| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Relative velocity: find resultant velocity (magnitude and/or direction) |
| Difficulty | Moderate -0.5 This appears to be a standard 2D vectors question from CAIE M2 involving relative velocity. The question structure suggests finding magnitudes and possibly directions of velocities, which is routine mechanics content requiring straightforward vector arithmetic and Pythagoras/trigonometry. Slightly easier than average as it's introductory mechanics rather than requiring complex problem-solving. |
| Answer | Marks | Guidance |
|---|---|---|
| \(Tx(4/5) - Tx(3/5) = 0.2g\) | M1 | Resolves vertically, 3 forces |
| \(T = 10\) | A1 | Maybe implied |
| \(Tx(4/5) + Tx(3/5) = 0.2v^2/(0.4 \times 3/5)\) | M1 | Resolves horizontally. N2L |
| \(v = 4.1(0) \text{ ms}^{-1}\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(Tx(4/5) = 0.2g\) | B1 | \(T = 2.5\) |
| \(Tx(3/5) = 0.2 \times^2 \sqrt{(0.4 \times 3/5)}\) | M1 | N2L horizontally, single force |
| \(\omega = 5.59 \text{ rads}^{-1}\) | A1 | [3] |
**(i)**
$Tx(4/5) - Tx(3/5) = 0.2g$ | M1 | Resolves vertically, 3 forces
$T = 10$ | A1 | Maybe implied
$Tx(4/5) + Tx(3/5) = 0.2v^2/(0.4 \times 3/5)$ | M1 | Resolves horizontally. N2L
$v = 4.1(0) \text{ ms}^{-1}$ | A1 | [5]
**(ii)**
$Tx(4/5) = 0.2g$ | B1 | $T = 2.5$
$Tx(3/5) = 0.2 \times^2 \sqrt{(0.4 \times 3/5)}$ | M1 | N2L horizontally, single force
$\omega = 5.59 \text{ rads}^{-1}$ | A1 | [3] | [8]\includegraphics{figure_5}
$A$ has velocity $\vec{x}$, there are velocities $\vec{x}$, $\vec{x}$, $\vec{v}$ and $\vec{v}$
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item $v$
\item $v$ and $v$
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2013 Q5}}