OCR MEI Further Mechanics Major 2022 June — Question 4 7 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeConical pendulum – horizontal circle in free space (no surface)
DifficultyStandard +0.3 This is a standard conical pendulum problem requiring resolution of forces and circular motion equations. The three parts involve straightforward application of T cos θ = mg, T sin θ = mrω², and KE = ½mv², with all values given. It's slightly easier than average as it's a textbook setup with no novel elements, though it does require correct geometric reasoning about the radius r = L sin θ.
Spec6.05c Horizontal circles: conical pendulum, banked tracks

\includegraphics{figure_4} The diagram shows a particle P, of mass 0.1 kg, which is attached by a light inextensible string of length 0.5 m to a fixed point O. P moves with constant angular speed 5 rad s\(^{-1}\) in a horizontal circle with centre vertically below O. The string is inclined at an angle \(\theta\) to the vertical.
  1. Determine the tension in the string. [3]
  2. Find the value of \(\theta\). [2]
  3. Find the kinetic energy of P. [2]

Question 4:
AnswerMarks Guidance
4(a) r =0.5sinθ
Tsinθ=0.1r(5)2
AnswerMarks
T =1.25(N)B1
M1
A1
AnswerMarks
[3]3.1b
3.3
AnswerMarks
1.1Where r is the radius of the horizontal circle
Use of N2L with T resolved (allow sin/cos confusion)
and a=mrω2with m,ωcorrect and any form for r
(allow just r)
AnswerMarks
(b)Tcosθ=0.1g
θ=38.4M1
A1
AnswerMarks
[2]3.3
1.1Resolve vertically (allow sin/cos confusion) – correct
number of terms (allow T for the tension) and
dimensionally consistent
θ=38.371715... or 0.66971276… (in radians)
AnswerMarks
(c)1
(0.1)r2(5)2
KE =
2
AnswerMarks
0.120(J)M1
A1
AnswerMarks
[2]1.1
1.11
Use of KE = mv2with correct m and v=5r
2
0.12042 – accept 0.121 from using θ=38.4or 0.118
from using θ=38
Question 4:
4 | (a) | r =0.5sinθ
Tsinθ=0.1r(5)2
T =1.25(N) | B1
M1
A1
[3] | 3.1b
3.3
1.1 | Where r is the radius of the horizontal circle
Use of N2L with T resolved (allow sin/cos confusion)
and a=mrω2with m,ωcorrect and any form for r
(allow just r)
(b) | Tcosθ=0.1g
θ=38.4 | M1
A1
[2] | 3.3
1.1 | Resolve vertically (allow sin/cos confusion) – correct
number of terms (allow T for the tension) and
dimensionally consistent
θ=38.371715... or 0.66971276… (in radians)
(c) | 1
(0.1)r2(5)2
KE =
2
0.120(J) | M1
A1
[2] | 1.1
1.1 | 1
Use of KE = mv2with correct m and v=5r
2
0.12042 – accept 0.121 from using θ=38.4or 0.118
from using θ=38
\includegraphics{figure_4}

The diagram shows a particle P, of mass 0.1 kg, which is attached by a light inextensible string of length 0.5 m to a fixed point O.

P moves with constant angular speed 5 rad s$^{-1}$ in a horizontal circle with centre vertically below O. The string is inclined at an angle $\theta$ to the vertical.

\begin{enumerate}[label=(\alph*)]
\item Determine the tension in the string. [3]
\item Find the value of $\theta$. [2]
\item Find the kinetic energy of P. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q4 [7]}}