Download 2006 VCAA Physics Exam 1 Solutions

2006 VCAA Physics Exam 1 Solutions
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SECTION A – Core
Area of study 1 – Motion in one and two dimensions
Q1
70 N
Q9 Relative to Mary, the ball has a horizontal velocity
component to the rear of the train. E.
Q10 Conservation of momentum:
6000× + 5.00 + M × 0 = (6000 + M )× + 0.098 , ∴ M = 3.0 × 10 5 kg.
Q11 Shuttle: impulse = change in momentum,
r
r
Fav × 20 = 6000× + 0.098 − 6000× + 5.00, ∴F av = − 1471 .
190 N
Average force on shuttle = 1.5 × 10 3 N.
F (driving force on the rear tyre
by the road)
Constant speed, no acceleration, ∴ Fnet = 0,
+
F + − 70+ − 190 = 0, ∴ F = 260 N, ∴driving force on the road
by the rear tyre is 260 N.
Q2
70 N
T
Constant speed, no acceleration, ∴ Fnet = 0,
∴T = 70 N.
+
T + − 70 = 0,
Q12 Total work done on Sam = heat energy generated due to
friction + increase in Sam’s gravitational potential energy.
22720 = 300 × L + 13720 , ∴ L = 30 m.
Q13 Increase in gravitational potential energy = mgh ,
∴13720 = 70 × 10 × h, ∴ h = 19.6 m.
Q14 Change in elastic potential energy = change in gravitational
1
potential energy, i.e. kx 2 = mgh,
2
1 2
∴ k 8 = 70 × 10 × 18, hence k = 3.9 × 10 2 Nm-1.
2
Q15 a = g ∝
Q3 When the cyclist stops pedalling, driving force = 0.
Work done by opposing forces = ∆E K ,
(70 + 190)x = 1 (90 + 40) × 6.0 2 , x = 9.0 .
2
Distance travelled during slowing = 9.0 m.
Q16 Since a = g in orbit, ∴
+
−
112
v2
= 60 ×
= 9.7 × 10 2 N
7.5
r
4π 2 r GM
= 2 ,
T2
r
4π 2 r 3
= 1.6 × 10 5 s.
GM
∴T =
Q4 Friction force F = weight force W = mg = 60 × 10 = 600 N
Q5 Fnet = ma, C = m
M
1
1
, ∴ a = 10 × × 2 = 4 ms-2.
2
1
10 ( 2 )
r
Area of study 2 – Electronics and photonics
Q1
Point
1
+
Q6 Fnet = 22+ 0.5 × 10= 17 N, resultant force = 17 N.
+
Fnet
17 +
=
= 34 ms-1 (constant upward acceleration),
m
0.5
u = 0, t = 1.5 s, s = ?
1
1
s = ut + at 2 = 0 + × 34 × 1.5 2 = 38.25 . Height = 38 m.
2
2
2
Q7 a =
3
Voltage (V)
20
1
× 20 = 4
1+ 4
20 − 8 mA × 1 kΩ = 12
Q2 ∆vOUT = −200 × ∆v IN = −200 × 50 mV = −10 V,
∆v R
+ 10
∴ ∆v R = +10 V. ∴ ∆iC =
=
= +0.01 A = +10 mAp-p.
RC
1000
C
Q8 Horizontal momentum = horizontal impulse
= 22 × 1.5 = 33 kgms-1.
p 33
= 66 ms-1.
After 1.5 s, horizontal velocity = =
m 0 .5
Vertical component: u = 0, a = −10, t = 1.5 , v = u + at = −15 .
∴speed = 15 2 + 66 2 = 68 ms-1
 15 
θ = tan −1   = 13 o .
 66 
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C
Q3 v IN is too large and clipping occurs. No change in period.
8 vOUT (V)
0
θ
4
time (ms)
− 12
1
Q4 A decoupler separates AC and DC components of signals in
a transistor amplifier. For a correctly biased transistor amplifier
only AC signals are fed to the input. To ensure this a capacitor
(CIN) acts as a decoupler to filter out the DC component from a
signal. Another capacitor (COUT) ensures only amplified AC
signals at the output. Capacitor CE filters out the AC component
across RE to maintain a constant biased DC voltage at the input.
Q5 Voltage across the 100-ohm resistor = 5.0 − 0.7 = 4.3 V,
V
4.3
∴I = =
= 0.043 A = 43 mA.
R 100
Q9 L ′ = L 1 −
∴1 −
v2
L′
v2
, ∴ = 1 − 2 = 99% = 0.99 ,
2
L
c
c
v2
= 0.9801 , ∴ v = 0.0199 × c = 4.23 × 10 7 ms-1.
c2
Q10 As the electron is accelerated to greater and greater speeds
close to c, its mass becomes larger and larger approaching
infinity. B
Q11 mc 2 = mo c 2 + E K .
C
Q6 Read from graph, 100Ω.
Q7 At T = 10 o C, RTHERM = 400 Ω, VTHERM = VOUT = 4.0 V,
∴ V R = 12 − 4.0 = 8.0 V.
Since
R
RTHERM
=
VR
VTHERM
, ∴R =
8.0
× 400 = 800 Ω.
4.0
Detailed study 2 – Investigating materials and their use in
structures
Q1 B
Q2 Polyethylene – it absorbs more energy before breaking
(larger area under σ-ε graph).
Q8 Modulation is a particular form of coding. Information
signal (sound signal) is used to modulate the light signal, which
is then transmitted through the fibre-optic cable. Demodulation
is the reverse process of modulation in which information
carried by the modulated light signal is recovered and decoded.
Q3 When placed under a tensile stress of 20 MPa,
∆L
ε=
≈ 2% = 0.02 (from graph) ∴ ∆L = 0.02 × 5.0 = 0.10 m.
L
Q9 At P, LED; at Q, LDR.
Q4 Young’s modulus =
SECTION B – Detailed studies
Detailed study 1 – Einstein’s special relativity
Q5 σ =
Q1 The experimenters did not detect any difference in the
speeds of two light beams (one parallel and one perpendicular to
the direction of the rotation of the Earth). One of the possible
explanations for the null result was that ether does not exist.
Q2 An inertial frame of reference is one in which Newton’s first
law (the law of inertia) is valid.
Q3 L is the proper length because it was measured by Kris who
was at rest in her rocket ship.
60
MPa = 3000 MPa.
0.02
F mg
=
,
A
A
σA 62 × 10 6 2.0 × 10 −4
∴m =
=
= 1.2 × 10 3 kg.
10
g
(
)(
)
Q6 Work = area under σ-ε graph × volume of material
1

