| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Velocity after impulse (direct calculation) |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation to find force from velocity (F=ma), then applying impulse-momentum theorem with vector addition. All steps are standard procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation3.03a Force: vector nature and diagrams3.03c Newton's second law: F=ma one dimension6.03e Impulse: by a force6.03f Impulse-momentum: relation |
2. A particle $P$ of mass 0.75 kg is moving under the action of a single force $\mathbf { F }$ newtons. At time $t$ seconds, the velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ of $P$ is given by
$$\mathbf { v } = \left( t ^ { 2 } + 2 \right) \mathbf { i } - 6 t \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of $\mathbf { F }$ when $t = 4$.\\
(5)
When $t = 5$, the particle $P$ receives an impulse of magnitude $9 \sqrt { } 2 \mathrm { Ns }$ in the direction of the vector $\mathbf { i } - \mathbf { j }$.
\item Find the velocity of $P$ immediately after the impulse.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2004 Q2 [9]}}