Question 12 - A-Level Maths

OCR MEI Further Mechanics Major 2019 June — Question 12 16 marks

Exam Board	OCR MEI
Module	Further Mechanics Major (Further Mechanics Major)
Year	2019
Session	June
Marks	16
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circular Motion 1
Type	Smooth ring on rotating string
Difficulty	Challenging +1.2 This is a multi-part Further Mechanics question involving circular motion with a conical pendulum setup. Parts (a)-(c) require standard resolution of forces and use of F=mrω² (routine for FM students), while part (d) requires recognizing that when the ring is free to slide, tensions must be equal, leading to a solvable equation. The geometry is given clearly, and the steps follow a predictable pattern for this topic. Slightly above average difficulty due to the multi-stage reasoning and FM content, but well within standard FM Major material.
Spec	3.03m Equilibrium: sum of resolved forces = 0 6.05c Horizontal circles: conical pendulum, banked tracks

\includegraphics{figure_12} The ends of a light inextensible string are fixed to two points A and B in the same vertical line, with A above B. The string passes through a small smooth ring of mass \(m\). The ring is fastened to the string at a point P. When the string is taut the angle APB is a right angle, the angle BAP is \(\theta\) and the perpendicular distance of P from AB is \(r\). The ring moves in a horizontal circle with constant angular velocity \(\omega\) and the string taut as shown in Fig. 12.

By resolving horizontally and vertically, show that the tension in the part of the string BP is \(m(r\omega^2\cos\theta - g\sin\theta)\). [6]
Find a similar expression, in terms of \(r\), \(\omega\), \(m\), \(g\) and \(\theta\), for the tension in the part of the string AP. [2]

It is given that AB = 5a and AP = 4a.

Show that \(16a\omega^2 > 5g\). [3]

The ring is now free to move on the string but remains in the same position on the string as before. The string remains taut and the ring continues to move in a horizontal circle.

Find the period of the motion of the ring, giving your answer in terms of \(a\), \(g\) and \(\pi\). [5]

Show mark scheme Show mark scheme source

Question 12:

Answer	Marks	Guidance
12	(a)	T cosT sinmg

AP BP

T sinT cosmr2

AP BP

T sin2mgsinT cos2mr2sin

BP BP

 

T m r2cosgsin

Answer	Marks
BP	M1*

A1

M1*

A1

M1dep*

A1

Answer	Marks
[6]	3.3

1.1

3.3

1.1

Answer	Marks
2.2a	Resolving vertically – correct number

of terms but allow sin/cos confusion

and sign errors

N2L horizontally – correct number of

terms – accept a for acceleration –

allow sin/cos confusion and sign

errors

Correctly eliminating T from their

AP

two equations

Answer	Marks
AG	Must be mg for the

weight – condone any

clear notation (even AP,

BP) for the tension in

the two strings – no

marks if T is used for

both strings

Or correct method for

solving simultaneous

equations for T

BP

Must show sufficient

working as AG

Answer	Marks	Guidance
12	(b)	T cos2mr2sinT sin2mgcos

AP AP

 

T m r2singcos

Answer	Marks
AP	M1

A1

Answer	Marks
[2]	1.1
1.1	Either substitutes given result into one

of their two equation from (a)

correctly to find T

AP

Correct answer with no working

Answer	Marks
scores both marks	Or re-starts and

eliminates T using

BP

given answer in (a)

