| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Angle change from impulse |
| Difficulty | Standard +0.3 This is a standard M3 impulse-momentum triangle problem requiring vector addition and basic trigonometry (cosine rule). The question guides students to use the triangle method, making it slightly easier than average. The calculation is straightforward once the triangle is set up correctly, with clean numerical values that work out neatly. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
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\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-2_385_741_269_701}
A particle $P$ of mass 0.5 kg is moving in a straight line with speed $6.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. An impulse of magnitude 2.6 N s applied to $P$ deflects its direction of motion through an angle $\theta$, and reduces its speed to $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). By considering an impulse-momentum triangle, or otherwise,\\
(i) show that $\cos \theta = 0.6$,\\
(ii) find the angle that the impulse makes with the original direction of motion of $P$.
\hfill \mbox{\textit{OCR M3 2009 Q1 [8]}}