paper 3 marksheme 2001

paper 3 marksheme 2001 - INTERNATIONAL BACCALAURAT...

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MARKSCHEME November 2001 PHYSICS Standard Level Paper 3 21 pages N01/430/S(3)M BACCALAUREATE BACCALAUR&AT INTERNATIONAL BACHILLERATO INTERNACIONAL
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OPTION A & MECHANICS EXTENSION [1] A1. (a) for two point masses, the gravitational force is proportional to the product of the masses &; & and inversely proportional to the square of the separation of two masses; [1] Accept formula version for full marks provided the symbols are defined. [2 max] (b) Sun Earth Moon arrow from centre of Moon towards centre of Earth; [1] arrow from centre of Moon towards centre of Sun; [1] Accept arrows if within half a radius of the centre. Correct idea but sloppy arrows should still gain credit. [2 max] (c) attempted use of 12 2 Gm m F r = for Earth / Moon AND Earth / Sun systems; [1] correct rearrangement; [1] ratio 2 2 23 11 2 30 8 5.98 10 1.5 10 1.99 10 3.84 10 S E S E r M M r  ×× =×= ×   to give ratio (no units); [1] 1 0.459 4.59 10 == × [3 max] (d) Moon would (still) orbit the Sun; [1] time of orbit the same j 1 year or orbital distance roughly the same as before; [1] any relevant extra detail or explanation or discussion of ideas; [1] e.g. gravitational attraction to the Sun is still present etc. Actual answer depends on the velocity of the Moon when the Earth was &removed±, giving an elliptical orbit. Award full marks to candidates who understand the principles but are unable to give a concise answer. N.B. reject the idea the Moon is kept in current orbit around the Sun as a result of its attraction to the Earth. [3 max] ± 6 ± N01/430/S(3)M
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A2. (a) one arrow towards the right along (or parallel to) the line of collision [1 max] Ignore magnitude of arrow and accept if shown on diagram 1. (b) attempt at application of conservation of momentum; [1] at least one correct resolving of velocity into appropriate components; [1] correct setting up of the equation along the original direction; [1] ( ) ( ) ( ) cos cos m v m 0.2 30 m 0.5 θ += ! correct setting up of the equation at right angles; [1] () sin sin mv m0 . 2 3 0 = ! correct answer for speed; [1] 1 0.34 ms v = correct answer for direction; [1] 17 = ! where is the angle between v and the line of collision [6 max] (c) correct calculation of KE before; [1] 2 before KE 0.5 0.2 0.5 0.025 J =×× = correct calculation of KE after; [1] ( ) 22 after KE 0.5 0.2 0.2 0.5 0.2 0.34 0.016 J +×× = correct answer; [1] Energy lost 0.025 0.016 0.009 J; =−= [3 max] & 7 & N01/430/S(3)M
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OPTION B & ATOMIC AND NUCLEAR PHYSICS EXTENSION B1. (a) use of [1] ; h p λ = to give [1] 24 1 6.6 10 kg m s ; −− × [2 max] (b) use of [1] 2 KE ; 2 p m = correct substitution; [1] to give [1] 17 2.4 10 J; × [3 max] (c) use of [1] KE ; e V = to give 150 V; [1] [2 max] (d) any graph that shows ! always decreasing as V increases&; [1] &that is non linear&; [1] &that does not cross either axis; [1] [3 max] Wavelength of beam Accelerating potential difference ± 8 ± N01/430/S(3)M
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B2. (a) reaction I : Carbon &15 decaying into Nitrogen &15; [1] since total mass of Carbon nucleus is greater; [1] Accept Carbon has smaller mass defect / binding energy per nucleon / OWTTE; accept answer without reasons or with incorrect reasons for [1] .
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paper 3 marksheme 2001 - INTERNATIONAL BACCALAURAT...

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