June 2022 Paper 3 Mechanics Q3
3. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors.]
A particle \(P\) of mass 4 kg is at rest at the point \(A\) on a smooth horizontal plane.
At time \(t = 0\), two forces, \(\mathbf{F}_1 = (4\mathbf{i} - \mathbf{j})\) N and \(\mathbf{F}_2 = (\lambda\mathbf{i} + \mu\mathbf{j})\) N, where \(\lambda\) and \(\mu\) are constants, are applied to \(P\)
Given that \(P\) moves in the direction of the vector \((3\mathbf{i} + \mathbf{j})\)
At time \(t = 4\) seconds, \(P\) passes through the point \(B\).
Given that \(\lambda = 2\)
| Scheme | Marks | AO |
|---|---|---|
| \((4\mathbf{i} - \mathbf{j}) + (\lambda\mathbf{i} + \mu\mathbf{j}) = (4 + \lambda)\mathbf{i} + (-1 + \mu)\mathbf{j}\) | M1 | 3.4 |
| Use ratios to obtain an equation in \(\lambda\) and \(\mu\) only | M1 | 2.1 |
| \(\dfrac{(4 + \lambda)}{(-1 + \mu)} = \dfrac{3}{1}\) or \(\dfrac{\frac{1}{4}(4 + \lambda)}{\frac{1}{4}(-1 + \mu)} = \dfrac{3}{1}\) | A1 | 1.1b |
| \(\lambda - 3\mu + 7 = 0\) * Allow \(0 = \lambda - 3\mu + 7\) but nothing else. | A1* | 1.1b |
| (4) |
Notes
Accept column vectors throughout
M1: Adding the two forces, \(\mathbf{i}\)’s and \(\mathbf{j}\)’s must be collected (or must be a single column vector) seen or implied
M1: Must be using ratios; Ignore an equation e.g. \((4 + \lambda)\mathbf{i} + (-1 + \mu)\mathbf{j} = 3\mathbf{i} + \mathbf{j}\) if they go on to use ratios.
However, if they write \(4 + \lambda = 3\) and \(-1 + \mu = 1\) then \(3(-1 + \mu) = 3\) so \(4 + \lambda = 3(-1 + \mu)\) with no use of a constant, it’s M0
They may use the acceleration, with a factor of \(\dfrac{1}{4}\) top and bottom, see alternative
Allow one side of the equation to be inverted
A1: Correct equation
A1*: Given answer correctly obtained. Must see at least one line of working, with the LH fraction ‘removed’.
| Scheme | Marks | AO |
|---|---|---|
| \(\lambda = 2 \Rightarrow \mu = 3\); Resultant force \(= (6\mathbf{i} + 2\mathbf{j})\) (N) | M1 | 3.1a |
| \((6\mathbf{i} + 2\mathbf{j}) = 4\mathbf{a}\) OR \(|(6\mathbf{i} + 2\mathbf{j})| = 4a\) | M1 | 1.1b |
| Use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) with \(\mathbf{u} = \mathbf{0}\), their \(\mathbf{a}\) and \(t = 4\) : Or they may integrate their \(\mathbf{a}\) twice with \(\mathbf{u} = \mathbf{0}\) and put \(t = 4\) : \(\mathbf{r} = \dfrac{1}{2} \times \dfrac{(6\mathbf{i} + 2\mathbf{j})}{4}4^2 = (12\mathbf{i} + 4\mathbf{j})\) | DM1 | 2.1 |
| \(\sqrt{12^2 + 4^2}\) | M1 | 1.1b |
| \(\sqrt{160},\ 2\sqrt{40},\ 4\sqrt{10}\) oe or 13 or better (m) | A1 | 1.1b |
| (5) | ||
| (9 marks) |
Alternative 1 for last two M marks
| Scheme | Marks |
|---|---|
| Use of \(s = ut + \dfrac{1}{2}at^2\), with \(u = 0\), their \(a\) and \(t = 4\) : \(s = \dfrac{1}{2} \times \sqrt{1.5^2 + 0.5^2} \times 4^2\) | DM1 |
| Use of Pythagoras to find mag of \(\mathbf{a}\) : \(a = \sqrt{1.5^2 + 0.5^2}\) | M1 |
Alternative 2 for last two M marks
| Scheme | Marks |
|---|---|
| Use of \(s = ut + \dfrac{1}{2}at^2\), with \(u = 0\), their \(a\) and \(t = 4\) : \(s = \dfrac{1}{2} \times \left(\dfrac{\sqrt{6^2 + 2^2}}{4}\right) \times 4^2\) | DM1 |
| Use of Pythagoras to find \(|(6\mathbf{i} + 2\mathbf{j})|\) : \(= \sqrt{6^2 + 2^2}\) | M1 |
Notes
Accept column vectors throughout
M1: Adding \(\mathbf{F}_1\) and \(\mathbf{F}_2\) to find the resultant force, \(\lambda\) and \(\mu\) must be substituted
N.B. M0 if they use \(\mu = 2\) coming from \(-1 + \mu = 1\) in part (a).
M1: Use of \(\mathbf{F} = 4\mathbf{a}\) Or \(|\mathbf{F}| = 4a\), where \(\mathbf{F}\) is their resultant. (including \(3\mathbf{i} + \mathbf{j}\))
This is an independent mark, so could be earned, for example, if they have subtracted the forces to find the ‘resultant’
N.B. M0 if only using \(\mathbf{F}_1\) or \(\mathbf{F}_2\)
DM1: Dependent on previous M mark for
Either: use of \(\mathbf{r} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) with \(\mathbf{u} = \mathbf{0}\), their \(\mathbf{a}\) and \(t = 4\) to produce a displacement vector
Or : integrate twice, with \(\mathbf{u} = \mathbf{0}\), their \(\mathbf{a}\) and \(t = 4\) to produce a displacement Vector
Or: use of \(s = ut + \dfrac{1}{2}at^2\) with \(u = 0\), their \(a\) and \(t = 4\) to produce a length
M1: Use of Pythagoras, with square root, to find the magnitude of their displacement vector, \(\mathbf{a}\) or \(\mathbf{F}\) (M0 if only using \(\mathbf{F}_1\) or \(\mathbf{F}_2\)) depending on which method they have used.
A1: cao