Physics 12
June 2000 Provincial Examination
ANSWER KEY / SCORING GUIDE
CURRICULUM:
Organizers
Sub-Organizers
1. Vector Kinematics in Two Dimensions
and
Dynamics and Vector Dynamics
2. Work, Energy and Power
and
Momentum
3. Equilibrium
4. Circular Motion
and
Gravitation
5. Electrostatics
6. Electric Circuits
7. Electromagnetism
A, B
C, D
E
F, G
H
I
J
K, L
M, N
O, P
PART A: Multiple Choice (each question worth TWO marks)
Q
K
C
CO
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
A
C
D
C
D
A
B
C
C
A
B
D
C
B
B
K
U
U
K
U
K
U
U
U
K
U
K
U
U
U
1
1
1
1
1
2
1
2
2
3
3
3
4
4
4
PLO
Q
K
C
CO
A1
B2
B5
C1, 3
D4, C2
E4, E5
E7
E3, 5
E10
F5, 6, E8
H3, 2
H4
H5, 11
I5, D4
I4
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
D
D
D
D
C
A
B
C
B
C
C
B
D
A
A
K
U
K
U
H
K
U
H
K
U
U
U
U
U
H
4
4
4
5
5
5
6
6
6
7
7
7
7
7
7
PLO
J3, 2
J1, 2
K4
L8
L6
M9
M11
M5, 6
P2
O3, 1, 2
O4
O6
P4
P9
O5, P1, M5, C4
Multiple Choice = 60 marks
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July 20, 2000
PART B: Written Response
Q
B
C
S
CO
1.
2.
3.
4.
5.
6.
7.
8
9.
1
2
3
4
5
6
7
8
9
U
U
U
U
U
U
U
H
H
7
7
7
9
7
7
7
5
4
1
2
3
4
5
6
7
1
7
PLO
C8, C3, E8
G3
H11
J9, J8
L6
M5, 7
P11
A10, B2
I5
Written Response = 60 marks
Multiple Choice = 60 (30 questions)
Written Response = 60 (9 questions)
EXAMINATION TOTAL = 120 marks
LEGEND:
Q = Question Number
CO = Curriculum Organizer
PLO = Prescribed Learning Outcome
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B = Score Box Number
K = Keyed Response
-2-
C = Cognitive Level
S = Score
July 20, 2000
1.
A 75 kg Olympic skier takes 20 s to reach a speed of 25 m s from rest while descending a
uniform 16° slope.
16 °
What is the coefficient of friction between the skis and the slope surface?
a=
=
∆v
∆t
25 m s
20 s
= 1. 25 m s2
ma = mg sin θ − µmg cos θ
a = g sin θ − µg cos θ
µ=
=
g sin θ − a
g cos θ
← 2 marks


 ← 3 marks

← 1 mark
9.8 m s2 ⋅ sin16° −1. 25 m s2
9.8 m s2 ⋅ cos16°
= 0.15
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(7 marks)
← 1 mark
-3-
July 20, 2000
2.
Two steel pucks are moving as shown in the diagram. They collide inelastically.
Before Collision
After Collision
m1 = 4.2 kg
v1 = 1.8 m s
30 °
θ
v ′ = 2. 3 m s
v2
m2 = 1.3 kg
Determine the speed and direction (angle θ ) of the 1.3 kg puck before the collision.
p1 = m1v1
p1
p2
θ
=7
.6
= 4. 2 × 1.8
kg
⋅m
30 °
pT′ = 12.7 kg ⋅ m s
s
= 7. 6 kg ⋅ m s
pT′ = mT v1′
= 5. 5 × 2.3
= 12. 7 kg ⋅ m s
(7 marks)





 ← 1 mark





Method 1:
Cosine Law:
p2 2 = ( pT′ ) + p12 − 2 pT′ p1 cos 30°
2
= 12. 72 + 7. 6 2 − 2 × 12. 7 × 7. 6 × cos 30°
p2 2 = 51. 9
p2 = 51. 9 = 7. 20 kg m s
v2 =
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← 3 marks
p2 7. 20 kg m s
=
= 5. 5 m s
m2
1.3 kg
← 1 mark
-4-
July 20, 2000
Sine Law:
sin θ sin 30°
=
7. 6
7. 2
sin θ =
7. 6 × sin 30°
7. 2
sin θ = 0. 528
θ = 32°





