| 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 -1.0 The question description is incomplete/corrupted, but based on the context (CAIE M2, Vectors Introduction & 2D, November 2013 Q3), this is likely a basic relative velocity problem involving vector addition/subtraction with simple notation. M2 vector questions at this level typically involve straightforward application of velocity triangle concepts with minimal algebraic manipulation. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.8v\frac{dv}{dx} = 4e^{-x} - 2.4x^2\) | M1 | N2L, terms different signs |
| \(v\frac{dv}{dx} = 5e^{-x} - 3x^2\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int v \, dv = \int (5e^{-x} - 3x^2)dx\) | M1 | Attempts integration |
| \(\frac{v^2}{2} = -5e^{-x} - 3x^3/3 (+c)\) | A1 | Accept \(c\) omitted |
| \(x = 0, v = 6\), hence \(c = 23\) | B1 | Or uses limits 0 and 2 |
| \(\frac{v^2}{2} = -5e^{-2} - 3x^2/3 + 23\) | M1 | Puts \(x = 2\) in \(v(x)\) expression |
| \(v = 5.35 \text{ ms}^{-1}\) | A1 | [5] |
**(i)**
$0.8v\frac{dv}{dx} = 4e^{-x} - 2.4x^2$ | M1 | N2L, terms different signs
$v\frac{dv}{dx} = 5e^{-x} - 3x^2$ | A1 | [2] | AG
**(ii)**
$\int v \, dv = \int (5e^{-x} - 3x^2)dx$ | M1 | Attempts integration
$\frac{v^2}{2} = -5e^{-x} - 3x^3/3 (+c)$ | A1 | Accept $c$ omitted
$x = 0, v = 6$, hence $c = 23$ | B1 | Or uses limits 0 and 2
$\frac{v^2}{2} = -5e^{-2} - 3x^2/3 + 23$ | M1 | Puts $x = 2$ in $v(x)$ expression
$v = 5.35 \text{ ms}^{-1}$ | A1 | [5] | $v = 5.352..$ | [7]$A$ has velocity $\vec{v}$, there are velocities $\vec{x}$, $\vec{v}$, $\vec{v}$ around point $O$, and velocity $\vec{v}$
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item —
\item $v$ and $v$
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2013 Q3}}