| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when moving in specific direction |
| Difficulty | Moderate -0.3 This is a straightforward M2 variable acceleration question requiring integration of vector components with initial conditions, then finding when j-component is zero. The integration is routine (polynomials), and finding the constant requires simple substitution. Part (b) adds minimal complexity by requiring recognition that parallel to i means j-component = 0. Slightly easier than average due to the mechanical nature of the calculus involved. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Integrate: \(\mathbf{v} = (t^3 - 2t^2)\mathbf{i} + (3t^2 - 5t)\mathbf{j} + \mathbf{C}\) | M1, A2 | At least 3 powers going up. Must be two separate component equations. \(-1\) each integration error: A1A1 one error, A0A0 two or more errors. Allow with no "\(+ \mathbf{C}\)" |
| \(t=3\): \(\mathbf{v} = 9\mathbf{i} + 12\mathbf{j} + \mathbf{C} = 11\mathbf{i} + 10\mathbf{j}\), so \(\mathbf{C} = 2\mathbf{i} - 2\mathbf{j}\) | DM1 | Substitute given values to find \(\mathbf{C}\). Dependent on previous M mark |
| \(\mathbf{v} = (t^3 - 2t^2 + 2)\mathbf{i} + (3t^2 - 5t - 2)\mathbf{j}\) | A1 | Correct velocity (any equivalent form) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Parallel to \(\mathbf{i} \Rightarrow 3t^2 - 5t - 2 = 0\) | M1 | Set \(\mathbf{j}\) component of their \(\mathbf{v}\) equal to zero and solve for \(t\). Correct answers imply method |
| \((3t+1)(t-2) = 0\), \(t = 2\) | A1 | Correct only. Ignore \(-\frac{1}{3}\) if present |
| \( | \mathbf{v} | = 8 - 8 + 2 = 2\ (\text{m s}^{-1})\) |
| A1 | Answer must be a scalar. Results from negative \(t\) must be rejected |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Integrate: $\mathbf{v} = (t^3 - 2t^2)\mathbf{i} + (3t^2 - 5t)\mathbf{j} + \mathbf{C}$ | M1, A2 | At least 3 powers going up. Must be two separate component equations. $-1$ each integration error: A1A1 one error, A0A0 two or more errors. Allow with no "$+ \mathbf{C}$" |
| $t=3$: $\mathbf{v} = 9\mathbf{i} + 12\mathbf{j} + \mathbf{C} = 11\mathbf{i} + 10\mathbf{j}$, so $\mathbf{C} = 2\mathbf{i} - 2\mathbf{j}$ | DM1 | Substitute given values to find $\mathbf{C}$. Dependent on previous M mark |
| $\mathbf{v} = (t^3 - 2t^2 + 2)\mathbf{i} + (3t^2 - 5t - 2)\mathbf{j}$ | A1 | Correct velocity (any equivalent form) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Parallel to $\mathbf{i} \Rightarrow 3t^2 - 5t - 2 = 0$ | M1 | Set $\mathbf{j}$ component of their $\mathbf{v}$ equal to zero and solve for $t$. Correct answers imply method |
| $(3t+1)(t-2) = 0$, $t = 2$ | A1 | Correct only. Ignore $-\frac{1}{3}$ if present |
| $|\mathbf{v}| = 8 - 8 + 2 = 2\ (\text{m s}^{-1})$ | DM1 | Substitute their $t$ to find $\mathbf{v}$. Dependent on previous M mark |
| | A1 | Answer must be a scalar. Results from negative $t$ must be rejected |
---2. At time $t$ seconds, where $t \geqslant 0$, a particle $P$ is moving on a horizontal plane with acceleration $\left[ \left( 3 t ^ { 2 } - 4 t \right) \mathbf { i } + ( 6 t - 5 ) \mathbf { j } \right] \mathrm { m } \mathrm { s } ^ { - 2 }$.
When $t = 3$ the velocity of $P$ is $( 11 \mathbf { i } + 10 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
Find
\begin{enumerate}[label=(\alph*)]
\item the velocity of $P$ at time $t$ seconds,
\item the speed of $P$ when it is moving parallel to the vector $\mathbf { i }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q2 [9]}}