 ← 2 marks




v2 = 5. 5 m s at 32°
Method 2: (one variation)
m1v1 cos 30° + m2 v2 cos θ = mT v ′
← 1 mark
4. 2 (1.8 ) cos 30° +1.3( v2 ) cos θ = ( 4. 2 + 1.3)( 2.3) ← 1 mark
v2 =
4. 69
cos θ
← 1 mark
m1v1 sin 30° + m2 v2 sin θ = 0
← 1 mark
4. 2 (1.8 )sin30° +1.3( v2 ) sin θ = 0
← 1 mark
v2 =
2. 91
sin θ




 ← 1 mark




4. 69 2. 91
=
cos θ sin θ
sin θ
2. 91
=
cos θ 4. 69
tan θ = 0. 618
θ = 32°
v2 =


 ← 1 mark


4. 69
cos 31.8
v2 = 5. 5 m s
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July 20, 2000
3.
A uniform 6.0 m-long boom has a mass of 55 kg. It is kept in position by a restraining cable
attached three-quarters of the way along the boom.
Cable
70 °
Boom
m = 150 kg
50 °
Hinge
What is the tension in this cable when the boom supports a 150 kg mass as shown?
(7 marks)
T
FL
?
Fb
About the hinge:
Στ = 0
← 1 mark
τ cable − τ mass − τ boom = 0
← 1 mark
( T cos 20 ) 43 ⋅ 6  − (150 ( 9.8 ) cos 50 )( 6 ) − ( 55( 9.8 ) cos 50 ) 12 ⋅ 6  = 0
← 4 marks
T = 1 600 N ← 1 mark
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July 20, 2000
4.
A space shuttle is placed in a circular orbit at an altitude of 3. 00 × 10 5 m above Earth’s
surface.
EARTH
a)
What is the shuttle’s orbital speed?
Fc = Fg
m
v 2 GMm
=
R
R2
GM
v2 =
R
v=
=
GM
R
6. 67 × 10 −11 × 5. 98 × 10 24
6. 68 × 10 6
v = 7. 73 × 103 m s
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(5 marks)
← 1 mark



 ← 2 marks



← 1 mark
← 1 mark
-7-
July 20, 2000
b)
The space shuttle is then moved to a higher orbit in order to capture a satellite.
Original orbit
New orbit
EARTH
The shuttle’s speed in this new higher orbit will have to be
p
4
p
p
greater than in the lower orbit.
less than in the lower orbit.
the same as in the lower orbit.
(Check one response.)
c)
(1 mark)
Using principles of physics, explain your answer to b).
(3 marks)
As the space shuttle moves further away from the earth’s centre the force of gravity acting
on the shuttle decreases. Since the centripetal force is provided by the force of gravity, it
must decrease as well. ← 2 marks
The smaller centripetal force generates a smaller centripetal acceleration ← 1 mark
which in turn requires a smaller orbital velocity.
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July 20, 2000
5.
a)
How much work is done in moving an electron from point X to point Y?
Ð
electron
Ð
X
Y
Ð
Q = − 5. 0 × 10 −6 C
0.50 m
1.00 m
W = ∆Ep


