| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 16 |
| 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 to show a stated result. Part (b) involves a similar calculation for the return journey with different friction direction. The trigonometry is simplified (3-4-5 triangle), and the method is routine for M2 students, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| resolve perp. to plane: \(R - mg\cos\alpha = 0 \quad \therefore R = mg(\frac{5}{13})\) | M1 A1 | |
| frictional force \(= \mu R = \frac{12}{35}mg\) | A1 | |
| work done against friction = loss in KE + gain in PE | M1 | |
| \(\frac{12}{35}mgd = \frac{1}{2}m(5.6)^2 - mgd\sin\alpha = 15.68m - \frac{4}{13}mgd\) | M2 A2 | |
| \(\frac{40}{35}gd = \frac{1}{2}(5.6)^2 \quad \therefore d = 1.4\) m | M1 A1 | |
| (b) work done against friction = loss in KE (as PE returns to initial value) | M2 A1 | |
| \(\frac{12}{35}mg \times 2.8 = \frac{1}{2}m(5.6^2 - v^2)\) | M2 A1 | |
| \(1.92g = 5.6^2 - v^2\) | M1 | |
| \(v^2 = 12.544 \quad \therefore v = 3.5\) ms\(^{-1}\) (2sf) | M1 A1 | (16) |
| Answer | Marks |
|---|---|
| Total | (75) |
**(a)** $m =$ mass of $P$; $d = AB$
resolve perp. to plane: $R - mg\cos\alpha = 0 \quad \therefore R = mg(\frac{5}{13})$ | M1 A1 |
frictional force $= \mu R = \frac{12}{35}mg$ | A1 |
work done against friction = loss in KE + gain in PE | M1 |
$\frac{12}{35}mgd = \frac{1}{2}m(5.6)^2 - mgd\sin\alpha = 15.68m - \frac{4}{13}mgd$ | M2 A2 |
$\frac{40}{35}gd = \frac{1}{2}(5.6)^2 \quad \therefore d = 1.4$ m | M1 A1 |
**(b)** work done against friction = loss in KE (as PE returns to initial value) | M2 A1 |
$\frac{12}{35}mg \times 2.8 = \frac{1}{2}m(5.6^2 - v^2)$ | M2 A1 |
$1.92g = 5.6^2 - v^2$ | M1 |
$v^2 = 12.544 \quad \therefore v = 3.5$ ms$^{-1}$ (2sf) | M1 A1 | (16)
---
**Total** | (75)7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ef2dd10c-5a3c-4868-af00-aaf7f2937d7e-5_495_604_214_580}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Figure 4 shows a particle $P$ projected from the point $A$ up the line of greatest slope of a rough plane which is inclined at an angle $\alpha$ to the horizontal where $\sin \alpha = \frac { 4 } { 5 } . P$ is projected with speed $5.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the coefficient of friction between $P$ and the plane is $\frac { 4 } { 7 }$.
Given that $P$ first comes to rest at point $B$,
\begin{enumerate}[label=(\alph*)]
\item use the Work-Energy principle to show that the distance $A B$ is 1.4 m .
The particle then slides back down the plane.
\item Find, correct to 2 significant figures, the speed of $P$ when it returns to $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [16]}}