OCR MEI Further Mechanics Minor 2021 November — Question 1 7 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2021
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeVerify dimensional consistency
DifficultyModerate -0.8 This is a straightforward dimensional analysis question requiring only routine application of standard techniques: stating dimensions of force (basic recall), verifying dimensional consistency (mechanical substitution), calculating percentage changes (simple arithmetic), and unit conversion (direct calculation). All parts are standard textbook exercises with no problem-solving insight required.
Spec6.01a Dimensions: M, L, T notation6.01c Dimensional analysis: error checking

1
  1. State the dimensions of force. The force \(F\) required to keep a car moving at constant speed on a circular track is given by the formula $$\mathrm { F } = \frac { \mathrm { mv } ^ { 2 } } { \mathrm { r } }$$ where
    It is proposed that a new unit of force, the trackforce (Tr), should be adopted in motor-racing. 1 Tr is defined as the amount of force required to accelerate a mass of 1 ton at a rate of 1 mile per hour per second. It is given that 1 ton \(= 1016 \mathrm {~kg}\) and 1 mile \(= 1609 \mathrm {~m}\).
  2. Determine the number of newtons that are equivalent to 1 Tr .

Question 1:
AnswerMarks Guidance
1(a) −2
MLTB1 1.2
[1]
AnswerMarks
(b)( −1 )2
M LT
[ ]=
RHS
AnswerMarks Guidance
LM1 3.4
and [r] = L and [v] = LT–1
AnswerMarks Guidance
A12.2a must see expanded
2 −2 −1 −2[2] −1 2
(c)= 𝑀𝑀2𝐿𝐿 𝑇𝑇 𝐿𝐿 =𝑀𝑀𝐿𝐿𝑇𝑇 =[𝐿𝐿𝐿𝐿𝐿𝐿]
1.1 ÷0 .9M1 1.1
0.9
AnswerMarks Guidance
=1.34444 so 34.4%A1 2.2b
[2]
AnswerMarks Guidance
(d)1016×1609÷602 M1
squared e.g for the M
mark 3
AnswerMarks Guidance
=454NA1 1.1
[2]
Question 1:
1 | (a) | −2
MLT | B1 | 1.2
[1]
(b) | ( −1 )2
M LT
[ ]=
RHS
L | M1 | 3.4 | Using given formula with [m] = M
and [r] = L and [v] = LT–1
A1 | 2.2a | must see expanded
2 −2 −1 −2 | [2] | −1 2
(c) | = 𝑀𝑀2𝐿𝐿 𝑇𝑇 𝐿𝐿 =𝑀𝑀𝐿𝐿𝑇𝑇 =[𝐿𝐿𝐿𝐿𝐿𝐿]
1.1 ÷0 .9 | M1 | 1.1 | Using cor ( r𝐿𝐿e𝑇𝑇ct fo ) rmula with 1.1 and
0.9
=1.34444 so 34.4% | A1 | 2.2b | 34.44444…
[2]
(d) | 1016×1609÷602 | M1 | 1.1 | Condone denominator which is not
squared e.g for the M
mark 3
=454N | A1 | 1.1 | 60 𝑜𝑜𝑜𝑜 60 | 454.0955…
[2]
1
\begin{enumerate}[label=(\alph*)]
\item State the dimensions of force.

The force $F$ required to keep a car moving at constant speed on a circular track is given by the formula

$$\mathrm { F } = \frac { \mathrm { mv } ^ { 2 } } { \mathrm { r } }$$

where

\begin{itemize}
  \item $m$ is the constant mass of the car,
  \item $v$ is the speed of the car,
  \item $r$ is the radius of the circular track.
\item Verify that the formula is dimensionally consistent.
\item Determine the percentage increase in force required to keep a car moving on a circular track if the speed of the car were to increase by $10 \%$ and if the track radius were to decrease by $10 \%$.
\end{itemize}

It is proposed that a new unit of force, the trackforce (Tr), should be adopted in motor-racing. 1 Tr is defined as the amount of force required to accelerate a mass of 1 ton at a rate of 1 mile per hour per second.

It is given that 1 ton $= 1016 \mathrm {~kg}$ and 1 mile $= 1609 \mathrm {~m}$.
\item Determine the number of newtons that are equivalent to 1 Tr .
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2021 Q1 [7]}}