| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Energy methods on slope |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question using SUVAT equations on a smooth slope (part a) and work-energy principle with friction (part b). Both parts follow routine procedures taught in the syllabus with no novel problem-solving required, making it slightly easier than average for M2 level. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.5g \times 2\sin 30 = \frac{1}{2} \times 0.5u^2 - \frac{1}{2} \times 0.5 \times 5^2\) | M1 A1 | |
| \(\frac{1}{4}u^2 = 0.5g + \frac{1}{2} \times 0.5 \times 5^2\) | ||
| \(u = 6.7\) m s\(^{-1}\) (accept 6.68) | DM1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 0.5g \cos 30\) | B1 | |
| \(F = 0.5g \cos 30 \times \mu\) | M1 | |
| Work done by friction \(= 1.5F\) | ||
| \(\frac{1}{2} \times 0.5 \times 5^2 = 1.5F + 0.5g \times 1.5\sin 30\) | M1 A1 A1 | |
| \(\mu = \frac{\frac{1}{2} \times 0.5 \times 5^2 - 0.5g \times 1.5\sin 30}{0.5g \cos 30 \times 1.5}\) | ||
| \(\mu = 0.40\) (accept 0.4 or 0.405) | A1 |
## Part (a):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $0.5g \times 2\sin 30 = \frac{1}{2} \times 0.5u^2 - \frac{1}{2} \times 0.5 \times 5^2$ | M1 A1 | |
| $\frac{1}{4}u^2 = 0.5g + \frac{1}{2} \times 0.5 \times 5^2$ | | |
| $u = 6.7$ m s$^{-1}$ (accept 6.68) | DM1 A1 | |
## Part (b):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $R = 0.5g \cos 30$ | B1 | |
| $F = 0.5g \cos 30 \times \mu$ | M1 | |
| Work done by friction $= 1.5F$ | | |
| $\frac{1}{2} \times 0.5 \times 5^2 = 1.5F + 0.5g \times 1.5\sin 30$ | M1 A1 A1 | |
| $\mu = \frac{\frac{1}{2} \times 0.5 \times 5^2 - 0.5g \times 1.5\sin 30}{0.5g \cos 30 \times 1.5}$ | | |
| $\mu = 0.40$ (accept 0.4 or 0.405) | A1 | |\includegraphics{figure_2}
A particle $P$ of mass 0.5 kg is projected from a point $A$ up a line of greatest slope $AB$ of a fixed plane. The plane is inclined at 30° to the horizontal and $AB = 2$ m with $B$ above $A$, as shown in Figure 2. The particle $P$ passes through $B$ with speed 5 m s$^{-1}$. The plane is smooth from $A$ to $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of projection. [4]
\end{enumerate}
The particle $P$ comes to instantaneous rest at the point $C$ on the plane, where $C$ is above $B$ and $BC = 1.5$ m. From $B$ to $C$ the plane is rough and the coefficient of friction between $P$ and the plane is $\mu$.
By using the work-energy principle,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $\mu$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q5 [10]}}