Question 4 - A-Level Maths

OCR Further Mechanics 2022 June — Question 4 8 marks

Exam Board	OCR
Module	Further Mechanics (Further Mechanics)
Year	2022
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Dimensional Analysis
Type	Find exponents with all unknowns
Difficulty	Standard +0.3 This is a standard dimensional analysis question with scaffolded parts. Part (a) requires understanding Newton's third law symmetry, parts (b-c) involve routine dimensional analysis to find exponents by equating dimensions, and part (d) applies the result to circular motion. While it requires multiple techniques, each step follows a well-established procedure taught in Further Mechanics courses. The scaffolding and clear structure make it slightly easier than average.
Spec	6.01a Dimensions: M, L, T notation 6.01b Units vs dimensions: relationship 6.01c Dimensional analysis: error checking 6.01d Unknown indices: using dimensions

4 When two objects are placed a distance apart in outer space each applies a gravitational force to the other. It is suggested that the magnitude of this force depends on the masses of both objects and the distance between them. Assuming that this suggestion is correct, it is further assumed that the magnitude of this force is given by a relationship of the form $$\mathrm { F } = \mathrm { Gm } _ { 1 } ^ { \alpha } \mathrm { m } _ { 2 } ^ { \beta } \mathrm { r } ^ { \gamma }$$ where

$F$ is the magnitude of the force
$m _ { 1 }$ and $m _ { 2 }$ are the masses of the two objects
$r$ is the distance between the two objects
$G$ is a constant.
1. Using a dimensional argument based on Newton's third law explain why $\alpha = \beta$.

It is given that the magnitude of the gravitational force is given by such a relationship and that $G = 6.67 \times 10 ^ { - 11 } \mathrm {~m} ^ { 3 } \mathrm {~kg} ^ { - 1 } \mathrm {~s} ^ { - 2 }$.

Write down the dimensions of $G$.

By using dimensional analysis, determine the values of $\alpha , \beta$ and $\gamma$. You are given that the mass of the Earth is $5.97 \times 10 ^ { 24 } \mathrm {~kg}$ and that the distance of the Moon from the Earth is $3.84 \times 10 ^ { 8 } \mathrm {~m}$. You may assume that the only force acting on the Moon is the gravitational force due to the Earth.

By modelling the Earth as stationary and assuming that the Moon moves in a circular orbit around the Earth, determine the period of the motion of the Moon. Give your answer to the nearest day.

Show mark scheme Show mark scheme source

Question 4:

Answer	Marks	Guidance
4	(a)	m and m have the same dimensions and NIII

1 2

says that the magnitude of the force that each

applies to the other is the same. So m and m

1 2

must be interchangeable which means that the

dependency on each must be the same and

Answer	Marks	Guidance
hence  = .	B1
[1]	2.4	Argument must be based on the

idea that the dimensions/units of

the quantities are equal and that

Answer	Marks
they are interchangeable	Must reference at least 2 of the 3

bold statements

Answer	Marks	Guidance
(b)	M–1L3T–2	B1
[1]	1.1	cao but condone change in order
of M, L and/or T	Don’t accept square brackets
(c)	[F] = MLT–2, [m] = M and [r] = L =>

i

M: 1 = –1 + 2

L: 1 = 3 + 

Answer	Marks
=>  =  = 1 and  = –2	M1

M1

A1

Answer	Marks
[3]	1.1

1.1

Answer	Marks
1.1	Setting up equation in  (and/or

) by considering mass

dimension

Ignore attempt to set up equation

involving time dimension

Setting up equation in  by

considering length dimension

Ignore attempt to set up equation

involving time dimension

or F = Gm m r–2

Answer	Marks
1 2	or 1 = –1 +  + 
(d)	Gm m r–2 = m r2

1 2 2

 = (6.6710–11  5.971024/(3.84108)3)

