| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Distance between two positions |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of velocity to find acceleration, then applying F=ma, followed by integration of velocity to find position. All steps are routine applications of standard techniques with no conceptual challenges or novel problem-solving required. Slightly easier than average due to its mechanical nature. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10f Distance between points: using position vectors3.03a Force: vector nature and diagrams3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\ddot{\mathbf{r}} = 6\mathbf{i} + (2t+3)\mathbf{j}\) | B1 | |
| \(\mathbf{F} = 0.4(6\mathbf{i} + 11\mathbf{j})\) | M1 | \(0.4 \times\) something obtained by differentiation, with \(t = 4\) |
| \( | \mathbf{F} | = \sqrt{(2.4^2 + 4.4^2)}\) |
| \(\approx 5.0\) | A1 | accept more accurate answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r} = (3t^2 + 4t)\mathbf{i} + \left(\frac{1}{3}t^3 + \frac{3}{2}t^2\right)\mathbf{j}\) \((+ \mathbf{C})\) | M1 | |
| Using boundary values: \(\mathbf{r} = (3t^2 + 4t - 3)\mathbf{i} + \left(\frac{1}{3}t^3 + \frac{3}{2}t^2 + 4\right)\mathbf{j}\) | A1 | |
| \(t = 4\): \(\mathbf{r} = 61\mathbf{i} + 49\frac{1}{3}\mathbf{j}\) | A1 | |
| \(OS = \sqrt{61^2 + 49\frac{1}{3}^2} \approx 78\) (m) | M1 A1 | accept more accurate answers |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\ddot{\mathbf{r}} = 6\mathbf{i} + (2t+3)\mathbf{j}$ | B1 | |
| $\mathbf{F} = 0.4(6\mathbf{i} + 11\mathbf{j})$ | M1 | $0.4 \times$ something obtained by differentiation, with $t = 4$ |
| $|\mathbf{F}| = \sqrt{(2.4^2 + 4.4^2)}$ | M1 | modulus of a vector |
| $\approx 5.0$ | A1 | accept more accurate answers |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = (3t^2 + 4t)\mathbf{i} + \left(\frac{1}{3}t^3 + \frac{3}{2}t^2\right)\mathbf{j}$ $(+ \mathbf{C})$ | M1 | |
| Using boundary values: $\mathbf{r} = (3t^2 + 4t - 3)\mathbf{i} + \left(\frac{1}{3}t^3 + \frac{3}{2}t^2 + 4\right)\mathbf{j}$ | A1 | |
| $t = 4$: $\mathbf{r} = 61\mathbf{i} + 49\frac{1}{3}\mathbf{j}$ | A1 | |
| $OS = \sqrt{61^2 + 49\frac{1}{3}^2} \approx 78$ (m) | M1 A1 | accept more accurate answers |
---4. A particle $P$ of mass 0.4 kg is moving under the action of a single force $\mathbf { F }$ newtons. At time $t$ seconds, the velocity of $P , \mathbf { v } \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by
$$\mathbf { v } = ( 6 t + 4 ) \mathbf { i } + \left( t ^ { 2 } + 3 t \right) \mathbf { j } .$$
When $t = 0 , P$ is at the point with position vector $( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m }$. When $t = 4 , P$ is at the point $S$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the magnitude of $\mathbf { F }$ when $t = 4$.
\item Calculate the distance $O S$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2005 Q4 [9]}}