| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Collision or meeting problems |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring integration of velocity to find position (standard technique), then solving simultaneous equations to show collision. The integration is simple polynomial work, and the collision proof just requires equating position vectors and checking consistency. Slightly above average difficulty due to vector notation and multi-part nature, but all techniques are routine for M2 students. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{p} = (2t^2 - 7t)\mathbf{i} - 5t\mathbf{j} + 3\mathbf{i} + 5\mathbf{j}\) | M1, M1 | |
| \(= (2t^2 - 7t + 3)\mathbf{i} + (5 - 5t)\mathbf{j}\) | A1+A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{q} = (2\mathbf{i} - 3\mathbf{j})t - 7\mathbf{i}\) | M1 A1 | |
| \(\mathbf{j}\): \(5 - 5t = -3t \Rightarrow t = 2.5\), equating and solving | M1 A1 | |
| At \(t = 2.5\), \(\mathbf{i}\): \(p_x = 2\times2.5^2 - 7\times2.5 + 3 = -2\); \(q_x = 2\times2.5 - 7 = -2\) | M1 | both |
| \(p_x = q_x \Rightarrow\) collision | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{i}\): \(2t^2 - 7t + 3 = 2t - 7 \Rightarrow 2t^2 - 9t + 10 = 0\); \(t = 2, 2.5\), equating and solving | M1 A1 | |
| At \(t = 2.5\), \(\mathbf{j}\): \(p_y = 5 - 5\times2.5 = -7.5\); \(q_y = -3\times2.5 = -7.5\) | M1 | both |
| \(p_y = q_y \Rightarrow\) collision | A1 | cso |
| *Ignore any working associated with \(t = 2\)* |
## Question 4:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{p} = (2t^2 - 7t)\mathbf{i} - 5t\mathbf{j} + 3\mathbf{i} + 5\mathbf{j}$ | M1, M1 | |
| $= (2t^2 - 7t + 3)\mathbf{i} + (5 - 5t)\mathbf{j}$ | A1+A1 | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{q} = (2\mathbf{i} - 3\mathbf{j})t - 7\mathbf{i}$ | M1 A1 | |
| $\mathbf{j}$: $5 - 5t = -3t \Rightarrow t = 2.5$, equating and solving | M1 A1 | |
| At $t = 2.5$, $\mathbf{i}$: $p_x = 2\times2.5^2 - 7\times2.5 + 3 = -2$; $q_x = 2\times2.5 - 7 = -2$ | M1 | both |
| $p_x = q_x \Rightarrow$ collision | A1 | cso |
**Alternative for (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{i}$: $2t^2 - 7t + 3 = 2t - 7 \Rightarrow 2t^2 - 9t + 10 = 0$; $t = 2, 2.5$, equating and solving | M1 A1 | |
| At $t = 2.5$, $\mathbf{j}$: $p_y = 5 - 5\times2.5 = -7.5$; $q_y = -3\times2.5 = -7.5$ | M1 | both |
| $p_y = q_y \Rightarrow$ collision | A1 | cso |
| *Ignore any working associated with $t = 2$* | | |
---4. At time $t$ seconds, the velocity of a particle $P$ is $[ ( 4 t - 7 ) \mathbf { i } - 5 \mathbf { j } ] \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0 , P$ is at the point with position vector $( 3 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m }$ relative to a fixed origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the position vector of $P$ after $t$ seconds, giving your answer in the form $( a \mathbf { i } + b \mathbf { j } ) \mathrm { m }$.
A second particle $Q$ moves with constant velocity $( 2 \mathbf { i } - 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0$, the position vector of $Q$ is $( - 7 \mathrm { i } ) \mathrm { m }$.
\item Prove that $P$ and $Q$ collide.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2004 Q4 [10]}}