OCR FM1 AS 2017 December — Question 2 4 marks

Exam BoardOCR
ModuleFM1 AS (Further Mechanics 1 AS)
Year2017
SessionDecember
Marks4
TopicDimensional Analysis
TypeDerive dimensions from formula
DifficultyModerate -0.8 This is a straightforward dimensional analysis problem requiring rearrangement of a given formula to isolate G, then substituting standard dimensions for force, mass, and distance. It's routine algebraic manipulation with no conceptual difficulty beyond knowing that [F] = MLT^-2, making it easier than average.
Spec6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking

2 The universal law of gravitation states that \(F = \frac { G m _ { 1 } m _ { 2 } } { r ^ { 2 } }\) where \(F\) is the magnitude of the force between two objects of masses \(m _ { 1 }\) and \(m _ { 2 }\) which are a distance \(r\) apart and \(G\) is a constant. Find the dimensions of \(G\).

AnswerMarks Guidance
\([G] = \frac{[F][r]^2}{[m_1][m_2]}\) or \([G] = \frac{[F][r]^2}{[m]^2}\)M1 Use of rearranged formula to find \([G]\)
\([r] = \text{L and } [m] = \text{M}\) \([F] = \text{M}[a] = \text{MLT}^{-2}\) So \([G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}\)B1, B1, A1 [4] soi; soi; Correct answer \(\text{M}^{-1}\text{L}^3\text{T}^{-2}\)
$[G] = \frac{[F][r]^2}{[m_1][m_2]}$ or $[G] = \frac{[F][r]^2}{[m]^2}$ | M1 | Use of rearranged formula to find $[G]$

$[r] = \text{L and } [m] = \text{M}$ $[F] = \text{M}[a] = \text{MLT}^{-2}$ So $[G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}$ | B1, B1, A1 [4] | soi; soi; Correct answer $\text{M}^{-1}\text{L}^3\text{T}^{-2}$

---
2 The universal law of gravitation states that $F = \frac { G m _ { 1 } m _ { 2 } } { r ^ { 2 } }$ where $F$ is the magnitude of the force between two objects of masses $m _ { 1 }$ and $m _ { 2 }$ which are a distance $r$ apart and $G$ is a constant.

Find the dimensions of $G$.

\hfill \mbox{\textit{OCR FM1 AS 2017 Q2 [4]}}