02138020/CAPE/MS 2016
CARIBBEAN
EXAMINATlONS
COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATIONS@
PHYSICS
UNITl_PAPER02
FINAI, MARK
SCHEME
MAY/JUNE 2016
'z-
o2t3ao2o/cAPE/Ms 2016
PHYS
ICS
UNITl_PAPER02
MARK SCHEME
Question
1
S.O:3.1,3.2,3.3,3.8
t,K
(a)
(i)
a
5
(ii
(b)
)
(i)
KC
xs
Acceleration is the rate of change of velocity
oR a : dvldt oR change of velocity divided by
elapsed ti:me
Unit: m s-'
Both definition and unit correct (1)
The Newton's Second Law:
. External Unba-l-anced force ,/ (resultant) (1)
. Produces an accel-erat'ion (rate of change of
momentum) directly proportionaf to the magnitude
of the unbalanced force in the di'rection of the
unbalanced force (1)
. E=ma (1)
Graph - See Page 3
AI1 7 Points correct
Plot:
5-6Pointscorrect
Lalce]. (1)
Sca].e (1)
Best ].ine (1)
t2t
(1)
1
(ii)
(iii)
The reaction time of the driver
(1)
Strai-ght line feature beyond 0'5s OR bet 0.5 & 3.5s
constant non-zero giadient
(1)
oR The slope of the graph between 0'5 & 3. 5s
OR
(iv)
Distance travelled : A-rea under graph (1) (1)
(Area of rectangle) + (Area of triangle)
: (15 , 0.5) + (l' 15 ' 3) (1)
i
1
SOI
2
4
: 7.5 + 22.5 : 30 m (1)
[acce1>t al.so area of TraPezium]
Eota]. 15 marks
6
4
5
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UNIT1-PAPER02
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Questi-on 1 cont'd
ISca]-e
il-l-:
I
r
- on x-axis
on y-axis
I :r
l.
4
2
:
cm q 1.O sec
96=2m-r
I.ll'.1
i :l
t
Labef
Scale
Plot
Best line
16
i
i
(1)
(f)
{2)
(1)
;14
5
d
0)
,
!)
8Lz
IO
8
:\
\
6
4
i--,
\i
\:
0.o
o.o
r.o
2.O
3.O
Tine (s)
rl
4.0
-4-
o2L3gO2O/cAPE/Ms 2016
PHYSICS
UNITl-PAPER02
MARK SCHEME
Question
2
S.o: 2.3, 2.5,2.7,2.9
uI(
(a)
(i)
7.6 mm (Read off from Figure
(ii
1B0o oR rI radians (accept antiphase or
)
liii
1
equi val ent ) (1)
1
V = f). (1)
1
SOI
:15x0.8:12ms-1(1)
(b)
(c)
1
between X and Y = 0o oR
radians (accept "in Phase") (1)
0
(i)
Phase difference
(j-i)
iii
Antinode - Point of maximum
displacement/amPlitude (1 )
Nod; - point of zero displacement/amplitude
: 2 (1)
Number of antinodes in Figure 3
(i)
Graph - See Page
(
xs
KC
1) (1)
(1)
5
points correct (2)
6 -.7 points correct (1)
A11
Plot
Lalce]- (1)
Scale (1); Best line(1)
(ii
)
Gradient
in wavel-ength) / (Change rn
= (Change
wave speed)
(0.4s - o.oo) /
G7
.s -
0.00)
o.oL2
(1)
Frequency : L / (gradi-ent of Plot)
=
-
L/
oa
o
.oL2
2
a
*
(1)
(Accept al so
a
Total 15 rnarks
corr ect
consistent
with
pl ot of v vs )' al ong
6
ana
5
4
ysl
s
q
}2L3AO2O /CAPE,/MS 201 6
PHYS
ICS
UNITl-PAPER02
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Questi-on 2 conL'd
liil:
- Scale:
Labcl
r - axis 2 cm =5 m s-r
- y - axis 4 cm =O.1m l:
..: :1.:....:l:.. I
I
:l
;.,.1
o.5
,-.1.--l t-
(
I
)
Plot a1I Points eortect (2) Best Line ( 1 )
--
-_:1
I
r ; ,,(
i
)
i
/
o.4
i
t
E
Ii
t-:- _ -
1l
tn
c
o
o
I
6
o.3
/
f'
o.2
/
{ {
/
I
L
{
I
.,1:
o.1
)
- -,
i
. i.,
-f-
-f
o
5
10
t5
20
2s
wave speed v/m s-l
30
3s
40
-5-
o 2L38O2O
/cAPE/Msl 201 6
PHYSTCS
UNITl_PAPER02
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Question
3
S.O: 6.6, 6-9, 6.10, 6.11
t,K
(a)
xs
Extension is directly proportional to the applied
force oR F = -kx (1) (accePt F = kx)
(i)
Provided that the elas ti-clProPortional
not exceeded (1)
(second mark cErll only be awarded if
correct)
(
KC
l-irn-it is
first
is
Hang a known mass from Lhe spring and measure the
ext.ension. (1)
ii)
Record mass and corresponding extension (1)
Repeat for 4 (rinim-rm) additional
masses (1)
use
PloL a graph of weight versus extension and
(1)
k
constant'.
