June 2023 Paper 3 Mechanics Q4
4. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors and position vectors are given relative to a fixed origin \(O\)]
A particle \(P\) is moving on a smooth horizontal plane.
The particle has constant acceleration \((2.4\mathbf{i} + \mathbf{j})\ \text{m s}^{-2}\)
At time \(t = 0\), \(P\) passes through the point \(A\).
At time \(t = 5\) s, \(P\) passes through the point \(B\).
The velocity of \(P\) as it passes through \(A\) is \((-16\mathbf{i} - 3\mathbf{j})\ \text{m s}^{-1}\)
The position vector of \(A\) is \((44\mathbf{i} - 10\mathbf{j})\) m.
At time \(t = T\) seconds, where \(T > 5\), \(P\) passes through the point \(C\).
The position vector of \(C\) is \((4\mathbf{i} + c\mathbf{j})\) m.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{v}_B = (-16\mathbf{i} - 3\mathbf{j}) + 5(2.4\mathbf{i} + \mathbf{j})\) | M1 | 3.4 |
| \(\mathbf{v}_B = (-4\mathbf{i} + 2\mathbf{j})\) | A1 | 1.1b |
| \(\sqrt{(-4)^2 + 2^2}\) | M1 | 3.1a |
| \(\sqrt{20} = 2\sqrt{5}\), 4.5 or better \((\text{m s}^{-1})\) | A1 | 1.1b |
| (4) |
Notes
Accept column vectors throughout
M1: Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) with \(t = 5\) to give an unsimplified \(\mathbf{v}_B\)
M0 if \(\mathbf{u} = \mathbf{0}\)
N.B. If using integration, they must get to the same stage i.e. have found the constant and put \(t = 5\)
M0 if they omit the constant altogether
A1: Correct \(\mathbf{v}_B\) with \(\mathbf{i}\)’s and \(\mathbf{j}\)’s collected
M1: Use of Pythagoras on their \(\mathbf{v}_B\) to give a magnitude (need the root)
A1: Must be positive
| Scheme | Marks | AO |
|---|---|---|
| Using \(A\) as the initial position: \(\mathbf{r}_C = \mathbf{v}_A t + \dfrac{1}{2}\mathbf{a}t^2 + \mathbf{r}_A\) where \(t = T\) \((4\mathbf{i} + c\mathbf{j}) = (-16\mathbf{i} - 3\mathbf{j})T + \dfrac{1}{2}(2.4\mathbf{i} + \mathbf{j})T^2 + (44\mathbf{i} - 10\mathbf{j})\) OR \(\begin{pmatrix}4\\c\end{pmatrix} = \begin{pmatrix}-16\\-3\end{pmatrix}T + \dfrac{1}{2}\begin{pmatrix}2.4\\1\end{pmatrix}T^2 + \begin{pmatrix}44\\-10\end{pmatrix}\) Equating \(\mathbf{i}\)-components, to give a quadratic equation in \(T\) only. Allow \(t\) instead of \(T\). N.B. Allow omission of 44 for this M mark. Also allow \(\pm 4\) but M0 if 4 is not used at all i.e. \(4 = -16T + \dfrac{1}{2} \times 2.4T^2\) scores M1A0A0 | M1 | 3.1a |
| \(4 = -16T + \dfrac{1}{2} \times 2.4T^2 + 44\) | A1 | 1.1b |
| \((T =)\ 10\) | A1 | 1.1b |
| (3) |
Alternative
| Scheme | Marks |
|---|---|
| ALTERNATIVE using \(B\) as the initial position: (The position vector of \(B\), \(\mathbf{r}_B\), should be \(-6\mathbf{i} - 12.5\mathbf{j}\) but no credit for finding this) \(\mathbf{r}_C = \mathbf{v}_B t + \dfrac{1}{2}\mathbf{a}t^2 + \mathbf{r}_B\) using their \(\mathbf{v}_B\) from (a) and their \(\mathbf{r}_B\) \((4\mathbf{i} + c\mathbf{j}) = (-4\mathbf{i} + 2\mathbf{j})t + \dfrac{1}{2}(2.4\mathbf{i} + \mathbf{j})t^2 + (-6\mathbf{i} - 12.5\mathbf{j})\) \(\begin{pmatrix}4\\c\end{pmatrix} = \begin{pmatrix}-4\\2\end{pmatrix}t + \dfrac{1}{2}\begin{pmatrix}2.4\\1\end{pmatrix}t^2 + \begin{pmatrix}-6\\-12.5\end{pmatrix}\) Equating \(\mathbf{i}\)-components, to give a quadratic equation in \(t\) only. Allow if they have \(T\) instead of \(t\). N.B. Allow omission of their \(-6\) or if they use 44 for this M mark. Also allow \(\pm 4\) but M0 if 4 is not used at all. e.g. \(4 = -4t + \dfrac{1}{2} \times 2.4t^2\) scores M1A0A0 | M1 |
| \(4 = -4t + \dfrac{1}{2} \times 2.4t^2 - 6\) | A1 |
| \(t = 5\) so \((T =)\ 10\) | A1 |
| (3) |
Notes
Accept column vectors throughout
M1: Equating components of \(\mathbf{i}\) to give an equation in \(T\) or \(t\) only.
N.B. (they could use integration to get to the same stage) for this M mark, they only need to be equating the \(\mathbf{i}\)-components, and receive no credit until they do so.
M0 if \(\mathbf{u} = \mathbf{0}\)
A1: A correct equation in \(T\) or \(t\) only (could be in \((T - 5)\) if using \(B\) as initial position)
A1: \(T = 10\)
| Scheme | Marks | AO |
|---|---|---|
| Equating \(\mathbf{j}\)-components, with their value of \(T\) or \(t\) substituted, to give an equation, which must have a square term, in \(c\) only. N.B. Allow \(\pm c\) in their equation. (N.B. Allow omission of \(-10\) or their \(-12.5\) for this M mark i.e. if using \(A\) as initial position \(c = (-3 \times 10) + \dfrac{1}{2} \times 1 \times 10^2\) scores M1M0A0 OR if using \(B\) as initial position \(c = (2 \times 5) + \dfrac{1}{2} \times 1 \times 5^2\) scores M1M0A0) | M1 | 2.1 |
| if using \(A\) as initial position \(c = (-3 \times 10) + \dfrac{1}{2} \times 1 \times 10^2 + (-10)\) N.B. Allow \(\pm c\) and/or \(\pm(-10)\) in their equation OR if using \(B\) as initial position \(c = (2 \times 5) + \dfrac{1}{2} \times 1 \times 5^2 + (-12.5)\) N.B. Allow \(\pm c\) and/or \(\pm(-12.5)\) in their equation | M1 | 1.1b |
| \(c = 10\) | A1 | 1.1b |
| (3) | ||
| (10 marks) |
Notes
Accept column vectors throughout
M1: Equating components of \(\mathbf{j}\) to give an equation in \(c\) only but allow omission of their initial position
M1: With their value of \(T\) or \(t\) and must include \(t = 0\) position (should be \(-10\) if using \(A\) OR their \(-12.5\) if using \(B\))
A1: cao