| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| 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 straightforward M2 work-energy question requiring application of energy conservation with friction on an inclined plane. Part (a) uses basic energy calculations (KE + PE changes), and part (b) applies work done by friction equals energy lost. Standard textbook exercise with clear setup and routine methods, slightly easier than average A-level due to direct application of formulas without problem-solving insight required. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Delta \text{KE} = \frac{1}{2} \times 3.5(12^2 - 8^2) (= 140)\) or KE at A, B correct separately | B1 M1 A1 | |
| \(\Delta \text{PE} = 3.5 \times 9.8 \times 14\sin 20° (\approx 164.238)\) or PE at A, B correct separately | DM1 A1 | [5] |
| \(\Delta E = \Delta \text{KE} + \Delta \text{PE} = 304, 300\) |
| Answer | Marks | Guidance |
|---|---|---|
| Using Work-Energy: \(F_r = \mu \times 3.5g \cos 20°\) | M1 A1 | |
| \(304.238 ... = F_r \times 14\) | ft their (a), \(F_r\) | |
| \(304.238 ... = \mu 3.5g \cos 20° \times 14\) | M1 A1 ft | |
| \(\mu \approx 0.674, 0.67\) | A1 cso | [5] [10] |
| Answer | Marks | Guidance |
|---|---|---|
| \(F_r = \mu \times 3.5g \cos 20°\) | M1 A1 | |
| \(v^2 = u^2 + 2as \Rightarrow 8^2 = 12^2 - 2a \times 14\) | ||
| \(\left(a = \frac{20}{7}\right) (2.857 ....)\) | ||
| N2L \(R \uparrow \text{: } \{their F_r\} - mg \sin 20° = ma\) | M1 A1 ft | |
| ft their \(F_r\) | ||
| Leading to \(\mu \approx 0.674\) or \(0.67\) | A1 | [5] |
**Part (a):**
$\Delta \text{KE} = \frac{1}{2} \times 3.5(12^2 - 8^2) (= 140)$ or KE at A, B correct separately | B1 M1 A1 |
$\Delta \text{PE} = 3.5 \times 9.8 \times 14\sin 20° (\approx 164.238)$ or PE at A, B correct separately | DM1 A1 | [5]
$\Delta E = \Delta \text{KE} + \Delta \text{PE} = 304, 300$ | |
**Part (b):**
Using Work-Energy: $F_r = \mu \times 3.5g \cos 20°$ | M1 A1 |
$304.238 ... = F_r \times 14$ | | ft their (a), $F_r$
$304.238 ... = \mu 3.5g \cos 20° \times 14$ | M1 A1 ft |
$\mu \approx 0.674, 0.67$ | A1 cso | [5] [10]
Alternative using N2L:
$F_r = \mu \times 3.5g \cos 20°$ | M1 A1 |
$v^2 = u^2 + 2as \Rightarrow 8^2 = 12^2 - 2a \times 14$ | |
$\left(a = \frac{20}{7}\right) (2.857 ....)$ | |
N2L $R \uparrow \text{: } \{their F_r\} - mg \sin 20° = ma$ | M1 A1 ft |
| | ft their $F_r$
Leading to $\mu \approx 0.674$ or $0.67$ | A1 | [5]3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2738ce4-4dc5-4cd1-ac3d-0c3fcf21ea71-04_511_922_260_511}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A package of mass 3.5 kg is sliding down a ramp. The package is modelled as a particle and the ramp as a rough plane inclined at an angle of $20 ^ { \circ }$ to the horizontal. The package slides down a line of greatest slope of the plane from a point $A$ to a point $B$, where $A B = 14 \mathrm {~m}$. At $A$ the package has speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at $B$ the package has speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, as shown in Figure 1. Find
\begin{enumerate}[label=(\alph*)]
\item the total energy lost by the package in travelling from $A$ to $B$,
\item the coefficient of friction between the package and the ramp.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2008 Q3 [10]}}