at $Z$.
\item Use an energy method to find the coefficient of friction between the child and $Y Z$, giving your answer as a fraction in its simplest form.\\
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 [10]}}