Physics Exam AK

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Instructions for Marking the Examination2
In order to determine what is expected of the students and to ensure a uniform understanding of
the evaluation tools, it is suggested that teachers in each school form a marking committee to
analyze the work of a sample of students.
Guidelines for correcting questions requiring
an explanation, a justification or a representation:
Analyze the student’s work and determine if it is appropriate.
• An explanation, a justification or a representation is appropriate if most of the elements of
the answer are correct and if appropriate terminology or symbolism is used.
• An explanation, a justification or a representation is partially appropriate if:
> Most of the elements of the answer are correctly indicated, but the terminology or
symbolism used is not appropriate.
Some elements of the answer are indicated, and some of the terminology or
symbolism used is appropriate.
• An explanation, a justification or a representation is inappropriate if most of the elements
of the answer are incorrect or missing or if the terminology or symbolism used is
inappropriate.
Guidelines for correcting questions requiring
the use of formal mathematical solutions:
Step 1
Analyze the work to understand the procedure used by the student, and then decide if the
procedure is appropriate or not.
A procedure is appropriate if most of the steps are relevant and could lead to the correct
answer.
A procedure is partially appropriate if the steps presented do not lead to the correct answer,
but include at least one step that is relevant and correct.
A procedure is inappropriate if none of the steps presented are relevant or if the student has
not shown any work.
Step 2
If the procedure is deemed appropriate or partially appropriate, then evaluate the answer. If
the answer is incorrect, identify the type of error(s) made.
An error is considered minor if it is an error in calculation or transcription, or if the unit of
measurement is incorrect or missing.
An error is considered major if a Jaw, rule or formula has been applied incorrectly.
No marks are allotted for a correct answer when the procedure used is inappropriate, or no
work is shown.
2. Adapted from: MELS, 555-47 0, Science and Technology, Marking Guide, June 2072, and provided as a recommendation.
Secondary 5 Physics (Guide) Page 6
WQSB, June 2017

Part B
Constructed-Response Questions 11 to 25
NOTE:
11.
The following examples of appropriate responses are guidelines and are notexhaustive. Teachers should use their professionaljudgement when correcting thisexam.
Example of an Appropriate Procedure
Given variables:
Distance from the fence (Megan) = 0.60 m
Distance to the mirror (Megan) = 2.0 m
Distance to the mirror (Thomas) = 5.0 m
1. Determine the angle of incidence and reflection:
Tan e — Distance to the fence (Megan)
— Distance to the mirror (Megan)
1 (0.60 m8 = Tan
( 2.0 m
8=16.7°
2. Determine the distance away from the fence Megan can see on the other side:
Distance to the fenceTan 8=
Distance to the mirror (Thomas)
Distance from the fence = Tan 16.7° x 5.0 m
Distance from the fence = 1 .5 m
Answer
The shortest distance from the fence where Thomas can stand in order for Megan to see him inthe mirror is t5 m.
NOTE: Accept response if student draws a scale diagram and finds correct answer.
Secondary 5 Physics (Guide)
WQSB, June 2017
Marking Scale
4 marks Appropriate procedure with a correct answer
3 marks Appropriate procedure with a minor error such as a calculation or transcription error, or anincorrect or missing unit of measure in the answer
2 marks Appropriate procedure with a major error such as an incorrect application of a law, formulaor rule — e.g. inverses the distances in the calculation of the angle
1 mark Partially appropriate or incomplete procedure — e.g. only determine angle of incidenceOrnarks appropnate procedure or did not show any work regardless of the answer
__
_
Page 8

12.
a) The length is 5.0 cm.
The height is 5.40 cm. NOTE: Accept 5.39 or 5.41.
Iirking Scale
—-
1 mark Correct height and length expressed
0rnarkshicorrectrneasurement or no response provided
b) Given variables:
—
1=5.0cm
h=5.4Ocm
1. Determine the angle along medium inteace e
/
=tan1(-’ /
1 (5.40 cm
O=tan
5.0cm47
2. Determine the angle of refraction
eR=9O—e
7 2
8R9047
eR—43
Answer
The angle of refraction is 430
Marking Scale
3 marks Appropriate procedure with correct answer
2 marks Appropriate procedure with transcription error
1 mark Partially appropriate procedure — e.g. incorrect tan function
0 marks Inappropriate procedure or did not show any work, regardless of the answer
Page 9.
WQSS, June 2077

