| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Standard +0.3 This is a straightforward kinematics question requiring differentiation of position to find velocity and acceleration, then applying F=ma. Part (a) involves finding velocity direction from components (standard arctangent), part (b) is direct application of Newton's second law, and part (c) requires solving a simple equation tan(45°)=1. All techniques are routine for A-level mechanics with no novel problem-solving required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration3.03d Newton's second law: 2D vectors |
11 In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in the directions east and north respectively.
A particle of mass 0.12 kg is moving so that its position vector $\mathbf { r }$ metres at time $t$ seconds is given by $\mathbf { r } = 2 t ^ { 3 } \mathbf { i } + \left( 5 t ^ { 2 } - 4 t \right) \mathbf { j }$.
\begin{enumerate}[label=(\alph*)]
\item Show that when $t = 0.7$ the bearing on which the particle is moving is approximately $044 ^ { \circ }$.
\item Find the magnitude of the resultant force acting on the particle at the instant when $t = 0.7$.
\item Determine the times at which the particle is moving on a bearing of $045 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 Q11 [9]}}