OCR Further Mechanics 2020 November — Question 4 15 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2020
SessionNovember
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeFind exponents with all unknowns
DifficultyStandard +0.3 This is a structured multi-part question that guides students through dimensional analysis and differential equations. Parts (a)-(c) involve routine dimensional analysis with clear setup, (d) is a standard separable differential equation with the result given, and (e)-(f) require simple limit analysis. While it covers multiple techniques, each step is scaffolded and uses standard A-level methods without requiring novel insight.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions
4 The resistive force, \(F\), on a sphere falling through a viscous fluid is thought to depend on the radius of the sphere, \(r\), the velocity of the sphere, \(v\), and the viscosity of the fluid, \(\eta\). You are given that \(\eta\) is measured in \(\mathrm { N } \mathrm { m } ^ { - 2 } \mathrm {~s}\).
  1. By considering its units, find the dimensions of viscosity. A model of the resistive force suggests the following relationship: \(\mathbf { F } = 6 \pi \eta ^ { \alpha } \mathbf { r } ^ { \beta } \mathbf { v } ^ { \gamma }\).
  2. Explain whether or not it is possible to use dimensional analysis to verify that the constant \(6 \pi\) is correct.
  3. Use dimensional analysis to find the values of \(\alpha , \beta\) and \(\gamma\). A sphere of radius \(r\) and mass \(m\) falls vertically from rest through the fluid. After a time \(t\) its velocity is \(v\).
  4. By setting up and solving a differential equation, show that \(\mathrm { e } ^ { - \mathrm { kt } } = \frac { \mathrm { g } - \mathrm { kv } } { \mathrm { g } }\) where \(\mathrm { k } = \frac { 6 \pi \eta \mathrm { r } } { \mathrm { m } }\). As the time increases, the velocity of the sphere tends towards a limit called the terminal velocity.
  5. Find, in terms of \(g\) and \(k\), the terminal velocity of the sphere. In a sequence of experiments the sphere is allowed to fall through fluids of different viscosity, ranging from small to very large, with all other conditions being constant. The terminal velocity of the sphere through each fluid is measured.
  6. Describe how, according to the model, the terminal velocity of the sphere changes as the viscosity of the fluid through which it falls increases.
Question 4:
AnswerMarks Guidance
4(a) [Force] = MLT–2
[Viscosity] = MLT–2L–2T = ML–1T–1M1
A1
AnswerMarks
[2]1.2
1.1
AnswerMarks Guidance
4(b) No it is not possible since a multiplicative constant
does not affect the dimensions of the quantity it is
AnswerMarks Guidance
multiplying.B1
[1]2.4 Or equivalent statement about
dimensional analysis only
applying to dimensioned
quantities
AnswerMarks Guidance
4(c) MLT–2 = (ML–1T–1)αLβ(LT–1)γ
From M, α = 1
L: 1 = β + γ – α, T: –2 = –α – γ
AnswerMarks
β = 1, γ = 1M1
B1
M1
A1
AnswerMarks
[4]3.3
1.1
1.1
AnswerMarks
1.1Setting up a dimensional
equation (using their [Force]
and [Viscosity])
Using their equation to derive
AnswerMarks
2 equations in β and γMLT–2 = Mα Lβ + γ – αT–α – γ
NB – B1 to allow this from M0
AnswerMarks Guidance
Alternative method 1M1
A1Using F = ma with two forces any differential expression for a
dv
mg−6πηrv=m
dt
dv ∫kdt
g =kv+ so Integrating Factor e =ekt
AnswerMarks Guidance
dtM1 Rearrange and find IF
yet
dv d ( )
gekt =kvekt + ekt = vekt
dt dt
vekt =∫gektdt = g ekt +c
k
g
t = 0, v = 0 => c=−
AnswerMarks Guidance
kM1 Substituting initial values to
find c
( )
vekt = g ekt − g ⇒ekt g −v = g
k k k k
g −v g−kv
⇒e −kt = k =
g g
AnswerMarks Guidance
kA1 AG. Must see some
intermediate working
AnswerMarks Guidance
Alternative method 2M1
A1Using F = ma with two forces any differential expression for a
dv
mg−6πηrv=m
dt
dv
g =kv+
AnswerMarks Guidance
dtM1 Rearrange and find CF and PI
yet
dv dv 1
CF: +kv=0⇒ =−kv⇒∫ dv=−∫kdt
dt dt v
⇒lnv=−kt+c⇒v= Ae −kt
g g
PI: try v=α⇒kα=g⇒α= ,so GS: v= Ae −kt +
k k
g g
t =0, v=0⇒0= A+ ⇒ A=−
AnswerMarks Guidance
k kM1 Substituting initial values to
find c
g g k  g  g−kv
So v=− e −kt + ⇒e −kt =  −v =
AnswerMarks Guidance
k k g k  gA1 AG. Must see some
intermediate working
[5]
M1
A1
Using F = ma with two forces
any differential expression for a
Rearrange and find IF
no need for k to be substituted
yet
Substituting initial values to
find c
AG. Must see some
intermediate working
M1
A1
Using F = ma with two forces
any differential expression for a
Rearrange and find CF and PI
no need for k to be substituted
yet
Substituting initial values to
find c
AG. Must see some
intermediate working
AnswerMarks Guidance
4(e) g−kv g
e −kt = so t → ∞, ⇒v =
T
AnswerMarks Guidance
g kB1 3.4
Alternative methodB1
dv mg g
=0⇒mg−6πηrv =0⇒v = =
dt T T 6πηr k
[1]
AnswerMarks Guidance
4(f) As the viscosity increases the terminal velocity
decreases...
...and as the viscosity tends to infinity the terminal
AnswerMarks
velocity tends to 0B1
B1
AnswerMarks
[2]2.2a
2.2a
Question 4:
4 | (a) | [Force] = MLT–2
[Viscosity] = MLT–2L–2T = ML–1T–1 | M1
A1
[2] | 1.2
1.1
4 | (b) | No it is not possible since a multiplicative constant
does not affect the dimensions of the quantity it is
multiplying. | B1
[1] | 2.4 | Or equivalent statement about
dimensional analysis only
applying to dimensioned
quantities
4 | (c) | MLT–2 = (ML–1T–1)αLβ(LT–1)γ
From M, α = 1
L: 1 = β + γ – α, T: –2 = –α – γ
β = 1, γ = 1 | M1
B1
M1
A1
[4] | 3.3
1.1
1.1
1.1 | Setting up a dimensional
equation (using their [Force]
and [Viscosity])
Using their equation to derive
2 equations in β and γ | MLT–2 = Mα Lβ + γ – αT–α – γ
NB – B1 to allow this from M0
Alternative method 1 | M1
A1 | Using F = ma with two forces | any differential expression for a
dv
mg−6πηrv=m
dt
dv ∫kdt
g =kv+ so Integrating Factor e =ekt
dt | M1 | Rearrange and find IF | no need for k to be substituted
yet
dv d ( )
gekt =kvekt + ekt = vekt
dt dt
vekt =∫gektdt = g ekt +c
k
g
t = 0, v = 0 => c=−
k | M1 | Substituting initial values to
find c
( )
vekt = g ekt − g ⇒ekt g −v = g
k k k k
g −v g−kv
⇒e −kt = k =
g g
k | A1 | AG. Must see some
intermediate working
Alternative method 2 | M1
A1 | Using F = ma with two forces | any differential expression for a
dv
mg−6πηrv=m
dt
dv
g =kv+
dt | M1 | Rearrange and find CF and PI | no need for k to be substituted
yet
dv dv 1
CF: +kv=0⇒ =−kv⇒∫ dv=−∫kdt
dt dt v
⇒lnv=−kt+c⇒v= Ae −kt
g g
PI: try v=α⇒kα=g⇒α= ,so GS: v= Ae −kt +
k k
g g
t =0, v=0⇒0= A+ ⇒ A=−
k k | M1 | Substituting initial values to
find c
g g k  g  g−kv
So v=− e −kt + ⇒e −kt =  −v =
k k g k  g | A1 | AG. Must see some
intermediate working
[5]
M1
A1
Using F = ma with two forces
any differential expression for a
Rearrange and find IF
no need for k to be substituted
yet
Substituting initial values to
find c
AG. Must see some
intermediate working
M1
A1
Using F = ma with two forces
any differential expression for a
Rearrange and find CF and PI
no need for k to be substituted
yet
Substituting initial values to
find c
AG. Must see some
intermediate working
4 | (e) | g−kv g
e −kt = so t → ∞, ⇒v =
T
g k | B1 | 3.4
Alternative method | B1
dv mg g
=0⇒mg−6πηrv =0⇒v = =
dt T T 6πηr k
[1]
4 | (f) | As the viscosity increases the terminal velocity
decreases...
...and as the viscosity tends to infinity the terminal
velocity tends to 0 | B1
B1
[2] | 2.2a
2.2a
4 The resistive force, $F$, on a sphere falling through a viscous fluid is thought to depend on the radius of the sphere, $r$, the velocity of the sphere, $v$, and the viscosity of the fluid, $\eta$. You are given that $\eta$ is measured in $\mathrm { N } \mathrm { m } ^ { - 2 } \mathrm {~s}$.
\begin{enumerate}[label=(\alph*)]
\item By considering its units, find the dimensions of viscosity.

