paper 3 2000 markscheme

paper 3 2000 markscheme - INTERNATIONAL BACCALAURAT...

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MARKSCHEME November 2000 PHYSICS Standard Level Paper 2 N00/430/S(2)M INTERNATIONAL BACCALAUREATE BACCALAURÉAT INTERNATIONAL BACHILLERATO INTERNACIONAL 16 pages
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SECTION A [1] A1. (a) Yes. Points lie on a straight line. [2] (b) [1] 2 34 2 3 32 3 3.6 y slope ( ) 3 10 y m 120 10 m k −− == = × × [1] If one pair of values used to calculate K instead of slope then [1] . [3] (c) 2 T 3 kR = [1] 3 R 2 T k = [1] 2 34 2 3 144 y 31 0 ym = × 33 3 480 10 m R m [1] 11 7.8 10 [3] A2. (a) Figure 3 Figure 2 Figure 1 Moving upward at constant speed Accelerating upward T mg T mg Free-body diagram Free-body diagram Physical system Upward force T exerted by cable or “crane”. Downward force mg exerted by “earth”. (Award marks as follows: [1] for unequal forces in Figure 2; [1] for equal forces in Figure 3; [1] for naming objects exerting the forces.) (b) Net Fm a = [1] 25000 20000 2000 a −= [2] [1] 2 5000 2.5 ms 2000 a continued… – 6 – N00/430/S(2)M
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Question A2 (c) continued [2] (c) (i) WFd so [1] PFv [1] 1 20000 0.5 10000 Js = (ii) PI V = [1] P I V = [2] A [1] 10000 25 400 == [1] (iii) Gone into (gravitational) potential energy (of the container). – 7 – N00/430/S(2)M
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[2] A3. (a) Prediction and reasoning both wrong. [1] Reply: free (conduction) electrons throughout wire all start moving essentially together when the switch is closed. So bulbs light simultaneously. No need for any electron from the battery to have reached the bulbs. [1] (Or answers to this effect.) [1] (b) All equal brightness. [3] (c) A and C will get brighter, because the equivalent resistance of the circuit is less, so the current is greater. [1] B will get dimmer: the circuit current increases so the PD across A and C increases ( [1] ) so PD across B decreases ( [1] ). (Note: It is incorrect reasoning to say that the current through B decreases since some of the current goes through D. Give [1] for this ‘local’ reasoning, against [2] for the ‘global’ reasoning above.) [3] (d) Several methods. One way is to look at resistor arrangement as a potential divider. PD across B/D is one fifth of 30V since the parallel resistance is of the whole. 1 5 So PD across B is V. [1] 30 6 5 = Original PD was 10 V. Now i.e. P proportional to . [1] 2 V P R = 2 V So new power is [1] (1W OK) 2 6 3 W 0.36 3 1.08 W 10  ×=× =   Another way would be to work out the resistance of a bulb (33.3 ) and then do a normal circuit calculation of equivalent resistances, currents, voltages and hence power.
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This note was uploaded on 09/30/2011 for the course CHEM 102 taught by Professor Tina during the Spring '11 term at Global.

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paper 3 2000 markscheme - INTERNATIONAL BACCALAURAT...

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