| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Angle change from impulse |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of impulse-momentum theorem and kinetic energy formula. Part (a) involves vector addition and finding an angle using dot product or components, while part (b) is direct calculation of KE before and after. The multi-step nature and vector work make it slightly below average difficulty rather than routine, but it requires no novel insight—just methodical application of standard techniques. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
\begin{enumerate}
\item A particle $P$ of mass 0.75 kg is moving with velocity $4 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it receives an impulse $( 6 \mathbf { i } + 6 \mathbf { j } ) \mathrm { Ns }$. The angle between the velocity of $P$ before the impulse and the velocity of $P$ after the impulse is $\theta ^ { \circ }$.
\end{enumerate}
Find\\
(a) the value of $\theta$,\\
(b) the kinetic energy gained by $P$ as a result of the impulse.
\hfill \mbox{\textit{Edexcel M2 2015 Q3 [8]}}