A model of the resistive force suggests the following relationship: $\mathbf { F } = 6 \pi \eta ^ { \alpha } \mathbf { r } ^ { \beta } \mathbf { v } ^ { \gamma }$.
\item Explain whether or not it is possible to use dimensional analysis to verify that the constant $6 \pi$ is correct.
\item Use dimensional analysis to find the values of $\alpha , \beta$ and $\gamma$.

A sphere of radius $r$ and mass $m$ falls vertically from rest through the fluid. After a time $t$ its velocity is $v$.
\item By setting up and solving a differential equation, show that $\mathrm { e } ^ { - \mathrm { kt } } = \frac { \mathrm { g } - \mathrm { kv } } { \mathrm { g } }$ where $\mathrm { k } = \frac { 6 \pi \eta \mathrm { r } } { \mathrm { m } }$.

As the time increases, the velocity of the sphere tends towards a limit called the terminal velocity.
\item Find, in terms of $g$ and $k$, the terminal velocity of the sphere.

In a sequence of experiments the sphere is allowed to fall through fluids of different viscosity, ranging from small to very large, with all other conditions being constant. The terminal velocity of the sphere through each fluid is measured.
\item Describe how, according to the model, the terminal velocity of the sphere changes as the viscosity of the fluid through which it falls increases.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2020 Q4 [15]}}
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