M2 June 2007 Q2
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 \(\mathbf{v}\) m s\(^{-1}\) of \(P\) is given by
\[\mathbf{v} = 3t^2\mathbf{i} + (1 - 4t)\mathbf{j}.\]Find
| Scheme | Marks |
|---|---|
| \(\mathbf{a} = \dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = 6t\mathbf{i} - 4\mathbf{j}\) | M1 A1 |
| (2) |
Notes
M1 Clear attempt to differentiate. Condone \(\mathbf{i}\) or \(\mathbf{j}\) missing.
A1 both terms correct (column vectors are OK)
| Scheme | Marks |
|---|---|
| Using \(\mathbf{F} = \tfrac{1}{2}\mathbf{a}\), sub \(t = 2\), finding modulus | M1, M1, M1 |
| e.g. 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) |
| (4) | |
| (6 marks) |
Notes
The 3 method marks can be tackled in any order, but for consistency on epen grid please enter as:
M1 \(\mathbf{F} = m\mathbf{a}\) (their \(\mathbf{a}\), (correct \(\mathbf{a}\) or following from (a)), not \(\mathbf{v}\). \(\mathbf{F} = \dfrac{1}{2}\mathbf{a}\)). Condone \(\mathbf{a}\) not a vector for this mark.
M1 subst \(t = 2\) into candidate’s vector \(\mathbf{F}\) or \(\mathbf{a}\) (\(\mathbf{a}\) correct or following from (a), not \(\mathbf{v}\))
M1 Modulus of candidate’s \(\mathbf{F}\) or \(\mathbf{a}\) (not \(\mathbf{v}\))
A1 CSO All correct (beware fortuitous answers e.g. from \(6t\mathbf{i} + 4\mathbf{j}\)) Accept 6.3, awrt 6.32, any exact equivalent e.g. \(2\sqrt{10}\), \(\sqrt{40}\), \(\dfrac{\sqrt{160}}{2}\)