=>  = 2.65 10–6 = 2 / T

=> T = awrt 2370000 so period is

2370000/(606024) = 27 days to the nearest

Answer	Marks
day	M1*

M1dep*

A1

Answer	Marks
[3]	3.3

3.4

Answer	Marks
1.1	Using NII with a correct form

for the centripetal acceleration

2𝜋𝑟 2𝜋

Using 𝑣 = or 𝜔 = with

𝑇 𝑇

Answer	Marks
values substituted in	4𝜋2(3.84×108)3

𝑇2 =

(6.67×10−11)(5.97×1024)

Question 4:
4 | (a) | m and m have the same dimensions and NIII
1 2
says that the magnitude of the force that each
applies to the other is the same. So m and m
1 2
must be interchangeable which means that the
dependency on each must be the same and
hence  = . | B1
[1] | 2.4 | Argument must be based on the
idea that the dimensions/units of
the quantities are equal and that
they are interchangeable | Must reference at least 2 of the 3
bold statements
(b) | M–1L3T–2 | B1
[1] | 1.1 | cao but condone change in order
of M, L and/or T | Don’t accept square brackets
(c) | [F] = MLT–2, [m] = M and [r] = L =>
i
M: 1 = –1 + 2
L: 1 = 3 + 
=>  =  = 1 and  = –2 | M1
M1
A1
[3] | 1.1
1.1
1.1 | Setting up equation in  (and/or
) by considering mass
dimension
Ignore attempt to set up equation
involving time dimension
Setting up equation in  by
considering length dimension
Ignore attempt to set up equation
involving time dimension
or F = Gm m r–2
1 2 | or 1 = –1 +  + 
(d) | Gm m r–2 = m r2
1 2 2
 = (6.6710–11  5.971024/(3.84108)3)
=>  = 2.65 10–6 = 2 / T
=> T = awrt 2370000 so period is
2370000/(606024) = 27 days to the nearest
day | M1*
M1dep*
A1
[3] | 3.3
3.4
1.1 | Using NII with a correct form
for the centripetal acceleration
2𝜋𝑟 2𝜋
Using 𝑣 = or 𝜔 = with
𝑇 𝑇
values substituted in | 4𝜋2(3.84×108)3
𝑇2 =
(6.67×10−11)(5.97×1024)

Show LaTeX source

4 When two objects are placed a distance apart in outer space each applies a gravitational force to the other. It is suggested that the magnitude of this force depends on the masses of both objects and the distance between them. Assuming that this suggestion is correct, it is further assumed that the magnitude of this force is given by a relationship of the form

$$\mathrm { F } = \mathrm { Gm } _ { 1 } ^ { \alpha } \mathrm { m } _ { 2 } ^ { \beta } \mathrm { r } ^ { \gamma }$$

where

\begin{itemize}
  \item $F$ is the magnitude of the force
  \item $m _ { 1 }$ and $m _ { 2 }$ are the masses of the two objects
  \item $r$ is the distance between the two objects
  \item $G$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Using a dimensional argument based on Newton's third law explain why $\alpha = \beta$.
\end{itemize}

It is given that the magnitude of the gravitational force is given by such a relationship and that $G = 6.67 \times 10 ^ { - 11 } \mathrm {~m} ^ { 3 } \mathrm {~kg} ^ { - 1 } \mathrm {~s} ^ { - 2 }$.
\item Write down the dimensions of $G$.
\item By using dimensional analysis, determine the values of $\alpha , \beta$ and $\gamma$.

You are given that the mass of the Earth is $5.97 \times 10 ^ { 24 } \mathrm {~kg}$ and that the distance of the Moon from the Earth is $3.84 \times 10 ^ { 8 } \mathrm {~m}$. You may assume that the only force acting on the Moon is the gravitational force due to the Earth.
\item By modelling the Earth as stationary and assuming that the Moon moves in a circular orbit around the Earth, determine the period of the motion of the Moon. Give your answer to the nearest day.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2022 Q4 [8]}}

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