M1 June 2014 Q5

5. A particle \(P\) of mass 0.5 kg is moving under the action of a single force \((3\mathbf{i} - 2\mathbf{j})\) N.

(a) Show that the magnitude of the acceleration of \(P\) is \(2\sqrt{13}\) m s\(^{-2}\). (4)

At time \(t = 0\), the velocity of \(P\) is \((\mathbf{i} + 3\mathbf{j})\) m s\(^{-1}\).

(b) Find the velocity of \(P\) at time \(t = 2\) seconds. (3)

Another particle \(Q\) moves with constant velocity \(\mathbf{v} = (2\mathbf{i} - \mathbf{j})\) m s\(^{-1}\).

(d) Show that at time \(t = 3.5\) seconds both particles are moving in the same direction. (3)

Mark scheme (a)

Scheme	Marks
\(\mathbf{F} = m\mathbf{a}\): \(3\mathbf{i} - 2\mathbf{j} = 0.5\mathbf{a}\)	M1
\(\mathbf{a} = 6\mathbf{i} - 4\mathbf{j}\)	A1
\(\|\mathbf{a}\| = \sqrt{6^2 + (-4)^2} = 2\sqrt{13}\) (m s\(^{-2}\)) **	M1A1
	(4)

Either:

First M1 for use of \(\mathbf{F} = m\,\mathbf{a}\)

First A1 for \(\mathbf{a} = 6\mathbf{i} - 4\mathbf{j}\)

Second M1 for \(a = \sqrt{(6^2 + (-4)^2)}\) (Allow \(\sqrt{(6^2 + 4^2)}\))

Second A1 for \(a = 2\sqrt{13}\) (ms\(^{-2}\)) Given answer

Or:

First M1 for \(F = \sqrt{(3^2 + (-2)^2)}\) (Allow \(\sqrt{(3^2 + 2^2)}\))

First A1 \(F = \sqrt{13}\)

Second M1 for \(\sqrt{13} = 0.5a\)

Second A1 for \(a = 2\sqrt{13}\) (ms\(^{-2}\)) Given answer

Mark scheme (b)

Scheme	Marks
\(\mathbf{v} = \mathbf{u} + \mathbf{a}t\): \(\mathbf{v} = (\mathbf{i} + 3\mathbf{j}) + 2(6\mathbf{i} - 4\mathbf{j})\)	M1A1 ft
\(= 13\mathbf{i} - 5\mathbf{j}\) m s\(^{-1}\)	A1
	(3)

M1 for \((\mathbf{i} + 3\mathbf{j}) + (2 \times\) their \(\mathbf{a})\)

First A1 ft for a correct expression

Second A1 for \(13\mathbf{i} - 5\mathbf{j}\); isw if they go on to find the speed

Mark scheme (c)

Scheme	Marks
Distance \(= 2\|\mathbf{v}\| = 2\sqrt{4 + 1} = 2\sqrt{5} = 4.47\) (m)	M1A1
	(2)

M1 for \(2\sqrt{(2^2 + (-1)^2)}\) or \(\sqrt{(4^2 + (-2)^2)}\)

A1 for \(2\sqrt{5}\) or \(\sqrt{20}\) or 4.5 or 4.47 or better

Mark scheme (d)

Scheme	Marks
When \(t = 3.5\), velocity of \(P\) is \((\mathbf{i} + 3\mathbf{j}) + 3.5(6\mathbf{i} - 4\mathbf{j}) = 22\mathbf{i} - 11\mathbf{j}\)	M1A1 ft
Given conclusion reached correctly. E.g. \(22\mathbf{i} - 11\mathbf{j} = 11(2\mathbf{i} - \mathbf{j})\)	A1
	(3)
	(12 marks)

M1 for \((\mathbf{i} + 3\mathbf{j}) + (3.5 \times\) their \(\mathbf{a})\), or possibly, their (b) + \((1.5 \times\) their \(\mathbf{a})\)

First A1 ft for a correct expression of form \(a\mathbf{i} + b\mathbf{j}\)

Second A1 for given conclusion reached correctly e.g. \(22\mathbf{i} - 11\mathbf{j} = 11(2\mathbf{i} - \mathbf{j})\) oe Given answer