| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 work-energy problem with friction on an inclined plane. Part (a) requires straightforward application of the work-energy principle with given values, while part (b) involves recognizing that friction reverses direction on the return journey. The trigonometry is simple (given tan α), and the problem follows a familiar textbook pattern with no novel insights required. Slightly easier than average due to clear structure and standard technique. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| KE at \(X = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 14^2\) | B1 | |
| GPE at \(Y = mgd\sin\alpha\left(= 2 \times g \times d \times \frac{7}{25}\right)\) | B1 | |
| Normal reaction \(R = mg\cos\alpha\) | B1 M1 | |
| Friction \(= \mu \times R = \frac{1}{8} \times 2g \times \frac{24}{25}\) | M1A1 | |
| Work-Energy: \(\frac{1}{2}mv^2 - mgd\sin\alpha = \mu \times R \times d\) or equivalent | ||
| \(196 = \frac{14gd}{25} + \frac{6gd}{25} = \frac{20gd}{25}\) | A1 | |
| \(d = 25\text{ m}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Work done \(= \mu \times R \times 2d = \frac{1}{8} \times 2g \times \frac{24}{25} \times 2d\) | ||
| Return to \(X\): \(\frac{1}{2}mv^2 = \frac{1}{2}m14^2 - \frac{1}{8} \times 2g \times \frac{24}{25} \times 50\) | M1A1 | |
| \(v = 8.9 \text{ ms}^{-1}\) (accept \(8.85 \text{ ms}^{-1}\)) | DM1A1 | |
| OR: \(2a = 2g\sin\alpha - F_{\max} = 2g \times \frac{7}{25} - \frac{6g}{25} = \frac{8g}{25}\) | M1A1 | |
| \(v^2 = (0+)2 \times a \times s = 8g;\ v = 8.9\) (accept \(8.85 \text{ ms}^{-1}\)) | DM1;A1 |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| KE at $X = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 14^2$ | B1 | |
| GPE at $Y = mgd\sin\alpha\left(= 2 \times g \times d \times \frac{7}{25}\right)$ | B1 | |
| Normal reaction $R = mg\cos\alpha$ | B1 M1 | |
| Friction $= \mu \times R = \frac{1}{8} \times 2g \times \frac{24}{25}$ | M1A1 | |
| Work-Energy: $\frac{1}{2}mv^2 - mgd\sin\alpha = \mu \times R \times d$ or equivalent | | |
| $196 = \frac{14gd}{25} + \frac{6gd}{25} = \frac{20gd}{25}$ | A1 | |
| $d = 25\text{ m}$ | | |
**Subtotal: (7)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Work done $= \mu \times R \times 2d = \frac{1}{8} \times 2g \times \frac{24}{25} \times 2d$ | | |
| Return to $X$: $\frac{1}{2}mv^2 = \frac{1}{2}m14^2 - \frac{1}{8} \times 2g \times \frac{24}{25} \times 50$ | M1A1 | |
| $v = 8.9 \text{ ms}^{-1}$ (accept $8.85 \text{ ms}^{-1}$) | DM1A1 | |
| OR: $2a = 2g\sin\alpha - F_{\max} = 2g \times \frac{7}{25} - \frac{6g}{25} = \frac{8g}{25}$ | M1A1 | |
| $v^2 = (0+)2 \times a \times s = 8g;\ v = 8.9$ (accept $8.85 \text{ ms}^{-1}$) | DM1;A1 | |
**Subtotal: (4) Total: [11]**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8e220b8a-46f1-4b9b-88a4-f032c7fbda50-11_501_1018_116_468}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle $P$ of mass 2 kg is projected up a rough plane with initial speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, from a point $X$ on the plane, as shown in Figure 4. The particle moves up the plane along the line of greatest slope through $X$ and comes to instantaneous rest at the point $Y$. The plane is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 7 } { 24 }$. The coefficient of friction between the particle and the plane is $\frac { 1 } { 8 }$.
\begin{enumerate}[label=(\alph*)]
\item Use the work-energy principle to show that $X Y = 25 \mathrm {~m}$.
After reaching $Y$, the particle $P$ slides back down the plane.
\item Find the speed of $P$ as it passes through $X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q7 [11]}}