| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 5 |
| Topic | Momentum and Collisions 1 |
| Type | Impulse from force-time graph |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics question requiring integration of a given force function to find impulse, then applying impulse-momentum theorem. While it involves exponential and polynomial integration (standard A-level techniques), it's a direct two-step application with no conceptual subtlety. Being Further Maths adds some difficulty, but the question structure is routine. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.06a Variable force: dv/dt or v*dv/dx methods |
3 A body, $Q$, of mass 2 kg moves in a straight line under the action of a single force which acts in the direction of motion of $Q$. Initially the speed of $Q$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t \mathrm {~s}$, the magnitude $F N$ of the force is given by
$$F = t ^ { 2 } + 3 \mathrm { e } ^ { t } , \quad 0 \leq t \leq 4 .$$
(i) Calculate the impulse of the force over the time interval.\\
(ii) Hence find the speed of $Q$ when $t = 4$.
\hfill \mbox{\textit{OCR Further Mechanics 2017 Q3 [5]}}