AQA Further Paper 3 Mechanics 2020 June — Question 3 2 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2020
SessionJune
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeDimensions of derivative or rate
DifficultyEasy -1.8 This is a straightforward dimensional analysis question requiring only basic substitution of known dimensions (speed = LT^{-1}, radius = L) into the given formula and simple algebraic manipulation to isolate ω. It's purely mechanical with no problem-solving or conceptual insight needed, making it significantly easier than average A-level questions.
Spec6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship

3 The speed, \(v\), of a particle moving in a horizontal circle is given by the formula \(v = r \omega\) where: \(v =\) speed \(r =\) radius \(\omega =\) angular speed.
Show that the dimensions of angular speed are \(T ^ { - 1 }\) [0pt] [2 marks]

Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\([v] = [r][\omega]\), so \(LT^{-1} = L[\omega]\)M1 (AO 1.1a) Forms dimensional analysis equation using the formula given
\([\omega] = T^{-1}\)R1 (AO 2.1) Completes a clear argument to show dimensions of angular speed are \(T^{-1}\)
## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v] = [r][\omega]$, so $LT^{-1} = L[\omega]$ | M1 (AO 1.1a) | Forms dimensional analysis equation using the formula given |
| $[\omega] = T^{-1}$ | R1 (AO 2.1) | Completes a clear argument to show dimensions of angular speed are $T^{-1}$ |

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3 The speed, $v$, of a particle moving in a horizontal circle is given by the formula $v = r \omega$ where:\\
$v =$ speed\\
$r =$ radius\\
$\omega =$ angular speed.\\
Show that the dimensions of angular speed are $T ^ { - 1 }$\\[0pt]
[2 marks]\\

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2020 Q3 [2]}}