| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find exponents with all unknowns |
| Difficulty | Standard +0.8 Part (a)(iii) requires setting up and solving a system of three simultaneous equations from dimensional analysis with all three exponents unknown, which is more demanding than standard dimensional analysis problems. The SHM problem in part (b) involves non-trivial application of energy conservation and integration, going beyond routine textbook exercises. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\rho] = \text{ML}^{-3}\) | B1 | |
| \([E] = [\rho v^2] = (\text{ML}^{-3})(\text{LT}^{-1})^2\) | M1 | Obtaining dimensions of \(E\) |
| Dimensions of Young's modulus are \(\text{ML}^{-1}\text{T}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E = \rho v^2 = 7800 \times 6100^2 = 2.90 \times 10^{11}\) (3 sf) | B1 | FT provided all powers are non-zero; No FT if derived units involved |
| Units are \(\text{kg m}^{-1}\text{s}^{-2}\) | B1 | OR \(\text{N m}^{-2}\) OR Pa |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{T}^{-1} = \text{L}^\alpha(\text{ML}^{-1}\text{T}^{-2})^\beta(\text{ML}^{-3})^\gamma\) | ||
| \(\beta = \frac{1}{2}\) | B1 | CAO |
| \(\gamma = -\frac{1}{2}\) | B1 | FT \(\gamma = -\beta\); Provided non-zero |
| \(\alpha - \beta - 3\gamma = 0\) | M1 | Equation from powers of L |
| \(\alpha = -1\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.04^2 = \omega^2(A^2 - 3.9^2)\) | M1 | Using \(v^2 = \omega^2(A^2 - x^2)\) |
| \(0.5^2 = \omega^2(A^2 - 6.0^2)\) | A1 | Both equations correct |
| \(\dfrac{A^2 - 15.21}{A^2 - 36} = 4.3264\) | M1 | Eliminating \(\omega\) or \(A\); \(A = 6.5\), \(\omega = 0.2\) |
| Amplitude (\(A\)) is \(6.5\) m | A1 | |
| Period \(\left(\dfrac{2\pi}{\omega}\right)\) is \(10\pi = 31.4\) s (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 6.5\sin 0.2t\) | B1 | FT for \(6.5\sin 0.2t\) or \(6.5\cos 0.2t\); OR \((v=)1.3\sin 0.2t\) etc |
| When \(x = 3.9\), \(t = 3.2175\) | M1 | Using \(x = 3.9\) or \(x = 6.0\) to find a time; OR using \(v = 1.04\) or \(v = 0.5\) |
| When \(x = 6.0\), \(t = 5.8800\) | M1 | Fully correct strategy for required time; \(4.6365 - 1.9740\) if cos is used |
| Time from X to Y is \(2.66\) s (3 sf) | A1 | CAO |
# Question 1:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\rho] = \text{ML}^{-3}$ | B1 | |
| $[E] = [\rho v^2] = (\text{ML}^{-3})(\text{LT}^{-1})^2$ | M1 | Obtaining dimensions of $E$ |
| Dimensions of Young's modulus are $\text{ML}^{-1}\text{T}^{-2}$ | A1 | |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E = \rho v^2 = 7800 \times 6100^2 = 2.90 \times 10^{11}$ (3 sf) | B1 | FT provided all powers are non-zero; No FT if derived units involved |
| Units are $\text{kg m}^{-1}\text{s}^{-2}$ | B1 | OR $\text{N m}^{-2}$ OR Pa |
## Part (a)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{T}^{-1} = \text{L}^\alpha(\text{ML}^{-1}\text{T}^{-2})^\beta(\text{ML}^{-3})^\gamma$ | | |
| $\beta = \frac{1}{2}$ | B1 | CAO |
| $\gamma = -\frac{1}{2}$ | B1 | FT $\gamma = -\beta$; Provided non-zero |
| $\alpha - \beta - 3\gamma = 0$ | M1 | Equation from powers of L |
| $\alpha = -1$ | A1 | CAO |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.04^2 = \omega^2(A^2 - 3.9^2)$ | M1 | Using $v^2 = \omega^2(A^2 - x^2)$ |
| $0.5^2 = \omega^2(A^2 - 6.0^2)$ | A1 | Both equations correct |
| $\dfrac{A^2 - 15.21}{A^2 - 36} = 4.3264$ | M1 | Eliminating $\omega$ or $A$; $A = 6.5$, $\omega = 0.2$ |
| Amplitude ($A$) is $6.5$ m | A1 | |
| Period $\left(\dfrac{2\pi}{\omega}\right)$ is $10\pi = 31.4$ s (3 sf) | A1 | |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 6.5\sin 0.2t$ | B1 | FT for $6.5\sin 0.2t$ or $6.5\cos 0.2t$; OR $(v=)1.3\sin 0.2t$ etc |
| When $x = 3.9$, $t = 3.2175$ | M1 | Using $x = 3.9$ or $x = 6.0$ to find a time; OR using $v = 1.04$ or $v = 0.5$ |
| When $x = 6.0$, $t = 5.8800$ | M1 | Fully correct strategy for required time; $4.6365 - 1.9740$ if cos is used |
| Time from X to Y is $2.66$ s (3 sf) | A1 | CAO |1
\begin{enumerate}[label=(\alph*)]
\item The speed $v$ of sound in a solid material is given by $v = \sqrt { \frac { E } { \rho } }$, where $E$ is Young's modulus for the material and $\rho$ is its density.
\begin{enumerate}[label=(\roman*)]
\item Find the dimensions of Young's modulus.
The density of steel is $7800 \mathrm {~kg} \mathrm {~m} ^ { - 3 }$ and the speed of sound in steel is $6100 \mathrm {~ms} ^ { - 1 }$.
\item Find Young's modulus for steel, stating the units in which your answer is measured.
A tuning fork has cylindrical prongs of radius $r$ and length $l$. The frequency $f$ at which the tuning fork vibrates is given by $f = k c ^ { \alpha } E ^ { \beta } \rho ^ { \gamma }$, where $c = \frac { l ^ { 2 } } { r }$ and $k$ is a dimensionless constant.
\item Find $\alpha , \beta$ and $\gamma$.
\end{enumerate}\item A particle P is performing simple harmonic motion along a straight line, and the centre of the oscillations is O . The points X and Y on the line are on the same side of O , at distances 3.9 m and 6.0 m from O respectively. The speed of P is $1.04 \mathrm {~ms} ^ { - 1 }$ when it passes through X and $0.5 \mathrm {~ms} ^ { - 1 }$ when it passes through Y.
\begin{enumerate}[label=(\roman*)]
\item Find the amplitude and the period of the oscillations.
\item Find the time taken for P to travel directly from X to Y .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2014 Q1 [18]}}