| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | When moving parallel to given vector |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard integration of velocity to find position, differentiation to find acceleration, and equating vector components to find when direction matches i+j. All techniques are routine with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(5T^2 - 12T + 15 = T^2 + 8T - 10\) | M1 | Parallel to \(\mathbf{i} + \mathbf{j}\) |
| \(\Rightarrow 4T^2 - 20T + 25 = 0\) | A1 | Correct quadratic in \(T\) |
| \(\Rightarrow T = \frac{5}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} = (10t-12)\mathbf{i} + (2t+8)\mathbf{j}\) | M1 | Correct differentiation (at least 2 powers going down by one) |
| \(= 18\mathbf{i} + 14\mathbf{j}\) | A1 | |
| \( | \mathbf{a} | = \sqrt{18^2 + 14^2}\) |
| \(= \sqrt{520} = 22.8\ (\text{m s}^{-2})\) | A1 | \(23\) or better e.g. \(2\sqrt{130}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{s} = \left(\frac{5}{3}t^3 - 6t^2 + 15t\right)\mathbf{i} + \left(\frac{1}{3}t^3 + 4t^2 - 10t\right)\mathbf{j}\) | M1 | Integrate (at least 2 powers going up by one) |
| A1 | At most one error | |
| A1 | All correct |
| Answer | Marks |
|---|---|
| \(= 36\mathbf{i} + 15\mathbf{j}\ \text{(m)}\) | A1 |
## Question 5(a):
$5T^2 - 12T + 15 = T^2 + 8T - 10$ | M1 | Parallel to $\mathbf{i} + \mathbf{j}$
$\Rightarrow 4T^2 - 20T + 25 = 0$ | A1 | Correct quadratic in $T$
$\Rightarrow T = \frac{5}{2}$ | A1 |
**[3]**
## Question 5(b):
$\mathbf{a} = (10t-12)\mathbf{i} + (2t+8)\mathbf{j}$ | M1 | Correct differentiation (at least 2 powers going down by one)
$= 18\mathbf{i} + 14\mathbf{j}$ | A1 |
$|\mathbf{a}| = \sqrt{18^2 + 14^2}$ | DM1 | Use of Pythagoras to find magnitude. Dependent on preceding M1
$= \sqrt{520} = 22.8\ (\text{m s}^{-2})$ | A1 | $23$ or better e.g. $2\sqrt{130}$
**[4]**
## Question 5(c):
$\mathbf{s} = \left(\frac{5}{3}t^3 - 6t^2 + 15t\right)\mathbf{i} + \left(\frac{1}{3}t^3 + 4t^2 - 10t\right)\mathbf{j}$ | M1 | Integrate (at least 2 powers going up by one)
| A1 | At most one error
| A1 | All correct
$= (45 - 54 + 45)\mathbf{i} + (9 + 36 - 30)\mathbf{j}$
$= 36\mathbf{i} + 15\mathbf{j}\ \text{(m)}$ | A1 |
**[4]**
---5. At time $t$ seconds, $t \geqslant 0$, a particle $P$ has velocity $\mathbf { v } \mathrm { ms } ^ { - 1 }$, where
$$\mathbf { v } = \left( 5 t ^ { 2 } - 12 t + 15 \right) \mathbf { i } + \left( t ^ { 2 } + 8 t - 10 \right) \mathbf { j }$$
When $t = 0 , P$ is at the origin $O$.\\
At time $T$ seconds, $P$ is moving in the direction of $( \mathbf { i } + \mathbf { j } )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $T$.
When $t = 3 , P$ is at the point $A$.
\item Find the magnitude of the acceleration of $P$ as it passes through $A$.
\item Find the position vector of $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2021 Q5 [11]}}