| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on slope then horizontal |
| Difficulty | Standard +0.3 This is a standard work-energy problem on an inclined plane with friction. Part (i) uses conservation of energy on a smooth surface (routine). Parts (ii-iii) require resolving forces and applying equations of motion on a rough incline—all standard M1 techniques with straightforward calculations. Slightly above average due to the multi-part nature and combining smooth/rough surfaces, but no novel insight required. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| PE loss \(= 2g(10 - 10 \times 0.28)\) | B1 | |
| \([\frac{1}{2} \times 2v^2 = 144]\) | M1 | For using \(\frac{1}{2}mv^2 =\) PE loss |
| Speed is \(12\ \text{ms}^{-1}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R = 2g \times 0.96\) | B1 | |
| \([2g \times 0.28 - 2g \times 0.96 \div 12 = 2a]\) | M1 | For using Newton's 2nd law |
| Acceleration is \(2\ \text{ms}^{-2}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([v^2 = 12^2 + 2 \times 2 \times 10]\) | M1 | For using \(v^2 = u^2 + 2as\) |
| Speed is \(13.6\ \text{ms}^{-1}\) | A1 [2] |
## Question 5(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| PE loss $= 2g(10 - 10 \times 0.28)$ | B1 | |
| $[\frac{1}{2} \times 2v^2 = 144]$ | M1 | For using $\frac{1}{2}mv^2 =$ PE loss |
| Speed is $12\ \text{ms}^{-1}$ | A1 [3] | |
## Question 5(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R = 2g \times 0.96$ | B1 | |
| $[2g \times 0.28 - 2g \times 0.96 \div 12 = 2a]$ | M1 | For using Newton's 2nd law |
| Acceleration is $2\ \text{ms}^{-2}$ | A1 [3] | |
## Question 5(iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[v^2 = 12^2 + 2 \times 2 \times 10]$ | M1 | For using $v^2 = u^2 + 2as$ |
| Speed is $13.6\ \text{ms}^{-1}$ | A1 [2] | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{01e73486-5a95-4e65-bf18-518d1adc7cfb-3_485_874_255_638}
The diagram shows the vertical cross-section $O A B$ of a slide. The straight line $A B$ is tangential to the curve $O A$ at $A$. The line $A B$ is inclined at $\alpha$ to the horizontal, where $\sin \alpha = 0.28$. The point $O$ is 10 m higher than $B$, and $A B$ has length 10 m (see diagram). The part of the slide containing the curve $O A$ is smooth and the part containing $A B$ is rough. A particle $P$ of mass 2 kg is released from rest at $O$ and moves down the slide.\\
(i) Find the speed of $P$ when it passes through $A$.
The coefficient of friction between $P$ and the part of the slide containing $A B$ is $\frac { 1 } { 12 }$. Find\\
(ii) the acceleration of $P$ when it is moving from $A$ to $B$,\\
(iii) the speed of $P$ when it reaches $B$.
\hfill \mbox{\textit{CAIE M1 2012 Q5 [8]}}