OCR Further Mechanics 2023 June — Question 1 8 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle: speed at specific point
DifficultyStandard +0.3 This is a standard circular motion problem requiring energy conservation (part a), centripetal acceleration formula (part b), and resolving weight tangentially (part c). All three parts use routine Further Maths mechanics techniques with straightforward application of formulas. The 'show that' in part (a) removes problem-solving difficulty, and the multi-part structure guides students through the solution methodically.
Spec6.02i Conservation of energy: mechanical energy principle6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration
One end of a light inextensible string of length \(0.8\) m is attached to a particle \(P\) of mass \(m\) kg. The other end of the string is attached to a fixed point \(O\). Initially \(P\) hangs in equilibrium vertically below \(O\). It is then projected horizontally with a speed of \(5.3\) m s\(^{-1}\) so that it moves in a vertical circular path with centre \(O\) (see diagram). \includegraphics{figure_1} At a certain instant, \(P\) first reaches the point where the string makes an angle of \(\frac{1}{3}\pi\) radians with the downward vertical through \(O\).
  1. Show that at this instant the speed of \(P\) is \(4.5\) m s\(^{-1}\). [3]
  2. Find the magnitude and direction of the radial acceleration of \(P\) at this instant. [3]
  3. Find the magnitude of the tangential acceleration of \(P\) at this instant. [2]
Question 1:
AnswerMarks Guidance
1(a) Initial KE = ½m5.32
energy equation after cancelling
the m’s
Energy at A = ½mv2 + mg0.8(1 – cos(π/3))
AnswerMarks Guidance
= ½m5.32M1 1.1
CoE: equating their energy (KE
+ PE) at the bottom with their
energy at A (KE + PE). At least
one PE must be non-zero.
Condone an incorrect height for
M1, but do not award A0. Allow
AnswerMarks
sin/ cos interchangeFor origin taken as 0 of PE
½mv2 - mg0.8cos(π/3))
= ½m5.32 -0.8mg
v2 = 5.32 – 20.8g(1 – 0.5) = 20.25
AnswerMarks Guidance
=> v = 20.25 = 4.5 so 4.5 ms–1 as requiredA1 1.1
attempting to solve for v2 or v
AnswerMarks
AGMust come from a correct
derivation of h
[3]
AnswerMarks
(b)4 .5 2
a =
AnswerMarks Guidance
r 0 .8M1 1.1
Use of a =
r
AnswerMarks Guidance
awrt 25.3 ms–2...A1 1.1
... towards OB1 1.2
that it is towards the centre,
which can rely on a clearly stated
angle. Allow “radially inwards”
AnswerMarks
or “centripetally”𝜋
Do not allow above horizontal
6
unless it specifies negative
horizontal axis. Do not allow
‘inwards’ on it’s own.
[3]
AnswerMarks
(c)𝜋
(−)𝑚𝑔𝑠𝑖𝑛 = 𝑚𝑎
𝑡
AnswerMarks Guidance
3M1 1.1
F = ma in the tangential
AnswerMarks
direction. Condone missing sign.Allow direct substitution into
𝜋
𝑔𝑠𝑖𝑛 = 𝑎
𝑡
3
So magnitude of tangential acceleration is awrt
AnswerMarks Guidance
8.49 ms–2...A1 1.1
4 .9 3 . or
AnswerMarks
2Do not allow -8.49 for final
magnitude
[2]
Question 1:
1 | (a) | Initial KE = ½m5.32 | B1 | 1.1 | 14.045m | Award if substitution for v seen in
energy equation after cancelling
the m’s
Energy at A = ½mv2 + mg0.8(1 – cos(π/3))
= ½m5.32 | M1 | 1.1 | DR
CoE: equating their energy (KE
+ PE) at the bottom with their
energy at A (KE + PE). At least
one PE must be non-zero.
Condone an incorrect height for
M1, but do not award A0. Allow
sin/ cos interchange | For origin taken as 0 of PE
½mv2 - mg0.8cos(π/3))
= ½m5.32 -0.8mg
v2 = 5.32 – 20.8g(1 – 0.5) = 20.25
=> v = 20.25 = 4.5 so 4.5 ms–1 as required | A1 | 1.1 | CoE: equating their energies and
attempting to solve for v2 or v
AG | Must come from a correct
derivation of h
[3]
(b) | 4 .5 2
a =
r 0 .8 | M1 | 1.1 | 2 v
Use of a =
r
awrt 25.3 ms–2... | A1 | 1.1 | Allow 405/16
... towards O | B1 | 1.2 | Need to have a clear indication
that it is towards the centre,
which can rely on a clearly stated
angle. Allow “radially inwards”
or “centripetally” | 𝜋
Do not allow above horizontal
6
unless it specifies negative
horizontal axis. Do not allow
‘inwards’ on it’s own.
[3]
(c) | 𝜋
(−)𝑚𝑔𝑠𝑖𝑛 = 𝑚𝑎
𝑡
3 | M1 | 1.1 | Resolving weight and using
F = ma in the tangential
direction. Condone missing sign. | Allow direct substitution into
𝜋
𝑔𝑠𝑖𝑛 = 𝑎
𝑡
3
So magnitude of tangential acceleration is awrt
8.49 ms–2... | A1 | 1.1 | 𝑔√3
4 .9 3 . or
2 | Do not allow -8.49 for final
magnitude
[2]
One end of a light inextensible string of length $0.8$ m is attached to a particle $P$ of mass $m$ kg. The other end of the string is attached to a fixed point $O$. Initially $P$ hangs in equilibrium vertically below $O$. It is then projected horizontally with a speed of $5.3$ m s$^{-1}$ so that it moves in a vertical circular path with centre $O$ (see diagram).

\includegraphics{figure_1}

At a certain instant, $P$ first reaches the point where the string makes an angle of $\frac{1}{3}\pi$ radians with the downward vertical through $O$.

\begin{enumerate}[label=(\alph*)]
\item Show that at this instant the speed of $P$ is $4.5$ m s$^{-1}$. [3]
\item Find the magnitude and direction of the radial acceleration of $P$ at this instant. [3]
\item Find the magnitude of the tangential acceleration of $P$ at this instant. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2023 Q1 [8]}}
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