June 2023 Paper 3 Mechanics Q3
3. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) has velocity \(\mathbf{v}\ \text{m s}^{-1}\) where
\[\mathbf{v} = (t^2 - 3t + 7)\mathbf{i} + (2t^2 - 3)\mathbf{j}\]Find
| Scheme | Marks | AO |
|---|---|---|
| \(7\mathbf{i} - 3\mathbf{j}\) seen or implied by Pythagoras | B1 | 1.1b |
| Use Pythagoras: \(\sqrt{7^2 + (-3)^2}\) | M1 | 3.1a |
| \(\sqrt{58}\), 7.6 or better \((\text{m s}^{-1})\) | A1 | 1.1b |
| (3) |
Notes
Allow column vectors throughout.
B1: cao
M1: Use of Pythagoras, including the square root, on a velocity vector at \(t = 0\)
A1: cao. Must come from a correct \(\mathbf{v}\).
| Scheme | Marks | AO |
|---|---|---|
| \(t^2 - 3t + 7 = 2t^2 - 3\) OR \(\dfrac{t^2 - 3t + 7}{2t^2 - 3} = \dfrac{1}{1} = 1\) | M1 | 2.1 |
| \(t = 2\) only | A1 | 1.1b |
| (2) |
Notes
Allow column vectors throughout.
M1: Equating \(\mathbf{i}\) and \(\mathbf{j}\) components of \(\mathbf{v}\) or a ratio of 1:1 to obtain a quadratic in \(t\) only.
If they use a constant, e.g. \(t^2 - 3t + 7 = k\) and \(2t^2 - 3 = k\), \(k\) must be eliminated to earn this mark.
N.B. M0 (since wrong working seen) if they write down
\(\mathbf{i} + \mathbf{j} = (t^2 - 3t + 7)\mathbf{i} + (2t^2 - 3)\mathbf{j}\)
OR \(\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}t^2 - 3t + 7\\2t^2 - 3\end{pmatrix}\)
OR \(t^2 - 3t + 7 = 1\) and \(2t^2 - 3 = 1\)
and then \(t^2 - 3t + 7 = 2t^2 - 3\)
A1: \(t = 2\)
N.B. Allow M1A1 for a correct trial and error method where they obtain \(\mathbf{v} = 5\mathbf{i} + 5\mathbf{j}\) when \(t = 2\) but M0 if they don’t get \(t = 2\)
| Scheme | Marks | AO |
|---|---|---|
| Differentiate \(\mathbf{v}\) wrt \(t\) to give a vector. | M1 | 3.1a |
| \((2t - 3)\mathbf{i} + 4t\mathbf{j}\) | A1 | 1.1b |
| (2) |
Notes
Allow column vectors throughout.
M1: At least one power decreasing by 1 in each component in their \(\mathbf{v}\)
(M0 if clearly dividing by \(t\))
Both \(\mathbf{i}\) and \(\mathbf{j}\) needed in their answer or a column vector
Allow recovery if the \(\mathbf{i}\) and \(\mathbf{j}\) disappear and then reappear.
A1: cao (must be a vector) isw e.g. if they find the magnitude or put \(t = 0\) or differentiate again
\(\mathbf{i}\)’s and \(\mathbf{j}\)’s do not need to be collected.
N.B. Allow M1A0 for \(2t - 3\mathbf{i} + 4t\mathbf{j}\)
| Scheme | Marks | AO |
|---|---|---|
| \(2t - 3 = 0\) | M1 | 3.1a |
| \(t = 1.5\) | A1 | 1.1b |
| (2) | ||
| (9 marks) |
Notes
Allow column vectors throughout.
M1: \(2t - 3 = 0\) or (their derivative of the \(\mathbf{i}\)-component of \(\mathbf{v}\)) = 0
N.B. M0 if they equate the derivative of both components of \(\mathbf{v}\) to zero.
A1: cao
N.B. Correct answer, with no working, can score both marks.