OCR MEI Further Mechanics Major 2020 November — Question 10 14 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2020
SessionNovember
Marks14
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and calculus-based kinematics. Part (a) combines energy methods with Newton's second law in circular motion; part (b) requires algebraic manipulation to derive a specific angular velocity formula; part (c) demands differentiation of angular velocity using the chain rule. The multi-step reasoning, integration of multiple mechanics concepts, and the need to work with angular quantities elevate this significantly above standard A-level, though it follows a relatively standard framework for Further Mechanics problems.
Spec6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration
\includegraphics{figure_10} Fig. 10 shows a small bead P of mass \(m\) which is threaded on a smooth thin wire. The wire is in the form of a circle of radius \(a\) and centre O. The wire is fixed in a vertical plane. The bead is initially at the lowest point A of the wire and is projected along the wire with a velocity which is just sufficient to carry it to the highest point on the wire. The angle between OP and the downward vertical is denoted by \(\theta\).
  1. Determine the value of \(\theta\) when the magnitude of the reaction of the wire on the bead is \(\frac{7}{5}mg\). [7]
  2. Show that the angular velocity of P when OP makes an angle \(\theta\) with the downward vertical is given by \(k\sqrt{\frac{g}{a}\cos\left(\frac{\theta}{2}\right)}\), stating the value of the constant \(k\). [4]
  3. Hence determine, in terms of \(g\) and \(a\), the angular acceleration of P when \(\theta\) takes the value found in part (a). [3]
Question 10:
AnswerMarks Guidance
10(a) =mg(2a)
At highest point, PE and KE = 0
1
At angle θ,PE =mga(1−cosθ), KE = mv2
AnswerMarks Guidance
2B1 1.1
θ- candidates may calculate initial speed
AnswerMarks
of P as 2 ga for this markPE zero at A
1
2mga=mga(1−cosθ)+ mv2
AnswerMarks Guidance
2M1* 3.3
number of terms
AnswerMarks Guidance
v2 =2ga(1+cosθ)A1 1.1
mv2
R−mgcosθ=
AnswerMarks Guidance
aM1* 3.3
condone a for accelerationAllow r for radius
R=mg(2+3cosθ)M1dep* 3.4
expression for R in terms of m, g and θ
7
mg(2+3cosθ)= mg⇒θ=...
AnswerMarks Guidance
2M1 1.1
Equate R with mg and solve to find θ
AnswerMarks
2Dependent on all
previous M marks
π
θ=
AnswerMarks Guidance
3A1 1.1
θ=60 
[7]
AnswerMarks Guidance
10(b) M1*
a2ω2 =2ga(1+cosθ)A1 1.1
θ
a2ω2 =4gacos2
 
AnswerMarks Guidance
2M1dep* 3.1a
1
cos2θ= (1+cos2θ)
2
4g θ g θ
ω2 = cos2   ⇒ω=2 cos 
AnswerMarks Guidance
a 2 a 2A1 2.2a
[4]
AnswerMarks Guidance
10(c) dω 1 g  θ
=−  2  sin ω
 
AnswerMarks Guidance
dt 2 a  2M1* 2.1
dω g π g π
=− sin  2 cos 
 
AnswerMarks Guidance
dt a  6 a  6M1dep* 3.4
expression for angular acceleration
 g 3
θ=−
AnswerMarks Guidance
2aA1 2.2a
[3]
Question 10:
10 | (a) | =mg(2a)
At highest point, PE and KE = 0
1
At angle θ,PE =mga(1−cosθ), KE = mv2
2 | B1 | 1.1 | Correct mechanical energy at either A or
θ- candidates may calculate initial speed
of P as 2 ga for this mark | PE zero at A
1
2mga=mga(1−cosθ)+ mv2
2 | M1* | 3.3 | Use of conservation of energy – correct
number of terms
v2 =2ga(1+cosθ) | A1 | 1.1
mv2
R−mgcosθ=
a | M1* | 3.3 | NII radially – correct number of terms –
condone a for acceleration | Allow r for radius
R=mg(2+3cosθ) | M1dep* | 3.4 | Substitute their expression for v2 to get
expression for R in terms of m, g and θ
7
mg(2+3cosθ)= mg⇒θ=...
2 | M1 | 1.1 | 7
Equate R with mg and solve to find θ
2 | Dependent on all
previous M marks
π
θ=
3 | A1 | 1.1 | Allow in degrees
θ=60 
[7]
10 | (b) | M1* | 3.4 | Use of v=rωwith r = a and their v
a2ω2 =2ga(1+cosθ) | A1 | 1.1 | oe
θ
a2ω2 =4gacos2
 
2 | M1dep* | 3.1a | Use of correct double-angle identity
1
cos2θ= (1+cos2θ)
2
4g θ g θ
ω2 = cos2   ⇒ω=2 cos 
a 2 a 2 | A1 | 2.2a | k =2
[4]
10 | (c) | dω 1 g  θ
=−  2  sin ω
 
dt 2 a  2 | M1* | 2.1 | Differentiate ω with respect to t
dω g π g π
=− sin  2 cos 
 
dt a  6 a  6 | M1dep* | 3.4 | Substitute their value of θ into their
expression for angular acceleration
 g 3
θ=−
2a | A1 | 2.2a
[3]
\includegraphics{figure_10}

Fig. 10 shows a small bead P of mass $m$ which is threaded on a smooth thin wire. The wire is in the form of a circle of radius $a$ and centre O. The wire is fixed in a vertical plane.

The bead is initially at the lowest point A of the wire and is projected along the wire with a velocity which is just sufficient to carry it to the highest point on the wire. The angle between OP and the downward vertical is denoted by $\theta$.

\begin{enumerate}[label=(\alph*)]
\item Determine the value of $\theta$ when the magnitude of the reaction of the wire on the bead is $\frac{7}{5}mg$. [7]
\item Show that the angular velocity of P when OP makes an angle $\theta$ with the downward vertical is
given by $k\sqrt{\frac{g}{a}\cos\left(\frac{\theta}{2}\right)}$, stating the value of the constant $k$. [4]
\item Hence determine, in terms of $g$ and $a$, the angular acceleration of P when $\theta$ takes the value found in part (a). [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q10 [14]}}
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