| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 10 |
| Topic | Dimensional Analysis |
| Type | Find exponents with all unknowns |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis problem requiring students to equate dimensions of M, L, T to find three exponents, followed by simple numerical verification using given data. The dimensional analysis is routine (easier than average), though part (ii)-(iii) adds minor problem-solving by checking consistency and identifying an error, keeping it slightly below average overall difficulty. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions |
| \(h ( \mathrm {~m} )\) | 0.40 | 2.50 | 3.60 |
| \(P ( \mathrm {~s} )\) | 1.27 | 2.17 | 3.81 |
| Answer | Marks | Guidance |
|---|---|---|
| \([P] = T\), \([m] = M\), \([h] = L\) | B1 | All three soi |
| \([g] = LT^{-2}\) | B1 | |
| M: \(\alpha = 0\) | B1 | |
| T: \(1 = -2\gamma \Rightarrow \gamma = -\frac{1}{2}\) | B1 | |
| L: \(0 = \beta + \gamma \Rightarrow \beta = \frac{1}{2}\) | B1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Values of \(\frac{P}{\sqrt{h}}\) are: 2.01, 1.37, 2.01 | M1 | Or equivalent calculations; (or graph plot/sketch) |
| so not consistent | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| 2.17 must be wrong | B1 | soi |
| \(1.27 \times \sqrt{\frac{2.50}{0.40}}\) or \(3.81 \times \sqrt{\frac{2.50}{3.60}}\) | M1 | oe, eg \(k = 6.286...\) or \(2.008\sqrt{g}\) |
| So an estimate of the correct value is 3.175 | A1 [3] | Allow 3sf answer 3.17 or 3.18 |
## (i)
$[P] = T$, $[m] = M$, $[h] = L$ | B1 | All three soi
$[g] = LT^{-2}$ | B1 |
M: $\alpha = 0$ | B1 |
T: $1 = -2\gamma \Rightarrow \gamma = -\frac{1}{2}$ | B1 |
L: $0 = \beta + \gamma \Rightarrow \beta = \frac{1}{2}$ | B1 [5] |
## (ii)
Values of $\frac{P}{\sqrt{h}}$ are: 2.01, 1.37, 2.01 | M1 | Or equivalent calculations; (or graph plot/sketch)
so not consistent | A1 [2] |
## (iii)
2.17 must be wrong | B1 | soi
$1.27 \times \sqrt{\frac{2.50}{0.40}}$ or $3.81 \times \sqrt{\frac{2.50}{3.60}}$ | M1 | oe, eg $k = 6.286...$ or $2.008\sqrt{g}$
So an estimate of the correct value is 3.175 | A1 [3] | Allow 3sf answer 3.17 or 3.18
---5 A simple pendulum consists of a small sphere of mass $m$ connected to one end of a light rod of length $h$. The other end of the rod is freely hinged at a fixed point. When the sphere is pulled a short distance to one side and released from rest the pendulum performs oscillations. The time taken to perform one complete oscillation is called the period and is denoted by $P$.\\
(i) Assuming that $P = k m ^ { \alpha } h ^ { \beta } g ^ { \gamma }$, where $g$ is the acceleration due to gravity and $k$ is a dimensionless constant, find the values of $\alpha , \beta$ and $\gamma$.
A student conducts an experiment to investigate how $P$ varies as $h$ varies. She measures the value of $P$ for various values of $h$, ensuring that all other conditions remain constant. Her results are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | l | l | l | }
\hline
$h ( \mathrm {~m} )$ & 0.40 & 2.50 & 3.60 \\
\hline
$P ( \mathrm {~s} )$ & 1.27 & 2.17 & 3.81 \\
\hline
\end{tabular}
\end{center}
(ii) Show that these results are not consistent with the answers to part (i).\\
(iii) The student later realises that she has recorded one of her values of $P$ incorrectly.
\begin{itemize}
\item Identify the incorrect value.
\item Estimate the correct value that she should have recorded.
\end{itemize}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q5 [10]}}