13.
Given variables:
“air = 1 .00
gass = = 1.55
8surface = 480
8 —8 —29°hydrocarbon — 3 —
1. Find angle of incidence 81
8 9Q01 — surface
8 =90°—48°
81 r=42°
2. Find angle of refraction in glass 82
n1 sin 8 n2 sin 82
82 = Sfl
sin
(1 J5Io
1(1.00sin42°
82 = sin
1.55
82 = 25.58°
3. Find index of refraction of hydrocarbon n3
Sifl 82 = sin 83
sine3
(1.55 sin 25.58°)
sin29°
n3 = 1.38143
n3 1.38, closest to hexane
Answer
Ramone has hexane in the glass jar.
Marking Scale
3 marks Appropriate procedure with a minor error such as a calculation or transcription error, or an
incorrect or missing unit of measure in the answer
2 marks Appropriate procedure with a major error such as an incorrect application of a law, formula
or rule — e.g. assumes 42° in glass jar
I mark Partially appropriate or incomplete procedure — e.g. determines angle of incidence
Page 10WQSB, June 2017

14.
Example of an Appropriate Set of Drawings
Diagram I — Converging Mirror
NOTE: Not all rays are needed to find the focal length.
IMarking Scale
2 marks Correct focal length and Correct ray diagram OR equation
1 mark Focal length is incorrect and some rays drawn correctly OR inappropriate use of the formula
O marks Incorrect ray diagram and focal length, or did not show any work
WQSB, June 2017
Diagram H — Convex (converging) Lens
NOTE: Not all rays are needed to find the virtual image.
rM
2 marks Correct image height and Correct ray diagram OR equation
1 mark Image height is incorrect and some rays drawn correctly OR inappropriate use of the
formula
0 marks Incorrect ray diagram and image height, or did not show any work
Page 11

15.
Given variables:
Concave mirror, therefore f— +6.0 cm
Virtual image (U0 and d have opposite signs)
Image is 2.5 times larger (M 2.5)
Step 1
Letd0=x, Letd1=-2.5x [oruseM=—-)
Step 2
1 11
U0 U f
1 1 1
=
x —2.5x 6.0cm
11_i
x — 2.5 x — 6.0 cm
2.5—1 1
2.5x 6.0cm
1.5 — 1
2.5x 6.0cm
(1.5) (6.0 cm)
U =x=
2.5
U0 = 3.6 cm in front of the mirror
d1 = —2.5 x= —2.5 (3.6 cm)= —9.0 cm
d1 = —9.0 cm or 9.0 cm behind the concave mirror
Answer
These virtual teeth will appear to be 9.0 cm from the mirror.
F
Marking Scale
or rule — e.g. did not recognize magnification of a virtual image as positive, OR did not
have a focal length of +6, used f= -6 instead.
1 mark Partially appropriate or incomplete procedure
WQSB, June 2017

16.
Given variables:
Ad=120m[NJ
M2=80m[EJ
M3=25m[S}
1. Find the total distance
MT _1213
MT =120m+80m+25m
MT =225m
2. Combine x and y components of the vectors
A = + +
MT 120 m EN]+ 80 m E]+ 25 m [s]
MT =95 m (N] +80 m [El
3. Find the_magnitude of the displacement
c = +
c=iJ(95m)2 +(80m)2
c=124m
4. Find the direction of the displacement
8=tan1
ad]
80 m
9 = tan
95m
9=40°
Answer
a) Emily’s total distance travelled is 225 m.
b) Emily’s total displacement is 124 m [N 400 E] OR 124 m [E 50° N].
IMarking Scale: FOUR MARKS IN TOTAL, BROKEN DOWN AS FOLLOWS:
1 mark Correct distance
2 marks Correct magnitude for displacement
WQSB, June 2017

