M5 June 2017 Q2

EdexcelOld spec10 marksVariable Force & Kinematics

2. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane and \(\mathbf{k}\) is a unit vector vertically upwards.]

A particle of mass 2 kg moves under the action of a constant gravitational force \(-19.6\mathbf{k}\) N. The particle is subject to a resistive force \(-\mathbf{v}\) newtons, where \(\mathbf{v}\) m s\(^{-1}\) is the velocity of the particle at time \(t\) seconds.

(a) By writing down an equation of motion of the particle, show that \(\mathbf{v}\) satisfies the differential equation \[\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} + 0.5\mathbf{v} = -9.8\mathbf{k}\] (2)

When \(t = 0\), \(\mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 11.6\mathbf{k})\)

(b) Find \(\mathbf{v}\) when \(t = \ln 4\) (8)

Mark scheme (a)

Scheme	Marks
\(2\dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = -19.6\mathbf{k} - \mathbf{v}\)	M1
\(\dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} + 0.5\mathbf{v} = -9.8\mathbf{k}\) PRINTED ANSWER	A1
	(2)

Notes

M1 for use of \(\mathbf{F} = m\mathbf{a}\) vertically with usual rules

A1 for PRINTED ANSWER

Mark scheme (b)

Scheme	Marks
Aux Eqn: \(m + 0.5 = 0\)	M1
\(\mathbf{v} = \mathbf{A}\mathrm{e}^{-0.5t}\) CF	A1
\(\mathbf{v} = -19.6\mathbf{k}\) PI	B1
\(\mathbf{v} = \mathbf{A}\mathrm{e}^{-0.5t} - 19.6\mathbf{k}\) GS	M1
\(t = 0,\ \mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 11.6\mathbf{k}) \Rightarrow \mathbf{A} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})\)	M1
\(\mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})\mathrm{e}^{-0.5t} - 19.6\mathbf{k}\)	A1
\(t = \ln 4,\ \mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})\dfrac{1}{2} - 19.6\mathbf{k}\)	M1
\(= (2\mathbf{i} - 3\mathbf{j} - 4\mathbf{k})\)	A1
	(8)
	(10 marks)

Notes

First M1 for forming an auxiliary equation

First A1 for a correct CF

B1 for a correct PI

Second M1 for \(\mathbf{v}\) = their CF + their PI

Third M1 for use of conditions to find the constant

Second A1 for a correct particular solution

Fourth M1 for putting \(t = \ln 4\) into their solution

Third A1 for correct answer

ALTERNATIVE using Integrating Factor

First M1 for multiplying through by an IF and integrating LHS

First A1 for \(\mathbf{v}\mathrm{e}^{0.5t} = \displaystyle\int -9.8\mathrm{e}^{0.5t}\,\mathrm{d}t\)

B1 for a correct IF seen

Second M1 for \(\mathbf{v} = -19.6\mathbf{k} + \mathbf{A}\mathrm{e}^{-0.5t}\) (Must have a constant vector)

Third M1 for use of conditions to find the constant

Second A1 for a correct particular solution

Fourth M1 for putting \(t = \ln 4\) into their solution

Third A1 for correct answer