AQA M3 2011 June — Question 2 5 marks

Exam BoardAQA
ModuleM3 (Mechanics 3)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDimensional Analysis
TypeFind exponents with all unknowns
DifficultyStandard +0.3 This is a standard dimensional analysis problem requiring students to equate dimensions of time with combinations of length, force, and mass per unit length. While it involves setting up and solving simultaneous equations from dimensional considerations, it follows a well-established algorithmic procedure taught explicitly in M3 with no novel insight required. Slightly above average difficulty due to the algebraic manipulation needed.
Spec6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions

2 The time, \(t\), for a single vibration of a piece of taut string is believed to depend on
the length of the taut string, \(l\),
the tension in the string, \(F\),
the mass per unit length of the string, \(q\), and
a dimensionless constant, \(k\),
such that $$t = k l ^ { \alpha } F ^ { \beta } q ^ { \gamma }$$ where \(\alpha , \beta\) and \(\gamma\) are constants.
By using dimensional analysis, find the values of \(\alpha , \beta\) and \(\gamma\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\([t] = \text{T}\), \([l] = \text{L}\), \([F] = \text{MLT}^{-2}\), \([q] = \text{ML}^{-1}\)B1 Dimensions of \(F\) and \(q\) stated or used correctly
\(\text{T} = \text{L}^\alpha \cdot (\text{MLT}^{-2})^\beta \cdot (\text{ML}^{-1})^\gamma\)M1 Setting up dimensional equation
M: \(0 = \beta + \gamma\)A1 Correct equation for M
L: \(0 = \alpha + \beta - \gamma\) Correct equation for L
T: \(1 = -2\beta\) Correct equation for T
\(\beta = -\dfrac{1}{2}\)A1 From T equation
\(\gamma = \dfrac{1}{2}\)A1 From M equation
\(\alpha = 1\)A1 From L equation: \(\alpha = \gamma - \beta = \frac{1}{2}+\frac{1}{2} = 1\)
# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $[t] = \text{T}$, $[l] = \text{L}$, $[F] = \text{MLT}^{-2}$, $[q] = \text{ML}^{-1}$ | B1 | Dimensions of $F$ and $q$ stated or used correctly |
| $\text{T} = \text{L}^\alpha \cdot (\text{MLT}^{-2})^\beta \cdot (\text{ML}^{-1})^\gamma$ | M1 | Setting up dimensional equation |
| M: $0 = \beta + \gamma$ | A1 | Correct equation for M |
| L: $0 = \alpha + \beta - \gamma$ | | Correct equation for L |
| T: $1 = -2\beta$ | | Correct equation for T |
| $\beta = -\dfrac{1}{2}$ | A1 | From T equation |
| $\gamma = \dfrac{1}{2}$ | A1 | From M equation |
| $\alpha = 1$ | A1 | From L equation: $\alpha = \gamma - \beta = \frac{1}{2}+\frac{1}{2} = 1$ |

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2 The time, $t$, for a single vibration of a piece of taut string is believed to depend on\\
the length of the taut string, $l$,\\
the tension in the string, $F$,\\
the mass per unit length of the string, $q$, and\\
a dimensionless constant, $k$,\\
such that

$$t = k l ^ { \alpha } F ^ { \beta } q ^ { \gamma }$$

where $\alpha , \beta$ and $\gamma$ are constants.\\
By using dimensional analysis, find the values of $\alpha , \beta$ and $\gamma$.

\hfill \mbox{\textit{AQA M3 2011 Q2 [5]}}