All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.

Information Report

Category:
## Creative Writing

Published:

Views: 3 | Pages: 4

Extension: PDF | Download: 0

Share

Related documents

Description

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 question paper for the…

Transcript

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 9702 PHYSICS 9702/22 Paper 2 (AS Structured Questions), maximum raw mark 60 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. ã Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 22 V πP r4 1 (a) = t 8Cl C = [π × 2.5 × 103 × (0.75 × 10–3)4] / (8 × 1.2 × 10–6 × 0.25) C1 = 1.04 × 10–3 N s m–2 A1 [2] (b) 4 × %r C1 %C = %P + 4 × %r + %V/t + %l = 2% + 5.3% + 0.83% + 0.4% (= 8.6%) A1 ∆C = ± 0.089 × 10–3 N s m–2 A1 [3] (c) C = (1.04 ± 0.09) × 10–3 N s m–2 A1 [1] 2 (a) (i) v2 = u2 + 2as = (8.4)2 + 2 × 9.81 × 5 C1 = 12.99 m s–1 (allow 13 to 2 s.f. but not 12.9) A1 [2] (ii) t = (v – u) / a or s = ut + ½at2 = (12.99 – 8.4) / 9.81 or 5 = 8.4t + ½ × 9.81t2 M1 t = 0.468 s A0 [1] (b) reasonable shape M1 suitable scale A1 correctly plotted 1st and last points at (0,8.4) and (0.88 – 0.96,0) with non-vertical line at 0.47 s A1 [3] (c) (i) 1. kinetic energy at end is zero so ∆KE = ½ mv2 or ∆KE = ½ mu2 – ½ mv2 C1 = ½ × 0.05 × (8.4)2 = (–) 1.8 J A1 [2] 2. final maximum height = (4.2)2 / (2 × 9.8) = (0.9 (m)) change in PE = mgh2 – mgh1 C1 = 0.05 × 9.8 × (0.9 – 5) C1 = (–) 2.0 J A1 [3] (ii) change is – 3.8 (J) B1 energy lost to ground (on impact) / energy of deformation of the ball / thermal energy in ball B1 [2] 3 (a) A body continues at rest or constant velocity unless acted on by a resultant (external) force B1 [1] (b) (i) constant velocity/zero acceleration and therefore no resultant force M1 no resultant force (and no resultant torque) hence in equilibrium A1 [2] (ii) component of weight = 450 × 9.81 × sin 12° (= 917.8) C1 tension = 650 + 450 g sin12° = (650 + 917.8) C1 = 1600 (1570) N A1 [3] © University of Cambridge International Examinations 2012 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 22 (iii) work done against frictional force or friction between log and slope M1 output power greater than the gain in PE / s A1 [2] 4 (a) total resistance = 20 (kΩ) C1 current = 12 / 20 (mA) or potential divider formula C1 p.d. = [12 / 20] × 12 = 7.2 V A1 [3] (b) parallel resistance = 3 (kΩ) C1 total resistance 8 + 3 = 11 (kΩ) C1 current = 12 / 11 × 103 = 1.09 × 10–3 or 1.1 × 10–3 A A1 [3] (c) (i) LDR resistance decreases M1 total resistance (of circuit) is less hence current increases A1 [2] (ii) resistance across XY is less M1 less proportion of 12 V across XY hence p.d. is less A1 [2] 5 (a) E = stress / strain B1 [1] (b) (i) 1. diameter / cross sectional area / radius 2. original length B1 [1] (ii) measure original length with a metre ruler / tape B1 measure the diameter with micrometer (screw gauge) B1 [2] allow digital vernier calipers (iii) energy = ½ Fe or area under graph or ½ kx2 C1 = ½ × 0.25 × 10–3 × 3 = 3.8 × 10–4 J A1 [2] (c) straight line through origin below original line M1 line through (0.25, 1.5) A1 [2] 6 (a) two waves travelling (along the same line) in opposite directions overlap/meet M1 same frequency / wavelength A1 resultant displacement is the sum of displacements of each wave / produces nodes and antinodes B1 [3] (b) apparatus: source of sound + detector + reflection system B1 adjustment to apparatus to set up standing waves – how recognised B1 measurements made to obtain wavelength B1 [3] (c) (i) at least two nodes and two antinodes A1 [1] (ii) node to node = λ / 2 = 34 cm (allow 33 to 35 cm) C1 c = fλ C1 f = 340 / 0.68 = 500 (490 to 520) Hz A1 [3] © University of Cambridge International Examinations 2012 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 22 7 (a) W = 1 and X = 0 A1 [1] Y=2 A1 [1] Z = 55 A1 [1] (b) explanation in terms of mass – energy conservation B1 energy released as gamma or photons or kinetic energy of products or em radiation B1 [2] © University of Cambridge International Examinations 2012

Recommended

5 pages

5 pages

4 pages

4 pages

4 pages

4 pages

4 pages

4 pages

5 pages

5 pages

4 pages

Related Search

We Need Your Support

Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks