| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find force using F=ma |
| Difficulty | Moderate -0.8 This is a straightforward application of differentiation to find acceleration from velocity, followed by using F=ma. Both parts require only direct recall of standard formulas with minimal calculation steps, making it easier than average for M2. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{a} = \frac{d\mathbf{v}}{dt} = 6t\mathbf{i} - 4\mathbf{j}\) | M1 A1 | Clear attempt to differentiate; condone i or j missing. Both terms correct (column vectors OK) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(\mathbf{F} = \frac{1}{2}\mathbf{a}\), sub \(t = 2\), finding modulus | M1, M1, M1 | M1: \(\mathbf{F}=m\mathbf{a}\) (their a, correct or following from (a)), not v; \(\mathbf{F}=\frac{1}{2}\mathbf{a}\). Condone a not a vector. M1: sub \(t=2\) into candidate's vector F or a. M1: modulus of candidate's F or a (not v) |
| At \(t=2\), \(\mathbf{a} = 12\mathbf{i} - 4\mathbf{j}\) | ||
| \(\mathbf{F} = 6\mathbf{i} - 2\mathbf{j}\) | ||
| \( | \mathbf{F} | = \sqrt{6^2 + 2^2} \approx 6.32\) N |
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{a} = \frac{d\mathbf{v}}{dt} = 6t\mathbf{i} - 4\mathbf{j}$ | M1 A1 | Clear attempt to differentiate; condone **i** or **j** missing. Both terms correct (column vectors OK) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $\mathbf{F} = \frac{1}{2}\mathbf{a}$, sub $t = 2$, finding modulus | M1, M1, M1 | M1: $\mathbf{F}=m\mathbf{a}$ (their **a**, correct or following from (a)), not **v**; $\mathbf{F}=\frac{1}{2}\mathbf{a}$. Condone **a** not a vector. M1: sub $t=2$ into candidate's vector **F** or **a**. M1: modulus of candidate's **F** or **a** (not **v**) |
| At $t=2$, $\mathbf{a} = 12\mathbf{i} - 4\mathbf{j}$ | | |
| $\mathbf{F} = 6\mathbf{i} - 2\mathbf{j}$ | | |
| $|\mathbf{F}| = \sqrt{6^2 + 2^2} \approx 6.32$ N | A1 (CSO) | CSO; beware fortuitous answers. Accept 6.3, awrt 6.32, any exact equivalent e.g. $2\sqrt{10}$, $\sqrt{40}$, $\frac{\sqrt{160}}{2}$ |
---2. A particle $P$ of mass 0.5 kg moves under the action of a single force $\mathbf { F }$ newtons. At time $t$ seconds, the velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ of $P$ is given by
$$\mathbf { v } = 3 t ^ { 2 } \mathbf { i } + ( 1 - 4 t ) \mathbf { j }$$
Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of $P$ at time $t$ seconds,
\item the magnitude of $\mathbf { F }$ when $t = 2$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q2 [6]}}