June 2019 Paper 3 Mechanics Q2
2. A particle, \(P\), moves with constant acceleration \((2\mathbf{i} - 3\mathbf{j})\ \text{m s}^{-2}\)
At time \(t = 0\), the particle is at the point \(A\) and is moving with velocity \((-\mathbf{i} + 4\mathbf{j})\ \text{m s}^{-1}\)
At time \(t = T\) seconds, \(P\) is moving in the direction of vector \((3\mathbf{i} - 4\mathbf{j})\)
At time \(t = 4\) seconds, \(P\) is at the point \(B\).
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{v} =)\ \mathbf{C} + (2\mathbf{i} - 3\mathbf{j})t\) | M1 | 3.1a |
| \((\mathbf{v} =)\ (-\mathbf{i} + 4\mathbf{j}) + (2\mathbf{i} - 3\mathbf{j})t\) | A1 | 1.1b |
| \(\dfrac{4 - 3T}{-1 + 2T} = \dfrac{-4}{3}\) oe | M1 | 3.1a |
| \(T = 8\) | A1 | 1.1b |
| (4) |
Notes
M1: Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\)
OR integration to give an expression of the form \(\mathbf{C} + (2\mathbf{i} - 3\mathbf{j})t\), where C is a non-zero constant vector
M0 if \(\mathbf{u}\) and \(\mathbf{a}\) are reversed
Condone use of \(\mathbf{a} = (2\mathbf{i} + 3\mathbf{j})\) for this M mark
A1: Any correct unsimplified expression seen or implied
M1: Correct use of ratios, using a velocity vector (must be using \(\dfrac{-4}{3}\)) to give equation in \(T\) only
M0 if they equate \(4 - 3T = -4\) and/or \(-1 + 2T = 3\) and therefore M0 if they then divide to produce their equation
A1: Correct only
N.B.
(i) Can score the second M1A1 if they get \(T = 8\), using a calculator to solve two simultaneous equations, but if answer is wrong, and no equation in \(T\) only, second M0
(ii) Can score M1A1 M1A1 if they get \(T = 8\), using trial and error, but if they don’t get \(T = 8\), can only score max M1A1M0A0
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{s} =)\ \mathbf{C}t + (2\mathbf{i} - 3\mathbf{j})\dfrac{1}{2}t^2\ (+\,\mathbf{D})\) | M1 | 3.1a |
| \((\mathbf{s} =)\ (-\mathbf{i} + 4\mathbf{j})t + \dfrac{1}{2}(2\mathbf{i} - 3\mathbf{j})t^2\ (+\,\mathbf{D})\) | A1 | 1.1b |
| \(AB = \sqrt{12^2 + 8^2}\) N.B. Beware you may see 4(2i – 3j) which leads to \(\sqrt{(8^2 + 12^2)}\) this is M0A0M0A0. | M1 | 3.1a |
| \(= 4\sqrt{13}\ (= 14.422051....)\) (m) | A1cso | 1.1b |
| (4) | ||
| (8 marks) |
Notes
M1: Use of \(\mathbf{s} = \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\) with \(\mathbf{a} = (2\mathbf{i} - 3\mathbf{j})\)
OR integration to give an expression of the form \(\mathbf{C}t + (2\mathbf{i} - 3\mathbf{j})\dfrac{1}{2}t^2\), where C is their non-zero constant vector from (a)
Condone use of \(\mathbf{a} = (2\mathbf{i} + 3\mathbf{j})\) for this M mark
OR any other complete method using vector suvat equations
A1: Correct unsimplified expression seen or implied
M1: Use of \(t = 4\) in their \(\mathbf{s}\) (which must be a displacement vector) and then Pythagoras with the root sign
N.B. This M mark can be implied by a correct answer, otherwise we need to see Pythagoras used, with the root sign, for the M mark.
A1cso: Any surd form or 14 or better