| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find force using F=ma |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of velocity to find acceleration, applying F=ma, and integration of velocity to find displacement. All steps are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration3.03d Newton's second law: 2D vectors |
\begin{enumerate}
\item A particle $P$ of mass 0.5 kg moves under the action of a single force $\mathbf { F }$ newtons. At time $t$ seconds, $t \geqslant 0 , P$ has velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, where
\end{enumerate}
$$\mathbf { v } = \left( 4 t - 3 t ^ { 2 } \right) \mathbf { i } + \left( t ^ { 2 } - 8 t - 40 \right) \mathbf { j }$$
(a) Find\\
(i) the magnitude of $\mathbf { F }$ when $t = 3$\\
(ii) the acceleration of $P$ at the instant when it is moving in the direction of the vector $- \mathbf { i } - \mathbf { j }$.
When $t = 1 , P$ is at the point $A$. When $t = 2 , P$ is at the point $B$.\\
(b) Find, in terms of $\mathbf { i }$ and $\mathbf { j }$, the vector $\overrightarrow { A B }$.\\
\hfill \mbox{\textit{Edexcel M2 2018 Q6 [14]}}