| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 6 |
| Topic | Circular Motion 1 |
| Type | Period or time for one revolution |
| Difficulty | Standard +0.3 This is a straightforward application of circular motion formulas (F = mv²/r or F = mω²r) with given values. Students must convert units (km to m, days to seconds) and calculate angular velocity, but the method is standard and requires no problem-solving insight. The 5-mark allocation reflects computational steps rather than conceptual difficulty, making it slightly easier than average. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 365 \times 24 \times 60 \times 60 \text{ s}\) and \(R = 1.5 \times 10^{11} \text{ m}\) | B1 | Time converted to s and radius to m (\(T = 31536000\) s); (\(\omega = 1.99 \times 10^{-7} \text{ s}^{-1}\)); (\(v = 2.99 \times 10^5 \text{ m s}^{-1}\)) |
| \(\omega = \frac{2\pi}{T}\) or \(v = \frac{2\pi R}{T}\) | M1 | Used (units of \(T\) could be wrong) |
| \(F = m R \omega^2\) or \(F = \frac{mv^2}{R}\) | M1 | Used (units could be wrong) |
| \(F = 5.97 \times 10^{24} \times 1.5 \times 10^{11} \times (1.99 \times 10^{-7})^2\) oe | M1 | FT their values, provided that there was a proper attempt at conversion of units |
| \(3.5 \times 10^{22}\) or \(3.6 \times 10^{22} \text{ N}\) | A1 | cao |
| Answer | Marks |
|---|---|
| [The force is directed] towards the sun | B1 |
## (i)
$T = 365 \times 24 \times 60 \times 60 \text{ s}$ and $R = 1.5 \times 10^{11} \text{ m}$ | B1 | Time converted to s and radius to m ($T = 31536000$ s); ($\omega = 1.99 \times 10^{-7} \text{ s}^{-1}$); ($v = 2.99 \times 10^5 \text{ m s}^{-1}$)
$\omega = \frac{2\pi}{T}$ or $v = \frac{2\pi R}{T}$ | M1 | Used (units of $T$ could be wrong)
$F = m R \omega^2$ or $F = \frac{mv^2}{R}$ | M1 | Used (units could be wrong)
$F = 5.97 \times 10^{24} \times 1.5 \times 10^{11} \times (1.99 \times 10^{-7})^2$ oe | M1 | FT their values, provided that there was a proper attempt at conversion of units
$3.5 \times 10^{22}$ or $3.6 \times 10^{22} \text{ N}$ | A1 | cao
## (ii)
[The force is directed] towards the sun | B1 |
---Assume that the earth moves round the sun in a circle of radius $1.50 \times 10^8$ km at constant speed, with one complete orbit taking 365 days. Given that the mass of the earth is $5.97 \times 10^{24}$ kg,
\begin{enumerate}[label=(\roman*)]
\item calculate the magnitude of the force exerted by the sun on the earth, giving your answer in newtons, [5]
\item state the direction in which this force acts. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q3 [6]}}