| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Find period from given information |
| Difficulty | Standard +0.3 Part (a) is a standard SHM problem requiring application of v_max = aω and v² = ω²(a²-x²) to find amplitude and period—routine for M3 students. Part (b) involves dimensional analysis which is methodical but straightforward once the technique is known. The multi-part structure and combination of SHM with dimensional analysis adds modest complexity, but all components are standard textbook exercises requiring no novel insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.01a Dimensions: M, L, T notation6.01d Unknown indices: using dimensions6.01e Formulate models: dimensional arguments |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega = 2.0 \times 10^{-6} \times \left(\frac{4.86 \times 10^{14}}{2500}\right)^{\frac{1}{2}} \times \left(\frac{30000}{50}\right)^{-\frac{1}{3}}\) | M1M1, A1 | For \(\left(\frac{4.86 \times 10^{14}}{2500}\right)^{\frac{1}{2}}\) and \(\left(\frac{30000}{50}\right)^{-\frac{1}{3}}\). Requires \(\beta \neq 0, \gamma \neq 0\). Correct equation for \(\omega\). FT if comparable. |
| OR | ||
| \(2.0 \times 10^{-6} = k \times G^{\frac{1}{3}} \times 2500^{-1} \times 50^{-\frac{1}{3}}\) | M1 | Requires \(\beta \neq 0\) and \(\gamma \neq 0\) |
| \(kG^{\frac{1}{3}} = 1.414 \times 10^{-5}\) | M1 | Condone the use of any value for \(G\) (including \(G = 1\)) |
| \(\omega = 1.414 \times 10^{-5} \times (4.86 \times 10^{14})^{\frac{1}{2}} \times 30000^{-\frac{1}{3}}\) | A1 | Correct equation for \(\omega\). FT if comparable. |
| Angular speed is \(6.0 \times 10^{-5}\) rad s\(^{-1}\) | A1 | CAO |
### Part (b)(iii)
| $\omega = 2.0 \times 10^{-6} \times \left(\frac{4.86 \times 10^{14}}{2500}\right)^{\frac{1}{2}} \times \left(\frac{30000}{50}\right)^{-\frac{1}{3}}$ | M1M1, A1 | For $\left(\frac{4.86 \times 10^{14}}{2500}\right)^{\frac{1}{2}}$ and $\left(\frac{30000}{50}\right)^{-\frac{1}{3}}$. Requires $\beta \neq 0, \gamma \neq 0$. Correct equation for $\omega$. FT if comparable. |
|---|---|---|
| **OR** | | |
|---|---|---|
| $2.0 \times 10^{-6} = k \times G^{\frac{1}{3}} \times 2500^{-1} \times 50^{-\frac{1}{3}}$ | M1 | Requires $\beta \neq 0$ and $\gamma \neq 0$ |
| $kG^{\frac{1}{3}} = 1.414 \times 10^{-5}$ | M1 | Condone the use of any value for $G$ (including $G = 1$) |
| $\omega = 1.414 \times 10^{-5} \times (4.86 \times 10^{14})^{\frac{1}{2}} \times 30000^{-\frac{1}{3}}$ | A1 | Correct equation for $\omega$. FT if comparable. |
| Angular speed is $6.0 \times 10^{-5}$ rad s$^{-1}$ | A1 | CAO |
|---|---|---|1
\begin{enumerate}[label=(\alph*)]
\item A particle P is executing simple harmonic motion, and the centre of the oscillations is at the point O . The maximum speed of P during the motion is $5.1 \mathrm {~ms} ^ { - 1 }$. When P is 6 m from O , its speed is $4.5 \mathrm {~ms} ^ { - 1 }$. Find the period and the amplitude of the motion.
\item The force $F$ of gravitational attraction between two objects of masses $m _ { 1 }$ and $m _ { 2 }$ at a distance $d$ apart is given by $F = \frac { G m _ { 1 } m _ { 2 } } { d ^ { 2 } }$, where $G$ is the universal gravitational constant.
\begin{enumerate}[label=(\roman*)]
\item Find the dimensions of $G$.
Three objects, each of mass $m$, are moving in deep space under mutual gravitational attraction. They move round a single circle with constant angular speed $\omega$, and are always at the three vertices of an equilateral triangle of side $R$. You are given that $\omega = k G ^ { \alpha } m ^ { \beta } R ^ { \gamma }$, where $k$ is a dimensionless constant.
\item Find $\alpha , \beta$ and $\gamma$.
For three objects of mass 2500 kg at the vertices of an equilateral triangle of side 50 m , the angular speed is $2.0 \times 10 ^ { - 6 } \mathrm { rad } \mathrm { s } ^ { - 1 }$.
\item Find the angular speed for three objects of mass $4.86 \times 10 ^ { 14 } \mathrm {~kg}$ at the vertices of an equilateral triangle of side 30000 m .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2013 Q1 [18]}}