 ← 2 marks

=
kQq kQq
−
r2
r1
=
9. 0 × 10 9 × −5. 0 × 10 −6 × −1. 60 × 10 −19 9. 0 × 10 9 × −5. 0 × 10 −6 × −1. 60 × 10 −19
−
1. 00
1. 50
= 2. 4 × 10 −15 J
b)
W
= 1. 5 × 10 4 V
q
← 2 marks
← 1 mark
What is the potential difference between point X and point Y?
∆V =
(5 marks)
(2 marks)
← 2 marks
Note: Both positive and negative answers will be accepted for b).
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July 20, 2000
6.
A current of 1. 50 A flows through the 40. 0 Ω resistor.
R1 = 6.0 Ω
I 3 = 1.50 A
R3 = 40.0 Ω
R2 = 10.0 Ω
V
R4 = 4.0 Ω
What is the potential difference of the power supply?
V3 = I3 R
= 1. 50 ( 40. 0 )
V3 = 60. 0 V
V2 = V3 = 60. 0 V
I2 =
V2 60. 0
=
= 6. 00 A
R2 10. 0
It = I3 + I2 = 1. 50 + 6. 00 = 7. 50 A
V1 = I1 R1
V1 = 7. 50 ( 6. 0 )
V1 = 45 V
V 4 = I4 R4
= 7. 50 ( 4. 0 )
V 4 = 30 V
Vt = Vb = V1 + V| | + V 4
= 45 + 60 + 30
Vb = 135 V
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(7 marks)



 ← 2 marks



← 1 mark
← 1 mark


 ← 1 mark




 ← 1 mark




 ← 1 mark


- 10 -
July 20, 2000
Alternate Solution:
V3 = I3 R
= 1. 50 ( 40. 0 )
V3 = 60. 0 V
V2 = V3 = 60. 0 V
I2 =
V2 60. 0
=
= 6. 00 A
R2 10. 0
← 1 mark
It = I3 + I2 = 1. 50 + 6. 00 = 7. 50 A
← 1 mark
1
1
=
1
1
1
1
+
+
R3 R2
40. 0 10. 0




 ← 2 marks




Rp =
= 8. 00 Ω
RT = 6. 0 Ω + 8. 0 Ω + 4. 0 Ω
= 18. 0 Ω
V0 = ( I0 )( RT )
= ( 7. 50 )(18. 0 ) = 135 V
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


 ← 2 marks




 ← 1 mark

- 11 -
July 20, 2000
7.
A transformer has 840 primary and 56 secondary windings. The primary coil is connected to a
110 V ac power supply which delivers a 0.30 A current to the transformer.
a)
Find the secondary voltage.
Vs
N
= s
Vp Np
Vs
56
=
110 840
V s = 7.3 V
b)



 ← 4 marks



Find the secondary current.
Ip
N
= s
Is
Np
0.30
56
=
Is
840
Is = 4. 5 A
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(4 marks)
(3 marks)



 ← 3 marks



- 12 -
July 20, 2000
8.
The data table shows the velocity of a car during a 5. 0 s interval.
a)
t (s)
0. 0
1. 0
2. 0
3. 0
4. 0
5. 0
v ( m s)
12
15
15
18
20
21
Plot the data and draw a best-fit straight line.
(2 marks)
25
v (m s )
20
15
10
5
0
0.0
1.0
2.0
3.0
4.0
5.0
t (s)
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July 20, 2000
b)
Calculate the area bounded by the graph and the time axis between t = 0. 0 s and
t = 5. 0 s.
(2 marks)
Area = 12 ( a + b ) c
≅ 85 m
c)
← 2 marks
What does this area represent?
(1 mark)
The area represents the distance (displacement) travelled by the car in the time
t = 0 to t = 5 s .
“Metres” only ←
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1
2
mark
- 14 -
July 20, 2000
9.
A mass is suspended by a string attached to a spring scale that initially reads 14 N as shown in
Diagram 1.
Spring scale
F = 14 N
Rope
Q
The mass is pulled to the side and then released as shown in Diagram 2.
Spring scale
F=?
Rope
Q
As the mass passes point Q, how will the reading on the spring scale compare to the previous
value of 14 N? Using principles of physics, explain your answer.
(4 marks)

mv 2 
The reading will be greater than 14 N.  by
← 1 mark
r 

Initially, the net force is zero, so the spring scale reads the weight of the mass. When
moving, there is a net (centripetal) force provided by the spring scale (tension in the rope)
which exceeds the weight (force of gravity) of the mass so that the mass goes in a vertical
circle. ← 3 marks
END OF KEY
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July 20, 2000