1 1 max c T will say the front strike occurred first d For T 100 m 1 For S 1

# 1 1 max c t will say the front strike occurred first

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[1] [1 max] (c) T will say the front strike occurred first. (d) For T: 100 m [1] For S: [1] 100 γ [3] [3 max] [1] = = 100 115 85 . m [2] [2 max] (e) For S: 85 m (Simultaneous marking of train and platform at ends, and train is length contracted according to S). [1] For T: [1] 85 m 85 74 m 1.15 γ = − = ± 15 ± M00/430/H(3)M
[12] [4] [4 max] G2. Space capsule (a) Experiment does not help distinguish. [1] Lee would explain acceleration as due to gravitational force on the hammer. [1] Anna would say that if the spaceship is accelerating, the floor of the capsule accelerates toward the hammer [1] , which just remains in the state of motion it had when released. [1] Hammer has not accelerated. (The above allocation is a guide: be flexible in awarding marks for explanations, looking for overall understanding.) [2] (b) Second experiment: (i) Photons would lose energy as they go up against the gravitational field, thus photon frequency would decrease slightly, i.e. red shift. [2] [2 max] (ii) Each wave front of light would be detected in a frame which was moving slightly faster than the previous one, so that frequency of wave fronts detected would be less, i.e shifted toward the red. [1] [1 max] (c) No. Equivalence principle [1] . (Actually no or yes! A planetary gravitational field is not uniform, though very nearly so across the capsule. So, if they released two objects simultaneously, one above the other, the distance between them would increase as they fell (slightly). No student is likely to think of this, but accept either answer no or yes with supporting statement.) [3] [3 max] (d) Anna is likely to be right. [1] If a spaceship had taken off from the planet, there would have been a stage where sensation increased during blastoff, only later diminishing. [2] ± 16 ± M00/430/H(3)M
G3. Decay in flight [1] [1 max] (a) Galilean: 05 . c + 0.7 c = 1.2 c (b) Relativistic: Relativistic velocity addition: u [1] 2 (u v) = 1 u v c ′+ + [1] = (0.7 c + 0.5 c) c 0.5 c c 2 1 0 7 + × F H G I K J . = F H G I K J 12 . c 1+ 0.35 = F H G I K J 12 135 . . c [1] = 089 . c [3] [3 max] (Essentially [1] for knowing the formula to apply, [1] for correct substitutions, [1] for calculations.) [2] [2 max] (c) [1] electron nucleus v v v = [1] 0.89 c 0.5 c 0.39 c = ± 17 ± M00/430/H(3)M
[12] H1. Images in a convex lens (a) Lens close to the page. [1] [1 max] (i) Anywhere to the right of the lens. [1] [2] [2 max] (ii) Right way up. Enlarged. Behind the lens. Further away than the page. Yes. (Mark as a whole, [2] for complete understanding, lose [1] per error: 2 errors gets zero.) [1] [1 max] (iii) No. [1] continued... ± 18 ± M00/430/H(3)M Letter ´iµ on page P P Lens printed page
Question H1 continued (b) Lens further from the page [4] [4 max] (i) (Image will be located just outside .) [4] P [1] [1 max] (ii) Eye located to the right of the image. [1] [2] [2 max] (iii) Upside down. Diminished. In front of lens. Nearer to her. No. (Mark as a whole, [2] for complete understanding, lose [1] per error: 2 errors gets zero.) [1] [1 max] (iv) Yes. [1] ± 19 ± M00/430/H(3)M P P Letter ´iµ on page Lens printed page
[8] H2. Double-slit interference [2] [2 max] (a) Light does arrive from both slits, but out of phase: by half a wavelength because of the path difference. [1] Crests of one coincide with troughs of the other, giving destructive interference. [1] continued...

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