Answer	Marks	Guidance
12	(c)	3 4 12

sin ,cos ,r a

5 5 5

12 4 3

T 0 a 2 g0

BP  5 5 5

Answer	Marks
48a2 15g16a2 5g	B1

M1

A1

Answer	Marks
[3]	1.1

3.1b

Answer	Marks
2.2a	Correct r and one of sin or cos correct

Setting T 0and substitute values

BP

Answer	Marks
AG	Allow = 0

Must show sufficient

working as AG

Answer	Marks	Guidance
12	(d)	   

m r2cosgsin m r2singcos

12 4 3 12 3 4

a2   g a2   g

5 5 5 5 5 5

35g 35g

2  or 

12a 12a

2

Period =



12a

2

Answer	Marks
35g	M1*

M1dep*

A1

M1dep*

A1

Answer	Marks
[5]	3.1b

3.4

1.1

1.2

Answer	Marks
1.1	Setting T T

AP BP

Substituting values and obtain an

equation in terms of a, g and 

Could be implied in their working

Dependent on both previous M marks

Answer	Marks
oe	

Condone use of for

the angular speed

Question 12:
12 | (a) | T cosT sinmg
AP BP
T sinT cosmr2
AP BP
T sin2mgsinT cos2mr2sin
BP BP
 
T m r2cosgsin
BP | M1*
A1
M1*
A1
M1dep*
A1
[6] | 3.3
1.1
3.3
1.1
1.1
2.2a | Resolving vertically – correct number
of terms but allow sin/cos confusion
and sign errors
N2L horizontally – correct number of
terms – accept a for acceleration –
allow sin/cos confusion and sign
errors
Correctly eliminating T from their
AP
two equations
AG | Must be mg for the
weight – condone any
clear notation (even AP,
BP) for the tension in
the two strings – no
marks if T is used for
both strings
Or correct method for
solving simultaneous
equations for T
BP
Must show sufficient
working as AG
12 | (b) | T cos2mr2sinT sin2mgcos
AP AP
 
T m r2singcos
AP | M1
A1
[2] | 1.1
1.1 | Either substitutes given result into one
of their two equation from (a)
correctly to find T
AP
Correct answer with no working
scores both marks | Or re-starts and
eliminates T using
BP
given answer in (a)
12 | (c) | 3 4 12
sin ,cos ,r a
5 5 5
12 4 3
T 0 a 2 g0
BP  5 5 5
48a2 15g16a2 5g | B1
M1
A1
[3] | 1.1
3.1b
2.2a | Correct r and one of sin or cos correct
Setting T 0and substitute values
BP
AG | Allow = 0
Must show sufficient
working as AG
12 | (d) |    
m r2cosgsin m r2singcos
12 4 3 12 3 4
a2   g a2   g
5 5 5 5 5 5
35g 35g
2  or 
12a 12a
2
Period =

12a
2
35g | M1*
M1dep*
A1
M1dep*
A1
[5] | 3.1b
3.4
1.1
1.2
1.1 | Setting T T
AP BP
Substituting values and obtain an
equation in terms of a, g and 
Could be implied in their working
Dependent on both previous M marks
oe | 
Condone use of for
the angular speed

Show LaTeX source

\includegraphics{figure_12}

The ends of a light inextensible string are fixed to two points A and B in the same vertical line, with A above B. The string passes through a small smooth ring of mass $m$. The ring is fastened to the string at a point P.

When the string is taut the angle APB is a right angle, the angle BAP is $\theta$ and the perpendicular distance of P from AB is $r$.

The ring moves in a horizontal circle with constant angular velocity $\omega$ and the string taut as shown in Fig. 12.

\begin{enumerate}[label=(\alph*)]
\item By resolving horizontally and vertically, show that the tension in the part of the string BP is
$m(r\omega^2\cos\theta - g\sin\theta)$. [6]

\item Find a similar expression, in terms of $r$, $\omega$, $m$, $g$ and $\theta$, for the tension in the part of the string AP. [2]
\end{enumerate}

It is given that AB = 5a and AP = 4a.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that $16a\omega^2 > 5g$. [3]
\end{enumerate}

The ring is now free to move on the string but remains in the same position on the string as before. The string remains taut and the ring continues to move in a horizontal circle.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the period of the motion of the ring, giving your answer in terms of $a$, $g$ and $\pi$. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2019 Q12 [16]}}

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