| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.3 This is a straightforward M2 question requiring integration of vector acceleration with initial conditions (part a) and applying impulse-momentum theorem (part b). Both parts use standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors3.02f Non-uniform acceleration: using differentiation and integration6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{v} = \int 2\mathbf{i} + 6t\mathbf{j}\ dt = 2t\mathbf{i} + 3t^2\mathbf{j}\ (+\mathbf{c})\) | M1 A1 | |
| \(\mathbf{c} = 2\mathbf{i} - 4\mathbf{j}\) | A1 | (3) |
| \(\mathbf{v} = (2t+2)\mathbf{i} + (3t^2-4)\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(t=2:\ \mathbf{v} = 6\mathbf{i} + 8\mathbf{j}\) | B1 | |
| \(3\mathbf{i} - 1.5\mathbf{j} = 0.5(\mathbf{v} - (6\mathbf{i}+8\mathbf{j}))\) | M1 A1 | |
| \(\Rightarrow \mathbf{v} = 12\mathbf{i} + 5\mathbf{j}\) | A1 | |
| \(\Rightarrow | \mathbf{v} | = \sqrt{12^2 + 5^2} = 13\ \text{m s}^{-1}\) |
# Question 6:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{v} = \int 2\mathbf{i} + 6t\mathbf{j}\ dt = 2t\mathbf{i} + 3t^2\mathbf{j}\ (+\mathbf{c})$ | M1 A1 | |
| $\mathbf{c} = 2\mathbf{i} - 4\mathbf{j}$ | A1 | (3) |
| $\mathbf{v} = (2t+2)\mathbf{i} + (3t^2-4)\mathbf{j}$ | | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t=2:\ \mathbf{v} = 6\mathbf{i} + 8\mathbf{j}$ | B1 | |
| $3\mathbf{i} - 1.5\mathbf{j} = 0.5(\mathbf{v} - (6\mathbf{i}+8\mathbf{j}))$ | M1 A1 | |
| $\Rightarrow \mathbf{v} = 12\mathbf{i} + 5\mathbf{j}$ | A1 | |
| $\Rightarrow |\mathbf{v}| = \sqrt{12^2 + 5^2} = 13\ \text{m s}^{-1}$ | M1 A1 | (6) **(9)** |
---6. At time $t$ seconds the acceleration, a $\mathrm { m } \mathrm { s } ^ { - 2 }$, of a particle $P$ relative to a fixed origin $O$, is given by $\mathbf { a } = 2 \mathbf { i } + 6 t \mathbf { j }$. Initially the velocity of $P$ is $( 2 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $P$ at time $t$ seconds.
At time $t = 2$ seconds the particle $P$ is given an impulse ( $3 \mathbf { i } - 1.5 \mathbf { j }$ ) Ns. Given that the particle $P$ has mass 0.5 kg ,
\item find the speed of $P$ immediately after the impulse has been applied.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [9]}}