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Kansbag Physics Papre1 Marking Scheme Form4 Question Paper

Kansbag Physics Papre1 Marking Scheme Form4

Course:Secondary Level

Exam Year:2009

PHYSICS PAPER 1
MARKING SCHEME
SECTION I
1) 1 Ht =100x960
84
= 1142.9cm or 11.43m
2) Scalar quantity has magnitude and no direction e.g. mass, volume ,density;
Vector quantity has both magnitude and direction e.g. weight, displacement, velocity
3) Force = Pressure xArea
=1x10x40x10N
= 400N
4) Motion is faster and more random
5) –Narrow bore
–Large bulb
6) – length of conductor
–cross section area
–type of material
–temperature gradient
7) Linear expansivities of clay and cement are very different but for stone and cement they are close.
Strain produced by the difference in the expansion of clay and cement is larger than that produced in stone and cement.
8) Air trapped between the two blankets increase its insulation.
9)

10) Moment = force x distance
= 250x0.24 = 60Nm
11) Unstable
12)

13) V.R =2
M.A =100N

100 = 75
E 100
2

E = 100 x100
2x75
= 66.7N
SECTION II
14) a) The extension of an elastic body is directly proportional to the applied force if elastic limit is not exceeded
b) i) 14.5cm
ii) Spring Constant =
Slope =
= 3.33N/cm
iii) Fmin = 40N

c) i) M.A. =
= 25
ii) V.R = Effort distance
Load distance
= 1000
5
= 200
iii) Efficiency =
=
= 12.5%

15) a) i) Ball is falling towards the ground
ii) 9moon =slope
=4.8
3
=1.6m
iii) Distance =area under graph
=1/2x3x4.8+1/2x2x3.3
=10.5m
iv) Mass of ball =2/10=0.2kg
Weight on moon =0.2x1.6 = 0.32N
b) i) V = u+at
=0+2x6
=12mls
ii) s = ut+ ½ at
=0x6+1/2x2x(6)2
=36m
16) a) Rate of change of momentum of a body is directly proportional to applies force and acts in the direction of the force.
b) v= 1
0.02 =50cm/s
v2 = 3.5
0.02 =175cm/s
A = v2–v1
t
=175-50
5x0.02
=125
0.1
=1250cm/s
c) Initial momentum = Final momentum
50,000 x0+100x0 = 50,000xv+100x600
V =100*600
50,000
=-1.2m/s
17) a) i) Warm water has lower density and rises.
ii) For faster conduction of heat energy to the water
iii) Black colour is a good absorber of radiation energy
iv) –Convection through air
–Radiation
b) For paraffin; Px20x60=0.1x800x20
P = 0.1x800x20
1200
(P = Power of heater in W )
For water; P*t = 0.1*4200*20
P= 0.1x4200x20
t
0.1x4200x20 =0.1x800x20
t 1200
T =0.1x4200x20x1200
0.1x800x20
= 6300 sec or 105 Mins or 1hr 45min

18) a) V1 =2x10m P1 =1.5x107Pa T1 =283K
V2= ? P2 =1x105Pa T2 =296K
P1V1 = P2V2
T1 T2
V2 = 1.5x10x2x10x296
283x1x10
=3.14x10m
=0.314m
b)
V(cm3) 20 25 30 35 40
T (oC) 0 68.3 136 205 273

20 = 25 20 = V2
273 t2 273 409

T2 = 25x273 V2 = 20x409
20
=341.3K or 68.3oC =29.96 = 30cm

20 =40 20 = V2
273 t2 273 478
T2 =273x40 V2 = 20x478
20 273
=546k or 273 oC = 35cm3

Kansbag Physics Papre1 Marking Scheme Form4 Question Paper

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