June 2019 Paper 3 Mechanics Q1
1. [In this question position vectors are given relative to a fixed origin \(O\)]
At time \(t\) seconds, where \(t \geqslant 0\), a particle, \(P\), moves so that its velocity \(\mathbf{v}\ \text{m s}^{-1}\) is given by
\[\mathbf{v} = 6t\mathbf{i} - 5t^{\frac{3}{2}}\mathbf{j}\]When \(t = 0\), the position vector of \(P\) is \((-20\mathbf{i} + 20\mathbf{j})\) m.
| Scheme | Marks | AO |
|---|---|---|
| Differentiate \(\mathbf{v}\) | M1 | 1.1a |
| \((\mathbf{a} =)\ 6\mathbf{i} - \dfrac{15}{2}t^{\frac{1}{2}}\mathbf{j}\) | A1 | 1.1b |
| \(= 6\mathbf{i} - 15\mathbf{j}\ (\text{m s}^{-2})\) | A1 | 1.1b |
| (3) |
Notes
N.B. Accept column vectors throughout and condone missing brackets in working but they must be there in final answers
M1: Use of \(\mathbf{a} = \dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\) with attempt to differentiate (both powers decreasing by 1)
M0 if \(\mathbf{i}\)’s and \(\mathbf{j}\)’s omitted and they don’t recover
A1: Correct differentiation in any form
A1: Correct and simplified.
Ignore subsequent working (ISW) if they go on and find the magnitude.
| Scheme | Marks | AO |
|---|---|---|
| Integrate \(\mathbf{v}\) | M1 | 1.1a |
| \((\mathbf{r} =)\ (\mathbf{r}_0) + 3t^2\mathbf{i} - 2t^{\frac{5}{2}}\mathbf{j}\) | A1 | 1.1b |
| \(= (-20\mathbf{i} + 20\mathbf{j}) + (48\mathbf{i} - 64\mathbf{j}) = 28\mathbf{i} - 44\mathbf{j}\) (m) | A1 | 2.2a |
| (3) | ||
| (6 marks) |
Notes
N.B. Accept column vectors throughout and condone missing brackets in working but they must be there in final answers
M1: Use of \(\mathbf{r} = \displaystyle\int \mathbf{v}\,\mathrm{d}t\) with attempt to integrate (both powers increasing by 1)
M0 if \(\mathbf{i}\)’s and \(\mathbf{j}\)’s omitted and they don’t recover
A1: Correct integration in any form. Condone \(\mathbf{r}_0\) not present
A1: Correct and simplified.