=  × 62 × 10 6 × 0.021 × 2.0 × 2.0 × 10 − 4 ≈ 260 J.
2

(
)
Q7 Steel rods are strong in tension and used to reinforce
concrete under tensile stress.
Q8 C
Q4 L ′ = L 1 −
2
v
, ∴L =
c2
L′
v2
1− 2
c
.
B
Q9 D
Q10
FB
8.0 m
Q5 Lee: 340 − 30 = 310 ms-1; Sung: 340 + 30 = 370 ms-1.
Q6 Light propagates through empty space with a definite
velocity c = 3.0 × 10 8 ms-1 independent of the velocity of the
source and/or observer. A
Q7 Dilated time =
2.2µs
c
)
1 − ( 0.995
c
2
= 22 µs
1.0 m
B
FA
5000 N
Net torque about point B = 0, + FA × 8.0+ − 5000 × 1.0 + FB × 0 = 0 ,
∴ FA = 625 N.
2
 0.995c 
Q8 Contracted height = 2627 1 − 
 = 262 m
 c 
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Q11 D
2
Detailed study 3 – Further electronics
V 2 12 2
=
= 0.60 W
R
240
Q1 P =
N S VS
18
, ∴ NS =
=
× 4800 = 360
240
N P VP
Q2
Q3 V peak = 18 2 ≈ 25.5 V, period =
1
1
=
= 0.02 s = 20 ms.
f 50
B
Q4 None of the sketches shown. Voltage between points 2a and
2b is the same as that between points 3a and 3b. See Q5.
Q5 The capacitor smooths the voltage. D
Q6 The voltage regulator keeps the output voltage at 12 V.
360
× 140 = 10.5 VRMS = 14.8 Vpeak.
4800
After rectification this peak voltage falls below the voltage
required by the voltage regulator to operate in the voltageregulating region, causing the voltage output of the power
supply to fall below 12 V and possibly the appearance of ripples.
Q7 Output of transformer =
Q8 A
Q9 Output current increases; output voltage remains the same.
Q10 Time constant = RC = 2000 × 10000 × 10 −6 = 20 s
Q11
Voltage(V)
10
8
6
4
2
Time(s)
0
0
10
20
30
40
50
60
Q12 After 10 s the capacitor discharges by 63% of 10 V,
i.e. 6.3 V. Voltmeter reading = 10 − 6.3 = 3.7 V.
Please inform [email protected] re conceptual,
mathematical and/or typing errors
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