17.
Find distance (mailbox to bus stop)
Ad = Ad of house to mailbox + Ad of house to bus stop
Ad =j_27 m[W]I+18 m[Efi= 45 m
2. Find Tom’s run to bus (distance that he covers)
At 11 seconds
v11 =0 mIs
At 17 seconds
v17 =9 m/s [E]or + 9m/s
Ad
= (v1 + v2 )At
2
Ad
= (v17 — v11)x (t17 — t11)
2
Ad=
(9m/s—O)x(17s—lls)
2
Ad=
(9m/sx6s)
2
Ad =27 m
3. Find time to bus stop
At17 seconds to bus stop
Ad to bus stop = 45 m —27 m =18 m
v =9 rn/s
At =?
Ad
V
At
At=
18m
— 9 m/s
Timéto teach bus stop
t=17 s+2s=19s
t=19s
/
4,
I’
At =2s
WQSB, June 2017
Page 14

:
Graph:
Velocity-Time Graph
Velocity
mis- _L_
Time (s):
-
Marking Scale
4 marks
3 marks
t2 marks
1 mark
0 marks
Correct calculations and graph with labels
Correct calculations and graph with a minor error — e.g. missing labels on graph
Correct calculations with a major error — e.g. line stops at 1 7 s or continues past 19 s
Partially appropriate response
Inappropriate_response or did not show any work
2 “ -
WQSB, June 2017
-5
i-t 1 18 20
Page 15

18.
Example of an Appropriate Procedure t. , Jil ..3ti ‘Q
Given variables:
ni1 =8.0 kg
m2=4Jkg L2)
Length of the ramp = 1.5 m
Ld mass, m1, travels = 1.5 m/2 = 0.75 m -
iy.
1. Balance the horizontal forces on each mass V
- I t. / , - O 1m1gsin9—T=m1a T—m2g=m2a .i i
I)
[%irQ151J2. Combine equations and solve for acceleration of the system -
1- !‘
m1g sin(e) — m2a — m2g = m1a
a
= g(m1 sin(9) — m2) V
LY
m1+m2
a = 9.8 m/s2 x
8.0 kg x sin35° — 4.0 kg
/8.Okg+4.Okg
a = 0.48 m/s2
3. Determine the time required for the mass, m1, to reach the bottom of the ramp
x =x. +v-t+—at2
0.75 m =0+0 (t) +i0.48}2
t = 1.77 seconds
Answer
The mass, m1, will reach the bottom of the ramp in 1.8 seconds.
Marking Scale
or rule — e.g. only determines the acceleration of the system
LOmarks appropate procedLwe or did not show any workegdless of the answer
WQSB, June 2017

19.
Given variables:
h=2.Om
v=28m/s
e = 600 above horizontal
Find horizontal and vertical components of initial velocity
v = vcos
v = (28 m/s)(cos 60°)
v. =14m/s
v=v1sine
v, = (28 m/s) (sin 60°)
v, = 24.2487 m/s
2. Solvefortime
v, = 24.25 m/s
ag = —9.8 mIs2
Ad=—2.0m
Ad=v. At+aAt2
‘‘
2
(—2.0 m)=(24.25 mls)At + 0.5 (_9.8m1s2)At2
0 = (—4.9 m/s)At2 + (24.25 m/s) At + (2.0 m)
_b±Jb2 4ac
At =
— —24.25 ± /(24.25)2 — 4 (4.9)(2.o m)
— 2(4.9)
__________
_
f ‘-S
=
At=5.Os
3. Solve for horizontal displacement
v=14m/s a=0
Atzz5.Os
Ad = v1At
Ad =(14 rn/s)(5.0 s)
Ad=ZOm
28 rn/s
vix
/
w
—
2&( :)
2a
J,JL{,Ls
Cc:)
——‘i2t(
5 ::4c
At
—24.25 ± 25.045
—9.8
-:
f
2,3
Z8
WQSB, June 2017
Page 17

19. (Cont’d)
Answer
The horizontal distance covered during this throw is 70 m.
Tiarking Scale
or rule — e.g. does not take /d into account, OR did not follow sign convention — e.g.
9.8 m/s2
1 mark Partially appropriate procedure e.g. only divides v1 into horizontal and vertical components I
0 marks inappropriate_procedure_or d not show any work, regardless of the answer
WQSB, June 2017

