Jan 2003 A Mark Scheme (scroll down for some hints)
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1 |
B |
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2 |
A |
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3 |
D |
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4 |
C |
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5 |
D |
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6 |
B |
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B |
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D |
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B |
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C |
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11 |
D |
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A |
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B |
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A |
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B |
1. greatest acceleration at maximum displacement - so when x=50mm
2. use max Ek = 1/2 m v2 when max v = 2pfA and f = 50/47
3. T2 = 2p square
root ( (L-600 x 10-3)/g) and T2 = 1/2 T
where T = 2p square root (L/g)
sub these two equations into each other and rearrange
to find L
4. c = fl therefore wavelength = 1.6km. 2km is 3p/2 phase diff which is the same as p/2
5. find f using c = fl this use this to calculate how many waves will be produced in 0.01 x 10-6s
6. use l = ws/D and rearrange for w
7. nl = d sinq put n = 2 and q = 45 to find l /d then use this value to find the smallest possible integer value if q = 90 (which is maximum diffraction angle through a grating)
8. E=1/2 QV and C = Q/V combine to find Energy, and then take 10% and then use potential energy, E = mgh
9.
10. 9 rev per minute = 9 / 60 rev /s. One rev = 2p so answer = 9 x 2p / 60