| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2020 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find position by integrating velocity |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question involving standard vector calculus techniques. Part (a) requires simple differentiation of velocity to find acceleration. Part (b) involves finding when the velocity is parallel to a given vector (ratio of components) and calculating speed—routine but multi-step. Part (c) requires integrating velocity to find position and showing it never returns to origin, which is algebraically straightforward. All parts use standard M2 techniques with no novel insight required, making this 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, acceleration3.02e Two-dimensional constant acceleration: with vectors |
# Question 5(a)
M1 Use of $a = \frac{dv}{dt}$: $a = 6ti + 2j$
A1 $t = 0 \Rightarrow a = 2j \text{ (m s}^{-2}\text{)}$
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# Question 5(b)
M1 Use of velocity parallel to $11i + j$. $11$ must be on the correct side
M1 $\frac{11}{10}(2t - 4) = (3t^2 - 4)$
DM1 $3t^2 - 22t + 40 = 0 \Rightarrow t = \frac{10}{3}, t = 4$. Select the larger root (dependent on the previous 2 M1 marks and on 2 positive roots)
A1 $v = 44i + 4j$, speed $= \sqrt{44^2 + 4^2} = 4\sqrt{122} \text{ (m s}^{-1}\text{)}$ or any equivalent simplified surd form. Condone if they find both speeds
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# Question 5(c)
M1 Use of $r = \int v \, dt$. Powers going up
A1 $r = (t^3 - 4t)i + (t^2 - 4t)j$
M1 Set $r = 0$ and solve for $t$. Consider both components
A1 $t^3 - 4t = 0 \Rightarrow t = 0, 2, (-2)$ and $t^2 - 4t = 0 \Rightarrow t = 0, 4$. The only common value is $t = 0$, so does not return to O.
**NB:** If a constant of integration is introduced, they must conclude it is equal to the zero vector
**NB:** Condone if they ignore $t = 0$. Do not need to see the roots. But do need to see the factorised form for each component if using this method
---5. A t time $t$ seconds ( $t \geqslant 0$ ), a particle $P$ has velocity $\mathbf { v m ~ s } ^ { - 1 }$, where
$$\mathbf { v } = \left( 3 t ^ { 2 } - 4 \right) \mathbf { i } + ( 2 t - 4 ) \mathbf { j }$$
When $t = 0 , P$ is at the fixed point $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ at the instant when $t = 0$
\item Find the exact speed of $P$ at the instant when $P$ is moving in the direction of the vector $( 11 \mathbf { i } + \mathbf { j } )$ for the second time.
\item Show that $P$ never returns to $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{c16c17b6-2c24-4939-b3b5-63cd63646b76-14_2658_1938_107_123}
\includegraphics[max width=\textwidth, alt={}, center]{c16c17b6-2c24-4939-b3b5-63cd63646b76-15_149_140_2604_1818}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2020 Q5 [10]}}