9702_s11_ms_21.pdf

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2011 question paper

for the guidance of teachers

9702 PHYSICS

9702/21

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 con j

unction with the question papers and the report on the

examination.

? Cambridge will not enter into discussions or correspondence in connection with these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

w w w .X t r e m e P

a p e r s .

c o m

1 (a) (i) metre rule / tape (not ‘rule’) B1 [1] (ii) micrometer (screw gauge) / digital caliper B1 [1]

(iii) ammeter and voltmeter / ohmmeter / multimeter on ‘ohm’ setting B1 [1]

(b) (i) resistivity = RA / L C1 = [7.5 × π × (0.38 × 10–3)2 / 4] / 1.75 M1

= 4.86 × 10–7

? m A0 [2] (ii) (uncertainty in R =) [0.2 / 7.5] × 100 = 2.7% and (uncertainty in L =) [3 / 1750] × 100 = 0.17% C1 (uncertainty in A =) 2 × (0.01 / 0.38) × 100 = 5.3 % C1 total = 8.13% C1 uncertainty = 0.395 × 10–7 (? m) A1 [4] (missing 2 factor in uncertainty in A, then allow max 3/4) (c) resistivity = (4.9 × 10–7 ± 0.4 × 10–7) ? m A1 [1]

2 (a) work done is the force × the distance moved / displacement in the direction of the

force

or work is done when a force moves in the direction of the force B1 [1] (b) component of weight = 850 × 9.81 × sin 7.5° C1 = 1090 N A1 [2] (use of incorrect trigonometric function, 0/2) (c) (i) Σ F = 4600 – 1090 = (3510) M1 deceleration = 3510 / 850 A1

= 4.1 m s –2

A0 [2]

(ii) v 2 = u 2 + 2as 0 = 252 + 2 × – 4.1 × s C1 s = 625 / 8.2 = 76 m A1 [2] (allow full credit for calculation of time (6.05 s) & then s )

(iii) 1. kinetic energy = ? mv 2 C1 = 0.5 × 850 × 252 = 2.7 × 105 J A1 [2] 2. work done = 4600 × 75.7 = 3.5 × 105 J A1 [1] (iv) difference is the loss in potential energy (owtte ) B1 [1]

3 (a)point where the weight of an object / gravitational force M1

may be considered to act A1 [2]

(b)product of the force and the perpendicular distance (to the pivot) B1 [1]

(c) (i) 1. sum / net / resultant force is zero B1

2. net / resultant moment is zero

]

sum of clockwise moments = sum of anticlockwise moments B1 [2]

(ii)W × 0.2 = 80 × 0.5 + 70 × 1.3 C1

= 40 + 91 C1

W= 655 N A1 [3]

(allow 2/3 for one error in distance but 0/3 if two errors)

(iii)move pivot to left (M1)

gives greater clockwise moment / smaller

moment (A1)

anticlockwise

or

move W to right (M1)

gives smaller anticlockwise moment (A1) [2]

4 (a) (i)stress is force / area B1 [1]

(ii)strain is extension / original length B1 [1]

(b)

(i)E = [F / A] ÷ [e / l C1

e = (25 × 1.7) / (5.74 × 10–8 × 1.6 × 1011) C1

e = 4.6 × 10–3 m A1 [3]

(ii)A becomes A/2 or stress is doubled B1

e∝l / A or substitution into full formula B1

[3

total extension increase is 4e A1

5 (a) (i)I = 12 / (

6 + 12) C1

minimum current = 0.67 A A1 [2]

(ii)correct start and finish points M1

correct shape for curve with decreasing gradient A1 [2]

(b)maximum current = 2.0 A A1

minimum current = 0 A1 [2]

(c) (i)smooth curve starting at (0,0) with decreasing gradient M1

end section not horizontal A1 [2]

(ii)full range of current / p.d. possible

or currents / p.d. down to zero

or brightness ranging from off to full brightness B1 [1]

6 (a) any two of: large number of molecules / atoms / particles molecules in random motion no intermolecular forces elastic collisions time of collisions much less than time between collisions volume of molecules much less than volume of containing vessel B1 + B1 [2] (b) molecules collide with the walls change in momentum of molecules implies force (on molecules) molecules exert equal and opposite force on wall pressure is averaging effect of many collisions (any three statements, 1 each ) B3 [3]

7 (a) when waves overlap / meet, (resultant) displacement is the sum of the individual

displacements B1 [1]

(b) (i) two (ball-type) dippers (M1) connected to the same vibrating source /motor (A1) or one wave source described (M1) with two slits (A1) [2]

(ii) lamp with viewing screen on opposite side of tank B1 means of freezing picture e.g. strobe B1 [2]

(c) (i) two correct lines labelled X B1 [1]

(ii) correct line labelled N B1 [1]