Standard +0.3 This is a standard M3 impulse-momentum problem requiring vector resolution in two perpendicular directions and solving simultaneous equations. While it involves multiple steps (resolving impulse, applying momentum conservation in two directions, solving for two unknowns), the approach is methodical and follows a well-practiced technique with no novel insight required. Slightly easier than average due to the straightforward setup and clean numerical values.
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\includegraphics[max width=\textwidth, alt={}, center]{08760a55-da6c-41f2-a88a-289ecc227f69-2_323_639_255_753}
A particle \(P\) of mass 0.4 kg is moving horizontally with speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when it receives an impulse of magnitude \(I \mathrm { Ns }\), in a direction which makes an angle \(( 180 - \theta ) ^ { \circ }\) with the direction of motion of \(P\). Immediately after the impulse acts \(P\) moves horizontally with speed \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The direction of motion of \(P\) is turned through an angle of \(60 ^ { \circ }\) by the impulse (see diagram). Find \(I\) and \(\theta\).
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\includegraphics[max width=\textwidth, alt={}, center]{08760a55-da6c-41f2-a88a-289ecc227f69-2_323_639_255_753}
A particle $P$ of mass 0.4 kg is moving horizontally with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it receives an impulse of magnitude $I \mathrm { Ns }$, in a direction which makes an angle $( 180 - \theta ) ^ { \circ }$ with the direction of motion of $P$. Immediately after the impulse acts $P$ moves horizontally with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The direction of motion of $P$ is turned through an angle of $60 ^ { \circ }$ by the impulse (see diagram). Find $I$ and $\theta$.
\hfill \mbox{\textit{OCR M3 2010 Q1 [7]}}