| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Moderate -0.3 This is a standard M2 mechanics question requiring routine application of work-energy principles and friction on an inclined plane. All parts follow predictable patterns: (a) uses mgh with the vertical height, (b) applies work-energy principle to find friction then uses F=μR, (c) repeats work-energy with different initial conditions. No novel problem-solving or geometric insight required—slightly easier than average due to straightforward setup and clear signposting of methods. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| PE lost \(= mgh = 2\times9.8\times5\sin30 = 49\) J | M1 | Condone sin/cos confusion |
| A1 (2) | Accept \(5g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(49 = \frac{1}{2}\times2\times4^2 + F_r\times5\) | M1 | Must be using work-energy. Must be using 5 for distance. Allow with their 49. Condone sign error(s) |
| A2 | -1 each error. Allow with their 49 | |
| \(F_r = 6.6\) N | A1 | Accept \(\frac{5g-16}{5}\) |
| \(N = 2g\cos30\) | B1 | 16.97 |
| \(\mu = \dfrac{F_r}{N} = \dfrac{6.6}{2g\cos30} = 0.389\) or \(0.39\) | M1 | Use of \(F = \mu N\) with their \(F\) & \(N\) |
| A1 (7) | 2 s.f. or 3 s.f. only |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(49 = \frac{1}{2}\times2\times v^2 - \frac{1}{2}\times2\times3^2 + F_r\times5\) | M1 | Work Energy equation. All terms required, must be dimensionally correct but condone sign error(s) |
| A1 | Correct equation for their \(F\) – any equivalent form | |
| \(49 + 9 - 33 = v^2\) | DM1 | Substitute and solve for \(v\). Dependent on preceding M1 |
| \(v = 5\) m s\(^{-1}\) | A1 (4) | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2g\sin30 - F = 2a\) | M1 | Using N2L with their \(F\). Condone sign error |
| (\(a = 1.6\)) | A1 | Correct equation (with their \(F\)) |
| \(v^2 = 9 + 2\times a\times5\) (= 25) | DM1 | Use of \(v^2 = u^2 + 2as\) with their \(a\). Dependent on preceding M1 |
| \(v = 5\) m s\(^{-1}\) | A1 (4) | — |
# Question 5:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| PE lost $= mgh = 2\times9.8\times5\sin30 = 49$ J | M1 | Condone sin/cos confusion |
| | A1 (2) | Accept $5g$ |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $49 = \frac{1}{2}\times2\times4^2 + F_r\times5$ | M1 | Must be using work-energy. Must be using 5 for distance. Allow with their 49. Condone sign error(s) |
| | A2 | -1 each error. Allow with their 49 |
| $F_r = 6.6$ N | A1 | Accept $\frac{5g-16}{5}$ |
| $N = 2g\cos30$ | B1 | 16.97 |
| $\mu = \dfrac{F_r}{N} = \dfrac{6.6}{2g\cos30} = 0.389$ or $0.39$ | M1 | Use of $F = \mu N$ with their $F$ & $N$ |
| | A1 (7) | 2 s.f. or 3 s.f. only |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $49 = \frac{1}{2}\times2\times v^2 - \frac{1}{2}\times2\times3^2 + F_r\times5$ | M1 | Work Energy equation. All terms required, must be dimensionally correct but condone sign error(s) |
| | A1 | Correct equation for their $F$ – any equivalent form |
| $49 + 9 - 33 = v^2$ | DM1 | Substitute and solve for $v$. Dependent on preceding M1 |
| $v = 5$ m s$^{-1}$ | A1 (4) | — |
### Alt(c):
| Working | Mark | Guidance |
|---------|------|----------|
| $2g\sin30 - F = 2a$ | M1 | Using N2L with their $F$. Condone sign error |
| ($a = 1.6$) | A1 | Correct equation (with their $F$) |
| $v^2 = 9 + 2\times a\times5$ (= 25) | DM1 | Use of $v^2 = u^2 + 2as$ with their $a$. Dependent on preceding M1 |
| $v = 5$ m s$^{-1}$ | A1 (4) | — |
> Watch out – there are a lot of incorrect ways of reaching a final answer of 5.5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{82cadc37-4cb0-455e-9531-e09ec0c19533-09_460_974_242_484}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A particle $P$ of mass 2 kg is released from rest at a point $A$ on a rough inclined plane and slides down a line of greatest slope. The plane is inclined at $30 ^ { \circ }$ to the horizontal. The point $B$ is 5 m from $A$ on the line of greatest slope through $A$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Find the potential energy lost by $P$ as it moves from $A$ to $B$.
The speed of $P$ as it reaches $B$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Use the work-energy principle to find the magnitude of the constant frictional force acting on $P$ as it moves from $A$ to $B$.
\item Find the coefficient of friction between $P$ and the plane.
The particle $P$ is now placed at $A$ and projected down the plane towards $B$ with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Given that the frictional force remains constant,
\end{enumerate}\item find the speed of $P$ as it reaches $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q5 [13]}}