| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 15 |
| 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 mechanics problem requiring work-energy principle application with friction on an inclined plane. While it involves multiple parts and careful handling of friction forces in both directions, the techniques are routine for M2 students: resolving forces, applying work-energy equation, and checking limiting friction. The trigonometry is straightforward (3-4-5 triangle), and the problem follows a predictable structure with no novel insights required. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R(\searrow),\ R = mg\cos\alpha = \frac{4}{5}mg\) | M1 A1 | |
| \(\frac{2}{5}mgd = \frac{1}{2}m \times 10^2 - mgd\sin\alpha\) | M1 A3 | Work-energy equation |
| \(OA = d = \dfrac{50}{g} = 5.10\ \text{m (3 s.f.)}\) | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| component of weight down plane \(= mg\sin\alpha = \dfrac{3mg}{5}\) | B1 | |
| limiting friction up \(= \dfrac{2mg}{5}\) | B1 | |
| \(\therefore\) slides down as \(\dfrac{3mg}{5} > \dfrac{2mg}{5}\) | M1 | (3) |
| Work done against friction = KE loss | ||
| \(2 \times \dfrac{2mg}{5} \times \dfrac{50}{g} = \frac{1}{2}m(10^2 - v^2)\) | M1 A3 | |
| \(v = \sqrt{20} = 4.47\ \text{m s}^{-1}\) | A1 | (5) (15) |
# Question 8:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R(\searrow),\ R = mg\cos\alpha = \frac{4}{5}mg$ | M1 A1 | |
| $\frac{2}{5}mgd = \frac{1}{2}m \times 10^2 - mgd\sin\alpha$ | M1 A3 | Work-energy equation |
| $OA = d = \dfrac{50}{g} = 5.10\ \text{m (3 s.f.)}$ | A1 | (7) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| component of weight down plane $= mg\sin\alpha = \dfrac{3mg}{5}$ | B1 | |
| limiting friction up $= \dfrac{2mg}{5}$ | B1 | |
| $\therefore$ slides down as $\dfrac{3mg}{5} > \dfrac{2mg}{5}$ | M1 | (3) |
| Work done against friction = KE loss | | |
| $2 \times \dfrac{2mg}{5} \times \dfrac{50}{g} = \frac{1}{2}m(10^2 - v^2)$ | M1 A3 | |
| $v = \sqrt{20} = 4.47\ \text{m s}^{-1}$ | A1 | (5) **(15)** |8. A particle $P$ is projected up a line of greatest slope of a rough plane which is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$. The coefficient of friction between $P$ and the plane is $\frac { 1 } { 2 }$. The particle is projected from the point $O$ with a speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and comes to instantaneous rest at the point $A$.
By Using the Work-Energy principle, or otherwise,
\begin{enumerate}[label=(\alph*)]
\item find, to 3 significant figures, the length $O A$.
\item Show that $P$ will slide back down the plane.
\item Find, to 3 significant figures, the speed of $P$ when it returns to $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q8 [15]}}