20.
Given variables:
j2
— — m,s
1
aMOOfl = — x — aEath
hmaximum = 2.50 m
1. Find aMOOfl
aMOOfl zJx_9.8m/s2
8Moon —1.63 rn/s2
2. Find v
Vf2 v2 + 2aM
Cm/s =v2 + 2(_9.8 rn/s2 )(2.5o m)
v zr7.00 rn/s
3. Find height on the Moon
Vf2 =v12 +2aAd
0 mis (7.00 mIs)2 + 2 (—1 .63 mIs) Ad
Ad =15.03m
Answer
Jane will reach a maximum height of 15 m while jumping on the trampoline on the Moon.
NOTE: Students may solve using energy transformation.
[Marking Scale
or rule — e.g. did not consider that final velocity is 0 m/s, or incorrect sign convention
1 mark Partially appropriate or incomplete procedure — e.g. only determines the acceleration on
the moon, or the initial velocity
WQSB, June 2017

21.
Given variables:
Weight of Rover on Earth = WRE = 8810 N
Weight of Rover on Mats = WRM = 3336 N
Weight of astronaut on Mars= WAM = 238 N
gE=9.8 N/kg
1. Find mass of rover (mR)
WRE = mRgE
_WAE_ 8810N
898.98kgmR — —
YE 9.8N/kg
62%
2. Find gravitational force on Mars (YM)
WRM mRgM
VV 3336N
3.711 N/kg
m 898.98 kg
3. Find mass of astronaut (mA)
WAM —mAgM
WAM 237N
m= = 64kg
YM 3.711N/kg
Answer
The mass of the astronaut is 64 kg.
33CtAJ ZA3
Marking Scale
or rule — e.g. only finds gravity of Mars
WQSB, June 2017
Page 20

22.
Given variables:
m=35kg
a =1.2 mIs2
Ff =12N
1. Find the net force in the direction of acceleration
et = ma
,et = (35 kg) (i .2 m/s2)
Fnet 42 N
2. Find the applied force in the direction of acceleration
Fnet Fappliedx + Ff
Fappliedx Fnet —
Fapphiedx = (42 N) — (12 N)
ppliedx 54 N
3. Find the applied force
F
— Fappliedx
applied
e
F
54N
applied
— cos 28
pplied = 61 N
Answer
Lucinda is applying a force of 61 N along the handle of the lawnmower.
MarkIng Scale
or rule — e.g. found the applied force in the direction of acceleration only
1 mark Partially appropriate or incomplete procedure — e.g. found F1
OmarksInappropriateprocedureorcdnotshowanywork,regaressoftheansw&
WQSB, June 2017

23.
Example of an Appropriate Drawing
Marking Scale
1 mark Opposite force vectors are the same length
___
___
___
___
___
_
irking Scale
3 marks Four correct force arrows with names and directions
2 marks Three correct force arrows with names and directions
1 mark Two correct force arrows with names and directions
0 marks One correct force arrow or no response given
________________________
Fnrmai
Fapplied
(engine)
-
Ffricton
(tires I air
resistance)
Fgravitationai
WQSB, June 2017 Page 22

Given variables:
k= 420 N/rn
Ax —12 cm
m=125g=0.125kg
Ahz:1.Om
Ad=1.4m
vJ
‘i2 H
& 1L
a. “
2( q- cc-S
OL
I
/
I
I 9
2’ LJ —) 3L-
_7 F
—) ti/)
24.
Exampie of an Appropriate Procedure
lilt
1. Find total mechanical energy at base of ramp
Esprjng = ‘/2k&2 = ‘/2(420 N/rn)(0. 12 rn)2 = 3.024 J
Etotai at base Espriiig =3024J
2. Find total mechanical energy at height of 1.0 m
mgAh= 0.125 kg (9.8 N/kg) (1.0 rn)= 1.225
Etotalatbase = Etotaiatim 1.0 in E + EKE
= Etotai at base —
3.024 J —1.225 J 1.799 J
FKE
1.799 =(0.125)v2
v/s V
Answer
The velocity of the block is 5.36 mIs at the top of the ramp.
Marking Scale
—
3 marks Appropriate procedure with a minor error such as a calculationör transcription error, or an
or rule — e.g. did not find potential energy
1 mark Partially appropriate or incomplete procedure — e.g. finds energy from spring
WQSB, June 2017
IL.
Page 23

25.
Example of an Appropriate Response
Largest < Potential Energy > SmaNest
A C B D
Marking Scale: FOUR MARKS IN TOTAL, BROKEN DOWN AS FOLLOWS:
> 0.5 mark per correct answer in the table
0
Greatest < Kinetic Energy > Smallest
B C A
WQSB, June 2017
Page 24

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