| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| 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 involving work-energy on an inclined plane with friction. Part (a) requires calculating the normal reaction force and friction force using given trigonometric ratios, then computing work done—straightforward application of formulas. Part (b) applies the work-energy principle with friction acting in the opposite direction. While it requires careful bookkeeping of energy terms and forces in both directions, it follows a well-established method with no novel insight required. Slightly easier than average due to the structured 'show that' in part (a) providing confirmation. |
| 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 | Mark | Guidance |
| \(F = \mu R = \frac{1}{5}mg\cos\alpha\) | B1 | Seen or implied |
| Work done = force × distance | M1 | Correct method for work done against friction |
| \(= \frac{1}{5}mg \times \frac{12}{13} \times d = \frac{12}{65}mgd\) | A1* | Obtain given answer from correct working |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work-energy equation | M1 | All terms required and dimensionally correct. Condone sign errors and sin/cos confusion |
| \(\frac{1}{2}mv^2 = mg \times d \times \frac{5}{13} - \frac{12}{65}mgd \left(= \frac{13}{65}mgd\right)\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| \(v = \sqrt{\frac{2gd}{5}}\) | A1 | Or exact equivalent e.g. \(\sqrt{\frac{26}{65}gd}\), \(\frac{1}{5}\sqrt{10gd}\). Accept \(0.63\sqrt{gd}\) or better |
| Total: (4) [7] |
# Question 1:
## Part 1a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \mu R = \frac{1}{5}mg\cos\alpha$ | B1 | Seen or implied |
| Work done = force × distance | M1 | Correct method for work done against friction |
| $= \frac{1}{5}mg \times \frac{12}{13} \times d = \frac{12}{65}mgd$ | A1* | Obtain given answer from correct working |
| **Total: (3)** | | |
## Part 1b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation | M1 | All terms required and dimensionally correct. Condone sign errors and sin/cos confusion |
| $\frac{1}{2}mv^2 = mg \times d \times \frac{5}{13} - \frac{12}{65}mgd \left(= \frac{13}{65}mgd\right)$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $v = \sqrt{\frac{2gd}{5}}$ | A1 | Or exact equivalent e.g. $\sqrt{\frac{26}{65}gd}$, $\frac{1}{5}\sqrt{10gd}$. Accept $0.63\sqrt{gd}$ or better |
| **Total: (4) [7]** | | |
---1.
\section*{Figure 1}
Figure 1
A particle of mass $m$ is held at rest at a point $A$ on a rough plane.\\
The plane is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 5 } { 12 }$\\
The coefficient of friction between the particle and the plane is $\frac { 1 } { 5 }$\\
The points $A$ and $B$ lie on a line of greatest slope of the plane, with $B$ above $A$, and $A B = d$, as shown in Figure 1.
The particle is pushed up the line of greatest slope from $A$ to $B$.
\begin{enumerate}[label=(\alph*)]
\item Show that the work done against friction as the particle moves from $A$ to $B$ is $\frac { 12 } { 65 } m g d$
The particle is then held at rest at $B$ and released.
\item Use the work-energy principle to find, in terms of $g$ and $d$, the speed of the particle at the instant it reaches $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2021 Q1 [7]}}