| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Collision or meeting problems |
| Difficulty | Moderate -0.3 This is a straightforward M2 kinematics question requiring integration of velocity to find position, setting components equal to zero for parallel vectors, and solving simultaneous equations for collision. All techniques are standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (content not fully shown) | B1 |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| (content not fully shown) | B1 | |
---4. At time $t$ seconds the velocity of a particle $P$ is $[ ( 4 t - 5 ) \mathbf { i } + 3 \mathbf { j } ] \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0$, the position vector of $P$ is $( 2 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m }$, relative to a fixed origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$ when the velocity of $P$ is parallel to the vector $\mathbf { j }$.
\item Find an expression for the position vector of $P$ at time $t$ seconds.
A second particle $Q$ moves with constant velocity $( - 2 \mathbf { i } + c \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0$, the position vector of $Q$ is $( 11 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m }$. The particles $P$ and $Q$ collide at the point with position vector ( $d \mathbf { i } + 14 \mathbf { j }$ ) m.
\item Find
\begin{enumerate}[label=(\roman*)]
\item the value of $c$,
\item the value of $d$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q4 [10]}}