# Physics Mark Scheme

**Topics:**GCE Advanced Level, Kinetic energy, General Certificate of Secondary Education

**Pages:**6 (975 words)

**Published:**May 7, 2013

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2007 question paper

9702 PHYSICS

9702/02 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination.

•

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

CIE is publishing the mark schemes for the May/June 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2

Mark Scheme GCE A/AS LEVEL – May/June 2007

Syllabus 9702

Paper 2 B2

1

(a) (i) all positions (accept 20, 40, 60, 80) marked to within ±5° positions are 40°, 70°, 90° and 102° (-1 for each error or omission) (ii) allow 107° → 113° (b) e.g. more sensitive at low volumes (do not allow reference to ‘accuracy’)

B1 B1

[3] [1]

2

(a) force per unit positive charge (on a small test charge) (b) field strength = (210/{1.5 × 10-2} =) 1.4 ×104 N C-1 (c) (i) acceleration = Eq / m = (1.4 × 104 × 1.6 × 10-19) / (9.1 × 10-31) = 2.5 × 1015 m s-2 (2.46 × 1015) towards positive plate / upwards (and normal to plate) (ii) time = 2.4 × 10-9 s (d) either vertical displacement after acceleration for 2.4 × 10-9 s = ½ × 2.46 × 1015 × (2.4 × 10-9)2 = 7.1 × 10-3 m (0.71 cm < 0.75 cm and) so will pass between plates i.e. valid conclusion based on a numerical value or 0.75 × 10-2 = ½ × 2.46 × 1015 × t2 t is time to travel ‘half-way across’ plates = 2.47 × 10-9 s (2.4 ns < 2.47 ns) so will pass between plates i.e. valid conclusion based on a numerical value

B1

[1]

A1 C1 C1 A1 B1 A1

[1]

[4] [1]

C1 A1 A1

[3]

(C1) (A1) (A1)

3

(a) mass / volume (ratio idea essential)

B1

[1]

(b) (i) mass = Ahρ (ii) pressure = force/area weight (of liquid)/force (on base) = Ahρg pressure = hρg (c) (i) ratio = 1600 or 1600:1 (ii) ratio = 3√1600 = 11.7 (allow 12)

B1 B1 B1 A0 A1 C1 A1

[1]

[2] [1]

[2]

© UCLES 2007

Page 3

Mark Scheme GCE A/AS LEVEL – May/June 2007

Syllabus 9702

Paper 2 B1 B1 B1 [1] [2]

(d) (i) density of solids and liquids are (about) equal (ii) strong forces: fixed volume rigid forces: retains shape / does not flow / little deformation (allow 1 mark for fixed volume, fixed shape) 4 (a) (i) (change in) potential energy = mgh = 0.056 × 9.8 × 16 = 8.78 J (allow 8.8) (ii) (initial) kinetic energy = ½mv2 = ½ × 0.056 × 182 = 9.07 J (allow 9.1) total kinetic energy = 8.78 + 9.07 = 17.9 J (b) kinetic energy = ½mv2 17.9 = ½ × 0.056 × v2 and v = 25(.3) m s-1 (c) horizontal velocity = 18 m s-1 (d) (i) correct shape of diagram (two sides of right-angled triangle with correct orientation) (ii) angle = 41° → 48° (allow trig. solution based on diagram) (for angle 38°→ 41° or 48°→ 51°, allow 1 mark) 5 (a) (i) vibrations (in plane) normal to direction of energy propagation (ii) vibrations in one direction (normal to direction of propagation)

C1 A1 C1 C1 A1 [2]

[3]

B1 B1

[1] [1]

B1 A2 [3]

B1 B1

[1] [1]

(b) (i) at (displacement) antinodes / where there are no heaps, wave has maximum amplitude (of vibration) B1 at (displacement) nodes/where there are heaps, amplitude of vibration is zero/minimum B1 dust is pushed to / settles at (displacement) nodes B1 (ii) 2.5λ = 39 cm v = fλ v = 2.14 × 103 × 15.6 × 10-2 = 334 m s-1 (allow 330, not 340) (c)...

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