| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done by constant force - vector setup |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard formulas: KE = ½mv² (with vector magnitude) and work-energy theorem. Part (a) is routine calculation, part (b) requires finding KE at two instants and taking the difference. No novel problem-solving or geometric insight needed, just careful arithmetic with vectors—slightly above average due to vector handling and two-step process. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v = 25\) \(\quad\) K.E. \(= \frac{1}{2} \times 1.8 \times 25^2 = 562.5\) J | B1 M1 A1 | |
| (b) New K.E. \(= \frac{1}{2} \times 1.8 \times 12.5^2 = 140.6\) J | M1 A1 A1 | |
| Work done \(=\) change in K.E. \(= 422\) J | M1 A1 | Total: 8 marks |
**(a)** $v = 25$ $\quad$ K.E. $= \frac{1}{2} \times 1.8 \times 25^2 = 562.5$ J | B1 M1 A1 |
**(b)** New K.E. $= \frac{1}{2} \times 1.8 \times 12.5^2 = 140.6$ J | M1 A1 A1 |
Work done $=$ change in K.E. $= 422$ J | M1 A1 | Total: 8 marks3. $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane. At a certain instant, a particle $P$ of mass 1.8 kg is moving with velocity $( 24 \mathrm { i } - 7 \mathrm { j } ) \mathrm { ms } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the kinetic energy of $P$ at this instant.\\
$P$ is now subjected to a constant retardation. After 10 seconds, the velocity of $P$ is $( - 12 \mathbf { i } + 3 \cdot 5 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$.
\item Calculate the work done by the retarding force over the 10 seconds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [8]}}