October 2020 Paper 3 Mechanics Q2
2. A particle \(P\) moves with acceleration \((4\mathbf{i} - 5\mathbf{j})\ \text{m s}^{-2}\)
At time \(t = 0\), \(P\) is moving with velocity \((-2\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-1}\)
At time \(t = 0\), \(P\) passes through the origin \(O\).
At time \(t = T\) seconds, where \(T \gt 0\), the particle \(P\) passes through the point \(A\).
The position vector of \(A\) is \((\lambda\mathbf{i} - 4.5\mathbf{j})\) m relative to \(O\), where \(\lambda\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) or integrate to give: \(\mathbf{v} = (-2\mathbf{i} + 2\mathbf{j}) + 2(4\mathbf{i} - 5\mathbf{j})\) | M1 | 3.1a |
| \((6\mathbf{i} - 8\mathbf{j})\ (\text{m s}^{-1})\) | A1 | 1.1b |
| (2) |
Notes
Accept column vectors throughout
M1: For any complete method to give a \(\mathbf{v}\) expression with correct no. of terms with \(t = 2\) used, so if integrating, must see the initial velocity as the constant.
Allow sign errors.
A1: Cao isw if they go on to find the speed.
| Scheme | Marks | AO |
|---|---|---|
| Solve problem through use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) or integration (M0 if \(\mathbf{u} = \mathbf{0}\)) Or any other complete method e.g use \(\mathbf{v} = \mathbf{u} + \mathbf{a}T\) and \(\mathbf{r} = \dfrac{(\mathbf{u} + \mathbf{v})T}{2}\): | M1 | 3.1a |
| \(-4.5\mathbf{j} = 2t\mathbf{j} - \dfrac{1}{2}t^2 5\mathbf{j}\) (\(\mathbf{j}\) terms only) | A1 | 1.1b |
| The first two marks could be implied if they go straight to an algebraic equation. | ||
| Attempt to equate \(\mathbf{j}\) components to give equation in \(T\) only \(\left(-4.5 = 2T - \dfrac{5}{2}T^2\right)\) | M1 | 2.1 |
| \(T = 1.8\) | A1 | 1.1b |
| (4) |
Notes
Accept column vectors throughout
M1: For any complete method to give a vector expression for \(\mathbf{j}\) component of displacement in \(t\) (or \(T\)) only, using \(\mathbf{a} = (4\mathbf{i} - 5\mathbf{j})\), so if integrating, RHS of equation must have the correct structure.
Allow sign errors.
A1: Correct \(\mathbf{j}\) vector equation in \(t\) or \(T\). Ignore \(\mathbf{i}\) terms.
M1: Must have earned 1st M mark.
Equate \(\mathbf{j}\) components to give equation in \(T\) (allow \(t\)) only (no \(\mathbf{j}\)’s) which has come from a displacement. Equation must be a 3 term quadratic in \(T\).
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| Solve problem by substituting their \(T\) value (M0 if \(T \lt 0\)) into the \(\mathbf{i}\) component equation to give an equation in \(\lambda\) only: \(\lambda = -2T + \dfrac{1}{2}T^2 \times 4\) | M1 | 3.1a |
| \(\lambda = 2.9\) or 2.88 or \(\dfrac{72}{25}\) oe | A1 | 1.1b |
| (2) | ||
| (8 marks) |
Notes
Accept column vectors throughout
M1: Must have earned 1st M mark in (b)
Complete method - must have an equation in \(\lambda\) only (no \(\mathbf{i}\)’s) which has come from an appropriate displacement.. (e.g M0 if \(\mathbf{a} = \mathbf{0}\) has been used)
Expression for \(\lambda\) must be a quadratic in \(T\)
A1: cao