spring
tf'"
the graph to determit'L
(accept aver age of f i ve( or mre) det ermi nat i ons )
Hang unknown werght from the spring and measure
.oti"=po*ding extension, x (1)
(b)
(i)
}vq
Determj-ne the welght of the rock using
mass(1)
the
and hence deternrine
: Area under graph
work done in extending spring
(1) sor
area
(nethod f or
: a.rea of A + area of rectangle + area between
of
curved Portion under AB and uPPermost side
squares
)
(c
:ount
AB
recLangle under
+(Lx B) + counting of squares (1)
:(a,rbxh)
sor
Calcu]-a
tl
the area
O.O29, 33) +
(0
016 , 33) + 49 squares (1)
1
1
1
sor
converting squares to q+ergy^
(5 x o.oo5)/25 : o 000lJ
1
=
(1)
"qlr"r"
So 49 squares = 49 x 0'0001 = o. o4eJ
Final cal culation and answer
: 0.48 J + 0.53 J +0.049J=1
i^7.d
(1)
06 + 0.02J
(any suitable,/acctrrate method used to arrive at
ttre final ans!iler within ttre tolerance
range will earn 4 rtarks)
(ii
6
Inel-astrc deformation:
1
1
-'7
-
02L38020 /CAPE,/MS 2O1 6
PHYSICS
UNITl-PAPER02
MARK SCHEME
a
applied Load > elastic Li-rtrit (1)
Therefore causes per:manent deformation (1)
Total 15 marks
5
4
6
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o2L38O 20 /CAPE,/MS 2016
ICS
PHYS
UNIT1_PAPER02
MARK SCHEME
Questi-on
4
UK
(a)
dr (1)
Upward
force, on l-ower surface (1)
(mrst rention pressure in order to get both of the
first 2 rmrks)
Upthrust is the upward force minus the dovmward
force (1)
L{hen upthrust is less than weight of obiect; obiect
will sink (1)
trdhen upthrust is egual to the weight of ob3ect' the
oblect will float (1)
(b)
KC
Downward force, on upper surface due to pressure at
Weight of water collected : 0'25 N (1)
(i)
(ii
Mass
)
of
HzO
:
0-
0-
H2O
000025
(1)
m3
1000
t
of
1
025 kg
1000 kglm3
5w:
l'. vo1 . of H2o = m= 0.025:
Volume
5
: volume of stone : o.ooo025
Density of stone :
(mass
of sLone) /
stone)
=
m3 (1)
3
(volume of
(0.1), / (0. 000025)
4000 kqm "
(1)
same in
The vol-ume of the liqurd displaced is the
and
water
both cases (when ornament is immersed in
(1)
when immersed in corn oil) '
(c) (i)
(ri
)
(rii)
than
Upthrust on ornament due to corn oil is less
(1)
that due to water
OR corn oi-I is fess dense than water
N
wt. of corn o-i1 displaced : (2'7 '1'87)N = 0'23
.. mass of corn oi-I displaced = 0'023 kq (1)
: (80 - 50) cm'
Volume of corn oil displaced
= 30 cm3 : 3Ox1O-6
(1)
1
1
I
m3
1
xs
-9-
o2t3ao
20
/cAPE/Msi
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PHYSICS
UNITl-PAPER02
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.'. Densi'ty of (r) corn oil :
Mass of corn oil : 0.023 = '761 kg m '
30"10-u
Vof . of corn oi-],
(comect units in order to obtain both rnarks)
Alternative
Of r
2
C
wei-ght of corn oj-1 d-isplaced = (2'7
1.87) =
0-23N
(1)
of corn oil displaced = (80 - 50) cm3 : 30x
(1)
10-6 m3
3]
below if presented in kgcm
ans.
