Question 9 - A-Level Maths

OCR MEI Further Mechanics Major 2019 June — Question 9 12 marks

Exam Board	OCR MEI
Module	Further Mechanics Major (Further Mechanics Major)
Year	2019
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circular Motion 2
Type	Vertical circle: tension at specific point
Difficulty	Challenging +1.2 This is a standard Further Maths circular motion problem requiring energy conservation and Newton's second law in the radial direction, followed by small angle approximation for SHM. The derivations are systematic with clear guidance on the form of answers, though it requires competent handling of multiple techniques across 4 parts totaling 12 marks.
Spec	1.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x 4.10f Simple harmonic motion: x'' = -omega^2 x 6.02i Conservation of energy: mechanical energy principle 6.05e Radial/tangential acceleration

\includegraphics{figure_9} A particle P of mass $m$ is joined to a fixed point O by a light inextensible string of length $l$. P is released from rest with the string taut and making an acute angle $\alpha$ with the downward vertical, as shown in Fig. 9. At a time $t$ after P is released the string makes an angle $\theta$ with the downward vertical and the tension in the string is $T$. Angles $\alpha$ and $\theta$ are measured in radians.

Show that $$\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 = \frac{2g}{l}\cos\theta + k_1,$$ where $k_1$ is a constant to be determined in terms of $g$, $l$ and $\alpha$. [4]
Show that $$T = 3mg\cos\theta + k_2,$$ where $k_2$ is a constant to be determined in terms of $m$, $g$ and $\alpha$. [3]

It is given that $\alpha$ is small enough for $\alpha^2$ to be negligible.

Find, in terms of $m$ and $g$, the approximate tension in the string. [2]
Show that the motion of P is approximately simple harmonic. [3]

Show mark scheme Show mark scheme source

Question 9:

Answer	Marks	Guidance
9	(b)	T mgcosml2

2g 

T mgcosml coscos

 

 l 

Answer	Marks
T 3mgcos2mgcos	M1*

M1dep*

A1

Answer	Marks
[3]	3.3

3.4

Answer	Marks
1.1	Applying N2L radially – correct

number of terms – allow r for l

M1 for substituting their v2 or 2

consistently – must be using l not r

k 2mgcos

Answer	Marks
2	For acceleration allow

v2

or a

l

But M0 if in terms of k

1

Answer	Marks	Guidance
9	(c)	coscos1
T mg	M1

A1

Answer	Marks
[2]	3.1a
2.2b	Setting cosand cosequal to 1 in

their expression for T

Answer	Marks	Guidance
9	(d)	dd2 2g d

2  dt      dt2     l sin   dt  

d2 g

2 2 

dt2 l

d2 2 gl1

2 where so motion is

dt2

Answer	Marks
approximately simple harmonic	M1*

M1dep*

A1

Answer	Marks
[3]	3.4

2.3

Answer	Marks
2.4	d

Differentiate 2 with respect to t

dt

or must appear on both sides before

cancelling

Cancel  terms and use small angle

approximation for sin

Answer	Marks
Must state that this is simple harmonic	Or use N2L tangentially

(correct number of

terms) e.g.

d2

mgsinml -

dt2

must be using 

Question 9:
9 | (b) | T mgcosml2
2g 
T mgcosml coscos
 
 l 
T 3mgcos2mgcos | M1*
M1dep*
A1
[3] | 3.3
3.4
1.1 | Applying N2L radially – correct
number of terms – allow r for l
M1 for substituting their v2 or 2
consistently – must be using l not r
k 2mgcos
2 | For acceleration allow
v2
or a
l
But M0 if in terms of k
1
9 | (c) | coscos1
T mg | M1
A1
[2] | 3.1a
2.2b | Setting cosand cosequal to 1 in
their expression for T
9 | (d) | dd2 2g d
2  dt      dt2     l sin   dt  
d2 g
2 2 
dt2 l
d2 2 gl1
2 where so motion is
dt2
approximately simple harmonic | M1*
M1dep*
A1
[3] | 3.4
2.3
2.4 | d
Differentiate 2 with respect to t
dt
or must appear on both sides before
cancelling
Cancel  terms and use small angle
approximation for sin
Must state that this is simple harmonic | Or use N2L tangentially
(correct number of
terms) e.g.
d2
mgsinml -
dt2
must be using 

Show LaTeX source

\includegraphics{figure_9}

A particle P of mass $m$ is joined to a fixed point O by a light inextensible string of length $l$. P is released from rest with the string taut and making an acute angle $\alpha$ with the downward vertical, as shown in Fig. 9.

At a time $t$ after P is released the string makes an angle $\theta$ with the downward vertical and the tension in the string is $T$. Angles $\alpha$ and $\theta$ are measured in radians.

\begin{enumerate}[label=(\alph*)]
\item Show that
$$\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 = \frac{2g}{l}\cos\theta + k_1,$$
where $k_1$ is a constant to be determined in terms of $g$, $l$ and $\alpha$. [4]

\item Show that
$$T = 3mg\cos\theta + k_2,$$
where $k_2$ is a constant to be determined in terms of $m$, $g$ and $\alpha$. [3]
\end{enumerate}

It is given that $\alpha$ is small enough for $\alpha^2$ to be negligible.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find, in terms of $m$ and $g$, the approximate tension in the string. [2]
\item Show that the motion of P is approximately simple harmonic. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2019 Q9 [12]}}

This paper (13 questions)

View full paper

Q1 5 Q2 4 Q3 5 Q4 6 Q5 7 Q6 7 Q7 8 Q8 11 Q9 12 Q10 8 Q11 14 Q12 16 Q13 17