Higher June 2023 Paper 4 Q5
5
(a) Write 0.003 86 in standard form. [1]
(b) The speed of sound is \(3.43 \times 10^{-1}\) km/s.
An object is travelling at the speed of sound.
An object is travelling at the speed of sound.
Work out how far the object travels in one day. [2]
(c) In a science fiction story, a spacecraft travelling faster than the speed of light is said to be travelling at ‘warp \(n\)’ where \(n\) is an integer.
Warp \(n\) is defined as \(n^3 \times\) the speed of light.
In the story, a spacecraft needs to travel from Earth to Neptune in less than 2 minutes.
- The speed of light is \(3.00 \times 10^5\) km/s.
- The distance from Earth to Neptune is \(4.41 \times 10^9\) km.
Find the smallest possible warp \(n\) at which the spacecraft can travel.
You must show your working. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(3.86 \times 10^{-3}\) | 1 | Condone trailing zeros | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 29 635[.2] or 29 640 or 29 600 | 2 | M1 for \(3.43 \times 10^{-1} \times 60^2 \times 24\) oe If 0 scored SC1 for 14 817.6, 14 818, 14 820 or 14 800 | Note: \(60^2 \times 24 = 86\,400\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 5 with correct working | 3 | M2 for \(\sqrt[3]{\frac{4.41 \times 10^9}{120 \times 3.00 \times 10^5}}\) oe OR M1 for \(\frac{4.41 \times 10^9}{3.00 \times 10^5}\) oe or \(\frac{4.41 \times 10^9}{120}\) oe or \(120 \times 3.00 \times 10^5\) oe If 0 scored SC1 for answer 5 with no or insufficient working | “Correct working” requires evidence of at least M1 or M2 if trials are used M2 implied by 4.966… or 4.97 or \(\sqrt[3]{122.5}\) or \(n^3 = 122.5\) condone use of a value \(119 \leqslant\) value \(\leqslant 120\) instead of 120 which should lead to 4.98 M1 may be seen within a larger calculation e.g. \(\frac{4.41 \times 10^9}{120 \times 3.00 \times 10^5}\) or \(\sqrt[3]{\frac{4.41 \times 10^9}{3.00 \times 10^5}}\) M1 implied by 14 700 or \(3.675 \times 10^7\) oe or \(3.6 \times 10^7\) oe or 122.5 use of trials M1 for each correct trial using integer values of \(n\) up to a maximum of M2 e.g. \(\frac{their\ 14700}{5^3}\) oe and 117 to 118 or 1.95 to 1.97 \(\frac{their\ 14700}{4^3}\) oe and 229[. ..] or 3.8[…] their 14 700 comes from \(3 \times 10^5\) and \(4.41 \times 10^9\) (see appendix for more examples) |
Appendix: Question 5(c)
Other trials, condone rot :
\(\frac{their\ 14700}{2^3}\) oe and 1837[.5…] or 30[.6…]
\(\frac{their\ 14700}{3^3}\) oe and 544[.4….] or 9[.07…]
\(\frac{their\ 14700}{6^3}\) oe and 68[.05…] or 1[.1…]
Or they might try to see how far they go at each warp speed in 2 minutes (rot at least 2 figures):
| Warp 2 : | \(2^3 \times 3 \times 10^5 \times 120 =\) | 288 000 000 |
| Warp 3 : | \(3^3 \times 3 \times 10^5 \times 120 =\) | 972 000 000 |
| Warp 4 : | \(4^3 \times 3 \times 10^5 \times 120 =\) | 2 304 000 000 |
| Neptune | 4 410 000 000 | |
| Warp 5 : | \(5^3 \times 3 \times 10^5 \times 120 =\) | 4 500 000 000 |
| Warp 6 : | \(6^3 \times 3 \times 10^5 \times 120 =\) | 7 776 000 000 |
Mark as trials so M1 each correct trial.