| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 6 |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Standard +0.3 This is a straightforward impulse-momentum vector problem requiring resolution in two perpendicular directions and basic trigonometry. The method is standard for Further Mechanics: apply conservation principles, resolve components, then use Pythagoras and inverse tan. While it's a Further Maths topic (making it slightly above average), the execution is mechanical with no conceptual subtlety. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Use of vector triangle for impulse and momentum | M1 | May be implied |
| \(I^2 = 7.2^2 + 9.6^2 - 2 \times 7.2 \times 9.6\cos 50°\) | M1 | Use of cosine rule |
| \(I = 7.43 \text{ (N s)}\) | A1 |
| Answer | Marks |
|---|---|
| Change in momentum \(\Rightarrow = 7.2\cos 50° - 9.6\) | M1 |
| \(I^2 = (7.2\cos 50° - 9.6)^2 + (7.2\sin 50°)^2\) | M1 |
| \(I = 7.43 \text{ (N s)}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\sin 50°}{7.43} = \frac{\sin \theta}{9.6}\) or \(\frac{\sin 50°}{7.43} = \frac{\sin \phi}{7.2}\) oe | M1 | e.g. use of cosine rule to find \(\theta\) or \(\phi\) |
| \(\sin \theta = 0.9903\) or \(\sin \phi = 0.7428\) oe | A1 | |
| Angle with original direction is \(132.0°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan \phi = \frac{5.515...}{4.971...}\) oe | M1 | Allow positive or negative value |
| \(\phi = 48\) oe | A1 | |
| Angle with original direction is \(132.0°\) | A1 |
## (i)
Use of vector triangle for impulse and momentum | M1 | May be implied
$I^2 = 7.2^2 + 9.6^2 - 2 \times 7.2 \times 9.6\cos 50°$ | M1 | Use of cosine rule
$I = 7.43 \text{ (N s)}$ | A1 |
**Alternative solution**
Change in momentum $\Rightarrow = 7.2\cos 50° - 9.6$ | M1 |
$I^2 = (7.2\cos 50° - 9.6)^2 + (7.2\sin 50°)^2$ | M1 |
$I = 7.43 \text{ (N s)}$ | A1 |
## (ii)
$\frac{\sin 50°}{7.43} = \frac{\sin \theta}{9.6}$ or $\frac{\sin 50°}{7.43} = \frac{\sin \phi}{7.2}$ oe | M1 | e.g. use of cosine rule to find $\theta$ or $\phi$
$\sin \theta = 0.9903$ or $\sin \phi = 0.7428$ oe | A1 |
Angle with original direction is $132.0°$ | A1 |
**Alternative solution**
$\tan \phi = \frac{5.515...}{4.971...}$ oe | M1 | Allow positive or negative value
$\phi = 48$ oe | A1 |
Angle with original direction is $132.0°$ | A1 |
---A particle of mass 0.8 kg is moving in a straight line on a smooth horizontal surface with constant speed $12 \text{ms}^{-1}$ when it is struck by a horizontal impulse. Immediately after the impulse acts, the particle is moving with speed $9 \text{ms}^{-1}$ at an angle of 50° to its original direction of motion (see diagram).
\includegraphics{figure_2}
Find
\begin{enumerate}[label=(\roman*)]
\item the magnitude of the impulse, [3]
\item the angle that the impulse makes with the original direction of motion of the particle. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q2 [6]}}