Physical Sciences Grade 12 P1 Western Cape Mock 2014 Memo

Downloaded from hlayiso.com Western Cape Government Education CAPE WINELANDS EDUCATION DISTRICT MEMORANDUM OF MARKING : September 2014 P.1 QUESTION 1 141 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 Cc D Cc D B D B B D B vv vv vv vv vv vv vv vv vv vv [20]

Downloaded from hlayiso.com 2 Question 2 2.1.1. 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 Vv and inversely proportional to the mass. W (2) 2.1.2 AX Foormal Friction —ea>a\— Fapplied Frension /Frope Fay =Fr + F AND NP Fr< Fo ov F gravity 5 forces , correctly labelled and sized ; 1 mark each (6) 2.4.3 truck :N= Fy = mg =(4500)(9,8) = 44 100 N Fx = HKN v 8820 = u (44 100) v u - (02) v (3) 2.1.4 bakkie: N=\Fy = mg=(1250)(9,8) =12250N v Fx = UkN (0,2)(12 250) ¥ = 2450N v (3) 2.1.5 Truck: Fret = Fapplied + Frriction + Ftension v 0 = 11 270 + (- 8 820) + Fiension v Ftension = 2450 N v (3)

Downloaded from hlayiso.com 3 2.2.1 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) 2.2.2 50 km.h-1 = 13,8889 m·s-1 ∑ pbefore = ∑ pafter m1v1i + m2v2i = m1v1f + m2v2f


(1250)(13,8889) + 0 = 0 + (4500)(v2) v2 = 3,86 m·s-1
forward (4) 2.2.3 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]

Downloaded from hlayiso.com 4 QUESTION 3 3.1 8 m·s-1
(downward) (1) 3.2 vf2 = vi2 + 2a∆x


2 0 = (8) + 2(-9,8)∆x ∆x = 3,27 m
∴ Height above ground = 3,27 + 1,73 = 5 m (4) 3.3 vf2 = vi2 + 2 a∆y
2 = 8 + 2(-9.8)(2,5)
-1 vf = 3,873 m·s
vf = vi + a∆t OR ∆y = (vi + vf) ∆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]

Downloaded from hlayiso.com 5 QUESTION 4 4.1. contact force — work done is dependent on the path taken vv (2) 4.2 force of air friction V Tension/ Force of rope = ¥ (2) 4.3 * Frension / Feabie / Frope Frriction Feravity Vv 1 mark for each correct force (3) 4.4 The net work done by an object V is equal to the change in its kinetic energy Vv OR The work done by the net force V is equal to the change in the objects kinetic energy ¥ 4.5 Wret Fret Axcos@ v AE v Yomi - Yam? Fnet(20)cos180° = %(50)(07) - %(50)(22) v Fre(-20) = -100N ma = +5 (0)a Y = 5 a = 0,1m-s*v upwards Vv (6) [15]

Downloaded from hlayiso.com 6 QUESTION 5 5.1 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
(2) 5.3.1 Doppler effect
(1) 5.3.2 Fl = v ± vl Fs
v ± vs
0,9 Fs = (340 - 0) Fs (340 + v)
v = 37,78 m.s-1
(4) [8] QUESTION 6 6.1 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). (2) 6.2
Shape
direction (2)

Downloaded from hlayiso.com 7 kQ 6.3 E=
r2 (9 × 10 9 )(100 × 10 −6 ) E=
(0,03 ) 2 E = 1 x 109 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.
(2) 6.5 force on C due to A : force on C due to B: FA = kQ1Q2
FB = kQ1Q2 d2
d2 = (9 x 10 )(100 x 10-6)2 9 = (9 x 109)(100 x 10-6)2 (0,06)2
(0,03)2
= 25 000 N
; right = 100 000 N
; right OR Distance is halved ∴ FB = 4 FA = 4 (25 000) = 100 000 N Fnet = FA + FB
= 25 000 + 100 000 = 125 000 N; right
(8) [17]

Downloaded from hlayiso.com 8 QUESTION 7 7.1 r A S V (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]

Downloaded from hlayiso.com 9 QUESTION 8
8.1 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 R1 VR1 = 10 V – 4 V = 6 V VR1 R R1 =
IR1 6 2
=
I IR1 = 3 A
(4) 8.3 emf = IR + Ir
OR Vlost = 12 – 10 = 2V 12 = 10 + 3r
rint = Vlost/I
r = ⅔ = 2/3
= 0,67 Ω
= 0,67 Ω
(3) 8.4 Curent through R2 resistor: V R= I 4
2= I I = 2A
For R3 I = 3 – 2 = 1A thus R = V/I = 4/1
= 4 Ω
(4) 8.5 Rcir will decreases
(as R3 is replaced with conducting wire) so current in circuit increases
(R2 is short-circuited) . Vlost will also increase
(because I increases and r is constant) ∴ V1 will decrease
[V across wire is 0 V] (less work is done across R3) (4) [16]

Downloaded from hlayiso.com 10 QUESTION 9 9.1 DC Motor
. It has a split-ring commutator.
(2) 9.2.1 Electromagnetic Induction [ACCEPT: Faradays Law]
(1) 9.2.2 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) 9.2.3 (3) 9.3.1 (3) 9.3.2 (5) [15]

Downloaded from hlayiso.com 11 QUESTION 10 10.1 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 + Ek (6) [10] TOTAL 150 MARKS