[accept
Volume
DensitY of corn oi:l
of corn oil x g)
(lieight of corn oil)
:
'76'1 kgm-3
(1)
ans
(unit must be stated to earn unit mark)
S.O: 4.L, 4-2
/ (VoI'
(1) unit
10
5
- 10-
o2l38O2O /CAPE,/MS 2O1 6
PHYSICS
UNITl_PAPER02
MARK SCHEME
Question
5
(rnathod rm:st be
practically feasible to earn ANY of
the first 5 m^arks)
Any suitable experiment to prove Snell's
ExamPle: -
(a)
Diagram
UK
KC
T'aw
(2t
1. A clearly labelled diagram showing (i) glass
b1ock, (ii) norma;I, (iii) incident ray' (iv)
refracted ray (v) emergent ray, (vi)angIe of
.incidence, i, (vii) angle of refraction, r '
Sheet
of,
PaPef
Normal- -
2
-
(subtract 1 mark for everl'rnissing conqponent)
(sr:btract 1 mark for conqglete J-isting of
apparatus but no t{i agram)
l2l
Procedure
trace'
ray
obtaining
of
method
Credible
Credible method of obtaining i and r.
4
Indicate at least 4 repet'itions of 2 &' 3
(subtract 1 mark for onrission of any steps)
5
(1)
Precaution
accuracy
the
Any sensible step taken Lo improve
of the resul-ts.
Anal vsis & Interpretation
5. Pfot a graph of sin i versus sin r
(b)
(i)
'7
(1)
't. A straight
line passing through the ori:gin
verifies Snell's Law :
(1)
(sin i/sin r) = constant
medium into
parti-cular
one
from
for lrght travelling
another (1)
sPeed of light in bl-ock A : Speed of light iq air
Refractrve index of
A
2
3r1o8 : L.67 x 1o-8 m'-t (1)
1.8
XS
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o2t3so2o /cAPE,/Ms 201
PHYSICS
UNIT.l _
PAPER 02
MARK SCHEME
Questi-on 5 cont'd
(j-i)
(1) fonmrla
Sin 72 : 1-B
Sin 0
Sin 0 = sin (72) /l.B : 0.53
(1) substn &
= O = 30o
ans
(ful I mrks f or exact ans\4er : 31 9 degrees; even if
appr ox' deni ed
(1) for:rtr]-a
Sin O. : rr2/\1
(1)su.bstn
& ans
= 1.40 .'. 0" = 51"
1-BO
(Accept exact ans \,1er : 51. 05 de gr ees )
At AC boundary 0 i (angle of incidence) : 22 + O
a
)
(iii)
(iv)
(22 + 30"\
2
1
= 52o
(1)
angle of 51" at AC I
52' is greater than the critical
j-s
internalJ-y
totally
light
and therefore the
(1)
AC.
at
reflected
Tota1 15 marks I
7
6
-72PHYS
02L3AO2O/cAPE/Ms 2016
ICS
UNITl_PAPER02
MARK SCHEME
Question
6
S.O: 3-2| 3.3t 3.1
t,K
(a)
short wavelength radiation (mainly visible)
passes through the earth atmosphere (1)
KC
enters and
This j-s absorbed bY the earth (land c sea) and wa-rms
them up (1)
The warm land and sea then re-radiates outwards ' (1)
is longer wavelength IR
radiation
The re-radiated
radiation (1)
A portion of the outgoing IR radiation is trapped' by
the greenhouse gases in the atmosphere' (1)
6
This traPPed radiation causes the temPerature of the
earLh's atmosPhere to j-ncrease' (1)
(b)
Area of glass window : L x
(2. B
(1)
B
2.0):
4.8
m2
1
2
KA AE
t
(1)
,
Ax
0.03 m (1) (unit conversion)
AX
=-0
96
)
x 4-B x (35 - 26) /0-03 (1)
1382 J s-t (1)
(c)
Heat inPut into room to cause 3"c rise in
temperature: 2
AT (1)
5.7 x 10s,
C
3
1.?'106J(1)
Time required for this temperature rl-se
1.
7,106
1382
2
(1)
: 1,230 seconds
= 20.5 minutes
(1)
xs