marscheme paper 2 2000 hl

# marscheme paper 2 2000 hl - INTERNATIONAL BACCALAURAT...

This preview shows pages 1–6. Sign up to view the full content.

MARKSCHEME November 2000 PHYSICS Higher Level Paper 2 N00/430/H(2)M INTERNATIONAL BACCALAUREATE BACCALAURÉAT INTERNATIONAL BACHILLERATO INTERNACIONAL 22 pages

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
SECTION A [1] A1. (a) Activity: the number of radioactive disintegrations per unit time. [2] (b) 0 () e t At A λ = [1] 0 ln ( ) ln A t =− Plot versus t [1] ln A (c) 0 1 2345 t / hr ln counts/sec R 1 2 [1] Yes consistent, since the points lie on a straight line in a semilog plot. [1] (d) Does not matter since R is proportional to A and hence graph will have the same behaviour characterised by the decay constant. (The semilog graph will have the same slope.) [2] (e) From (b) above, slope of semilog plot is . From graph, slope is . 1 1.5 0.375 hr 4 = [1] (f) Half-life: the time required for the activity to drop to half. (g) 0 e t A = , when , i.e. . [1] 0 2 A A = t τ = 1 e 2 t = [2] So , or . [1] ln 2 t =+ ln 2 = [1] (h) hours. ln 2 0.693 1.85 0.375 == = – 6 – N00/430/H(2)M
A2. (a) Figure 2 Figure 1 T M Mg m T mg M m Free-body Diagrams Physical System [3] (Award marks as follows: [1] for drawing two opposing forces on each block; [1] for drawing T less than Mg on M. [1] for drawing T greater than mg on m.) [2] (b) (i) [1] Tm gm a −= [1] Mg T Ma Block m Block M (ii) Add the two equations to eliminate T : , so [1] Mg mg Ma ma −=+ () Mm g Mm a + [1] gM m a = + To find T substitute a in first equation: [1] Mg M m Mg T + [4] [1] . 2 Mmg T = + continued… – 7 – N00/430/H(2)M

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Question A2 continued [3] (c) (i) If M >> m, then M will essentially fall freely, with m having little effect on it. Thus predict accleration g. [1] Block m will then be accelerated upward at g, which requires that T 2 mg. [2] [2] (ii) g if M >> m [1] () gM m a Mm = + 2 mg if M >> m [1] 2 Mmg T = + – 8 – N00/430/H(2)M
[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.) [1] (d) (i) 3 W. [3] (ii) 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 1 5 whole. 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] 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/30/2011 for the course CHEM 102 taught by Professor Tina during the Spring '11 term at Global.

### Page1 / 18

marscheme paper 2 2000 hl - INTERNATIONAL BACCALAURAT...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online