| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - horizontal motion with resistance (no driving force) |
| Difficulty | Moderate -0.8 This is a straightforward application of the work-energy principle with standard values. Part (a) requires only the kinetic energy formula KE = ½mv², and part (b) uses work done = force × distance = change in KE. Both are direct substitutions with no problem-solving insight needed, making this easier than average but not trivial due to the two-part structure. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{KE lost is } \frac{1}{2} \times 2.5 \times 8^2 = 80 \text{ (J)}\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Work energy: \(80 = R \times 20\), \(R = 4\) | M1 A1 ft A1 | (3 marks) [5 marks total] |
| Answer | Marks | Guidance |
|---|---|---|
| N2L: \(R = 2.5 \times 1.6 = 4\) | M1 A1 ft A1 | (3 marks) |
**Part (a)**
$\text{KE lost is } \frac{1}{2} \times 2.5 \times 8^2 = 80 \text{ (J)}$ | M1 A1 | (2 marks)
**Part (b)**
Work energy: $80 = R \times 20$, $R = 4$ | M1 A1 ft A1 | (3 marks) [5 marks total]
Fits their (a)
**Alternative to (b)**
$0^2 = 8^2 - 2 \times a \times 20 \Rightarrow a = (-)1.6$
N2L: $R = 2.5 \times 1.6 = 4$ | M1 A1 ft A1 | (3 marks)
Fits their $a$
---A parcel of mass 2.5 kg is moving in a straight line on a smooth horizontal floor. Initially the parcel is moving with speed 8 m s$^{-1}$. The parcel is brought to rest in a distance of 20 m by a constant horizontal force of magnitude $R$ newtons. Modelling the parcel as a particle, find
\begin{enumerate}[label=(\alph*)]
\item the kinetic energy lost by the parcel in coming to rest, [2]
\item the value of $R$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2008 Q1 [5]}}