| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Loss of energy in collision |
| Difficulty | Standard +0.3 This is a straightforward M2 question requiring differentiation of position vectors to find velocity, then calculating kinetic energies at two time points. The calculus is routine (polynomial differentiation), and the energy calculation is direct application of KE = ½mv². Slightly above average difficulty only due to vector notation and multi-step nature, but no conceptual challenges or novel problem-solving required. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v = \frac{dr}{dt} = (2t - 3)i + \frac{1}{3}t^2j\) | M1 A1 | |
| When \(t = 0\), \(v = -3i \text{ ms}^{-1}\) | A1 | |
| (b) At \(t = 2\), \(v = i + 2j\) \(\therefore | v | = \sqrt{1^2 + 2^2} = \sqrt{5}\) |
| KE lost \(= \frac{1}{2}(3)(3^2 - 5) = 6J\) | M1 A1 | (8 marks) |
**(a)** $v = \frac{dr}{dt} = (2t - 3)i + \frac{1}{3}t^2j$ | M1 A1 |
When $t = 0$, $v = -3i \text{ ms}^{-1}$ | A1 |
**(b)** At $t = 2$, $v = i + 2j$ $\therefore |v| = \sqrt{1^2 + 2^2} = \sqrt{5}$ | M2 A1 |
KE lost $= \frac{1}{2}(3)(3^2 - 5) = 6J$ | M1 A1 | (8 marks)2. A particle $P$ of mass 3 kg moves such that at time $t$ seconds its position vector, $\mathbf { r }$ metres, relative to a fixed origin, $O$, is given by
$$\mathbf { r } = \left( t ^ { 2 } - 3 t \right) \mathbf { i } + \frac { 1 } { 6 } t ^ { 3 } \mathbf { j }$$
where $\mathbf { i }$ and $\mathbf { j }$ are perpendicular horizontal unit vectors.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $P$ when $t = 0$.
\item Find the kinetic energy lost by $P$ in the interval $0 \leq t \leq 2$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q2 [8]}}