A2 October 2020 Paper 1 Q10

OCR ACurrent spec13 marksFirst Order Differentials

10 A particle of mass 0.5 kg is initially at point \(O\). It moves from rest along the \(x\)-axis under the influence of two forces \(F_1\) N and \(F_2\) N which act parallel to the \(x\)-axis. At time \(t\) seconds the velocity of the particle is \(v\,\mathrm{m\,s^{-1}}\).
\(F_1\) is acting in the direction of motion of the particle and \(F_2\) is resisting motion.

In an initial model

\(F_1\) is proportional to \(t\) with constant of proportionality \(\lambda \gt 0\),
\(F_2\) is proportional to \(v\) with constant of proportionality \(\mu \gt 0\).

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

(b) Solve the differential equation in part (a), giving the particular solution for \(v\) in terms of \(t\), \(\lambda\) and \(\mu\). [7]

You are now given that \(\lambda = 2\) and \(\mu = 1\).

In a refined model

\(F_1\) is constant, acting in the direction of motion with magnitude 2 N,
\(F_2\) is as before with \(\mu = 1\).

(d) Write down a differential equation for the refined model. [1]

(e) Without solving the differential equation in part (d), write down what will happen to the velocity in the long term according to this refined model. [1]

Mark scheme (a)

Scheme	Marks	AO
\(F = ma\) (\(F_1\) is in the direction of motion and \(F_2\) is resisting motion) \(F = \lambda t - \mu v\)	M1	3.3
\(\Rightarrow \dfrac{1}{2}\dfrac{\mathrm{d}v}{\mathrm{d}t} = \lambda t - \mu v\) AG	A1	2.1
	[2]

Notes

M1: Use of Newton II with constants of proportionality. \(F = ma\) must be seen

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{1}{2}\dfrac{\mathrm{d}v}{\mathrm{d}t} = \lambda t - \mu v\) \(\Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} + 2\mu v = 2\lambda t\)	M1*	1.1a
I.F. \(\mathrm{e}^{2\mu t}\)	A1	1.1
\(\Rightarrow \mathrm{e}^{2\mu t}\dfrac{\mathrm{d}v}{\mathrm{d}t} + \mathrm{e}^{2\mu t}2\mu v = 2\lambda t\mathrm{e}^{2\mu t}\)	M1dep	3.1a
\(\Rightarrow \dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{2\mu t}v\right) = 2\lambda t\mathrm{e}^{2\mu t}\)	A1	1.1
\(\Rightarrow \mathrm{e}^{2\mu t}v = \displaystyle\int 2\lambda t\,\mathrm{e}^{2\mu t}\,\mathrm{d}t\)	M1dep	3.1a
\(\Rightarrow \mathrm{e}^{2\mu t}v = 2\lambda\left(\dfrac{1}{2\mu}t\mathrm{e}^{2\mu t} - \dfrac{1}{4\mu^2}\mathrm{e}^{2\mu t}\right) + c\)	A1	1.1
Given that \(t = 0, v = 0 \Rightarrow c = \dfrac{2\lambda}{4\mu^2}\) \(\Rightarrow v = \dfrac{\lambda}{\mu}t - \dfrac{\lambda}{2\mu^2} + \dfrac{\lambda}{2\mu^2}\mathrm{e}^{-2\mu t}\) oe	A1	3.4
	[7]

Notes

M1*: Attempt to find IF

M1dep: Getting DE in correct form

M1dep: Attempt integration by parts

Alternative method

Scheme	Marks	AO
\(\dfrac{\mathrm{d}v}{\mathrm{d}t} + 2\mu v = 2\lambda t\)	M1	E
\(AE : m + 2\mu = 0\) \(\Rightarrow m = -2\mu \Rightarrow CF = A\mathrm{e}^{-2\mu t}\)	M1 A1	E E
\(PI : v = at + b\)	M1	C
\(\Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} = a \Rightarrow a + 2\mu at + 2\mu b = 2\lambda t\) \(\therefore a = \dfrac{\lambda}{\mu},\ b = -\dfrac{a}{2\mu} = -\dfrac{\lambda}{2\mu^2}\)	M1 A1	C C
\(\therefore GS : v = \dfrac{\lambda}{\mu}t - \dfrac{\lambda}{2\mu^2} + A\mathrm{e}^{-2\mu t}\)	A1	A
	[7]

Mark scheme (c)

Scheme	Marks	AO
\(v = \dfrac{\lambda}{\mu}t - \dfrac{\lambda}{2\mu^2} + \dfrac{\lambda}{2\mu^2}\mathrm{e}^{-2\mu t}\) \(\lambda = 2, \mu = 1 \Rightarrow v = 2t - 1 + \mathrm{e}^{-2t}\) When \(t\) is large, \(\mathrm{e}^{-2t}\) is very small so \(v \approx 2t - 1\)	M1 A1	3.4 3.3
	[2]

Notes

M1: Consider the behaviour of the exponential function in their equation from (b) soi

A1: or \(v \approx 2t\)

Scheme	Marks	AO
\(\dfrac{1}{2}\dfrac{\mathrm{d}v}{\mathrm{d}t} = 2 - v\) oe	B1	3.5c
	[1]