| 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 -0.3 This appears to be an incomplete question fragment showing only notation setup for a relative velocity problem. Based on the M2 mechanics context and typical CAIE structure, this would likely involve standard vector addition/subtraction for relative motion, which is routine for M2 students but requires more application than pure recall. |
| Answer | Marks | Guidance |
|---|---|---|
| \(V(\text{vert}) = 14\sin60° - 1.8g\) | B1 | \(-5.8756..\) |
| \(V^2 = (-)5.8756^2 + (14\cos60°)^2\) | M1 | |
| \(V = 9.14 \text{ ms}^{-1}\) | A1 | \(9.1391..\) |
| \(\tan\theta = \frac{(-)5.8756}{(14\cos60°)}\) | M1 | |
| \(\theta = 40(0)°\) below horizontal | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2 = (14\sin60°)t - gt^2/2\) | M1 | \(-2 = ut - gt^2/2\) used vertically |
| \(5t^2 - 12.124t - 2 = 0\) | M1 | Solves correct 3 term quadratic |
| \(t = 2.58 \text{ s}\) | A1 | [3] |
**(i)**
$V(\text{vert}) = 14\sin60° - 1.8g$ | B1 | $-5.8756..$
$V^2 = (-)5.8756^2 + (14\cos60°)^2$ | M1 |
$V = 9.14 \text{ ms}^{-1}$ | A1 | $9.1391..$
$\tan\theta = \frac{(-)5.8756}{(14\cos60°)}$ | M1 |
$\theta = 40(0)°$ below horizontal | A1 | [5]
**(ii)**
$-2 = (14\sin60°)t - gt^2/2$ | M1 | $-2 = ut - gt^2/2$ used vertically
$5t^2 - 12.124t - 2 = 0$ | M1 | Solves correct 3 term quadratic
$t = 2.58 \text{ s}$ | A1 | [3] | [8]$A$ has velocity $\vec{v}$
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item $C$ with velocity $\vec{v}$
\item $C$
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2013 Q4}}