CAPE WINELANDS EDUCATION DISTRICT

Transcription

1 1 CAPE WINELANDS EDUCATION DISTRICT MEMORANDUM OF MARKING : September 2014 P.1 QUESTION C 1.2 D 1.3 C 1.4 D 1.5 B 1.6 D 1.7 B 1.8 B 1.9 D 1.10 B [20]

2 2 Question When a net force acts on an object, the object will accelerate in the direction of the net force. This acceleration is directly proportional to the net force and inversely proportional to the mass. (2) F friction F tension /F rope F app F T + F f AND F T < F f 5 forces, correctly labelled and sized ; 1 mark each (6) truck : N = F g = mg = (4 500)(9,8) = N F k = µ k N = µ (44 100) µ = 0,2 (3) bakkie: N = F g = mg = (1 250)(9,8) = N F k = µ k N = (0,2)(12 250) = N (3) Truck: F net = F applied + F friction + F tension 0 = ( ) + F tension F tension = 2450 N (3)

3 Total linear momentum in an isolated system is conserved (constant). OR In an isolated system the total (linear) momentum before collision equals the total (linear) momentum after collision. (2) km.h -1 = 13,8889 m s -1 p before = p after m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (1250)(13,8889) + 0 = 0 + (4500)(v 2 ) v 2 = 3,86 m s -1 forward (4) If the rope snaps the bakkie will according to Newton s First Law, continue to move forward due to inersia and collide with the truck.. With the solid bar it will remain a fixed distance behind the truck. Also as the truck stops, the bakkie will experience a net force in opposite direction of motion and also stop. (2) [25]

4 4 QUESTION m s -1 (downward) (1) v f = v i + 2a x 0 = (8) 2 + 2(-9,8) x x = 3,27 m Height above ground = 3,27 + 1,73 = 5 m (4) 3.3 v f 2 = v i a y = (-9.8)(2,5) v f = 3,873 m s -1 v f = v i + a t OR y = (v i + v f ) t 2 (-3,873) = (8) + (-9,8) t 2,5 = (8 3,873) t 2 t = 1,21 s t = 1,21 s (5) 3.4 Position (m) 3,27 Time (s) Correct max value on y-axis Start and end at zero Shape Correctly marked x and y axis (5) [15]

5 5 QUESTION contact force work done is dependent on the path taken (2) 4.2 force of air friction Tension/ Force of rope (2) 4.3 F tension / F cable / F rope 1 mark for each correct force (3) 4.4 The net work done by an object is equal to the change in its kinetic energy OR The work done by the net force is equal to the change in the objects kinetic energy 4.5 W net = E k F net xcosø = ½mv 2 f 2 - ½mv i F net (20)cos180 0 = ½(50)(0 2 ) - ½(50)(2 2 ) F net (- 20) = -100 N ma = + 5 (50) a = 5 a = 0,1 m s -2 upwards (6) [15] (2)

6 6 QUESTION Doppler flow meter (1) 5.2 Transmitter sends sound which is reflected off a red blood cell Reflected sound has different frequency so velocity of blood cells can be measured Doppler effect (1) F l = v ± v l F s v ± v s 0,9 F s = (340-0) F s (340 + v) v = 37,78 m.s -1 (4) [8] (2) QUESTION An electric field is a region of space in which an electric charge experiences a force. (The direction of the electric field at a point is the direction that a positive test charge would move if placed at that point). 6.2 (2) Shape direction (2)

7 7 kq 6.3 E = 2 r 9 (9 10 )( E = 2 (0,03) 6 ) E = 1 x 10 9 N C -1 (3) 6.4 The electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between their centres. 6.5 force on C due to A : force on C due to B: (2) F A = kq 1 Q 2 F B = kq 1 Q 2 d 2 d 2 = (9 x 10 9 )(100 x 10-6 ) 2 = (9 x 10 9 )(100 x 10-6 ) 2 (0,06) 2 (0,03) 2 = N ; right = N ; right OR Distance is halved F B = 4 F A = 4 (25 000) = N F net = F A + F B = = N; right (8) [17]

8 8 QUESTION r S V A (2) 7.2 Solve simultaneously for E and r using any two sets of readings. E = IR + Ir E = V + Ir E = 10,2 + 1,5 r (l) E = 9,6 + 2 r (ll) (ll) (l): 0 = - 0,6 + 0,5 r OR 10,2 + 1,5 r = 9,6 + 2 r r = 1,2 Ω Substitute r = 1,2 Ω into (ll): [OR into equation (I)] E = 9,6 + 2(1,2) = 12 V (Consider answers where learners use a graph to get to the answer) (7) [9]

9 9 QUESTION The emf of a cell is the total energy supplied, per coulomb of charge, by the chemical reaction occurring in the cell. OR The emf of a cell is the total work done by a cell in moving charges through a circuit. (1) 8.2 For R 1 V R1 = 10 V 4 V = 6 V V R1 R R1 = IR1 6 2 = I I R1 = 3 A (4) 8.3 emf = IR + Ir OR V lost = = 2V 12 = r r int = V lost /I r = ⅔ = 2/3 = 0,67 Ω = 0,67 Ω (3) 8.4 Curent through R 2 resistor: V R = I 4 2 = I I = 2 A For R 3 I = 3 2 = 1A thus R = V/I = 4/1 = 4 Ω (4) 8.5 R cir will decreases (as R 3 is replaced with conducting wire) so current in circuit increases (R 2 is short-circuited). V lost will also increase (because I increases and r is constant) V 1 will decrease [V across wire is 0 V] (less work is done across R 3 ) (4) [16]

10 10 QUESTION DC Motor. It has a split-ring commutator. (2) Electromagnetic Induction [ACCEPT: Faradays Law] (1) These connect the coil to the brushes (external circuit) contact or Allows electrical contact between coil and conducting wires or Ensures free rotation or Ensures that AC current is produced in the external circuit. [Any one] (1) (3) (3) [15] (5)

11 11 QUESTION C ejected electrons B cathode / zinc metal (2) 10.2 The photoelectric effect is the process whereby electrons are ejected from a metal surface when light of suitable frequency is incident on that surface. (2) 10.3 E = W 0 + E k [10] (6) TOTAL 150 MARKS