A2 June 2019 Paper 2 Q5

OCR ACurrent spec11 marksFirst Order Differentials

5 A particle of mass 2 kg moves along the \(x\)-axis. At time \(t\) seconds the velocity of the particle is \(v\,\mathrm{m\,s^{-1}}\).

The particle is subject to two forces.

One acts in the positive \(x\)-direction with magnitude \(\frac{1}{2}t\,\mathrm{N}\).
One acts in the negative \(x\)-direction with magnitude \(v\,\mathrm{N}\).

(a) Show that the motion of the particle can be modelled by the differential equation \[\frac{\mathrm{d}v}{\mathrm{d}t} + \frac{1}{2}v = \frac{1}{4}t.\] [1]

The particle is at rest when \(t = 0\).

(b) Find \(v\) in terms of \(t\). [5]

When \(t = 2\) the force acting in the positive \(x\)-direction is replaced by a constant force of magnitude \(\frac{1}{2}\,\mathrm{N}\) in the same direction.

(d) Refine the differential equation given in part (a) to model the motion for \(t \geqslant 2\). [1]

(e) Use the refined model from part (d) to find an exact expression for \(v\) in terms of \(t\) for \(t \geqslant 2\). [3]

Mark scheme (a)

Scheme	Marks	AO
\(F = ma \Rightarrow \dfrac{1}{2}t - v = 2\dfrac{\mathrm{d}v}{\mathrm{d}t} \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{1}{2}v = \dfrac{1}{4}t\)	B1	3.3
	[1]

Notes

B1: AG. Intermediate step as shown or both sides halved

Mark scheme (b)

Scheme	Marks	AO
IF \(= \mathrm{e}^{\int \frac{1}{2}\mathrm{d}t} = \mathrm{e}^{\frac{1}{2}t}\)	B1	1.1
\(\mathrm{e}^{\frac{1}{2}t}\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{1}{2}\mathrm{e}^{\frac{1}{2}t}v = \dfrac{\mathrm{d}}{\mathrm{d}t}\left(v\mathrm{e}^{\frac{1}{2}t}\right) = \dfrac{1}{4}t\mathrm{e}^{\frac{1}{2}t}\)	M1	1.1
\(\displaystyle v\mathrm{e}^{\frac{1}{2}t} = \frac{1}{2}t\mathrm{e}^{\frac{1}{2}t} - \frac{1}{2}\int \mathrm{e}^{\frac{1}{2}t}\,\mathrm{d}t = \frac{1}{2}t\mathrm{e}^{\frac{1}{2}t} - \mathrm{e}^{\frac{1}{2}t}\ (+c)\)	A1	1.1
\(0 = -1 + c \Rightarrow c = 1\)	M1	3.4
\(v = \dfrac{1}{2}t - 1 + \mathrm{e}^{-\frac{1}{2}t}\)	A1	1.1
	[5]

Notes

B1: Or \(\mathrm{e}^{\frac{1}{2}t + c}\) or \(A\mathrm{e}^{\frac{1}{2}t}\)

M1: Multiplying both sides by IF and writing new LHS as an exact derivative

M1: Use initial conditions to determine \(c\). Allow if using a solution of correct form with wrong (non-zero) coefficients

Alternate method

Scheme	Marks
AE: \(\lambda + \frac{1}{2} = 0,\ \lambda = -\frac{1}{2}\) CF: \((v =)A\mathrm{e}^{-\frac{1}{2}t}\)	B1
Trial function: \(v = at + b,\ \frac{\mathrm{d}v}{\mathrm{d}t} = a\) \(a + \dfrac{1}{2}(at + b) = \dfrac{1}{4}t\) \(a = \dfrac{1}{2},\ b = -1\)	M1 A1
GS: \((v =)A\mathrm{e}^{-\frac{1}{2}t} + \frac{1}{2}t - 1\) \(v = 0, t = 0\) gives \(A = 1\)	M1
\(v = \mathrm{e}^{-\frac{1}{2}t} + \dfrac{1}{2}t - 1\)	A1
	[5]

M1: Substitutes boundary condition into general solution to find \(A\). Allow if using a solution of correct form with wrong (non-zero) coefficients

Scheme	Marks	AO
\(t = 2 \Rightarrow v = \mathrm{e}^{-1}\) (so velocity is \(\mathrm{e}^{-1}\,\mathrm{m\,s^{-1}}\))	B1	3.4
	[1]

Scheme	Marks	AO
change the RHS of DE in part (a) from \(\frac{1}{4}t\) to \(\frac{1}{4}\). oe	B1	3.5c
	[1]

Mark scheme (e)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(v\mathrm{e}^{\frac{1}{2}t}\right) = \dfrac{1}{4}\mathrm{e}^{\frac{1}{2}t} \Rightarrow \displaystyle v\mathrm{e}^{\frac{1}{2}t} = \int \frac{1}{4}\mathrm{e}^{\frac{1}{2}t}\,\mathrm{d}t\) \(v\mathrm{e}^{\frac{1}{2}t} = \dfrac{1}{2}\mathrm{e}^{\frac{1}{2}t} + c\). oe	B1	2.2a
\(\mathrm{e}^{-1}\mathrm{e}^{1} = \dfrac{1}{2}\mathrm{e}^{1} + c \Rightarrow c = 1 - \dfrac{1}{2}\mathrm{e}\)	M1	3.3
\(v = \dfrac{1}{2} + \left(1 - \dfrac{1}{2}\mathrm{e}\right)\mathrm{e}^{-\frac{1}{2}t}\)	A1	1.1
	[3]

Notes

SC1 \(v\mathrm{e}^{\frac{1}{2}t} = \frac{N}{2}\mathrm{e}^{\frac{1}{2}t} + c\) oe

M1: substitute their boundary condition from (c) into correct GS to find \(c\)

Or \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{1}{2}v = \dfrac{1}{4}\) rearranged to \(\displaystyle\int \frac{2\,\mathrm{d}v}{1 - 2v} = \frac{1}{2}\int \mathrm{d}t = \frac{1}{2}t\) giving \(-\ln(1 - 2v) + c' = \dfrac{1}{2}t\); \(c' = 1 + \ln(1 - 2\mathrm{e}^{-1})\)