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SEMESTER 1

STAV Publishing 2007

PHYSICSUnit 3

Trial ExaminationSOLUTIONS BOOK

© STAV Publishing 2007 1 Physics Unit 3 Solutions

AREA 1 – MOTION IN ONE AND TWO DIMENSIONS

Q Answer Solution

1 338 N F = m a = 75 × 4.5 = 337.5 N

2 66°

The blocks exert an angled force as shown. The vertical component is equal to Craig's weight. The horizontal force is from Q1.

tan θ = 5.337

750 θ = tan−1 ⎟⎠⎞

⎜⎝⎛

5.337750 = 65.77°

3

θ block

9 m s−1u = 0, a = 4.5, t = 2, v = ?

v = u + a t = 0 + 4.5 × 2 = 9

4 12.1 s

for the first 2 seconds, s = u t + ½ a t2 = 0 + 0.5 × 4.5 × 22 = 9 m

for the remaining time, Craig covers 100 − 9 = 91 m

s = 91, u = 9, v = 9, a = 0, t = ?

s = u t + ½ a t2

91 = 9 t + 0 ∴ t = 91 ÷ 9 = 10.11 seconds

total time = 2 + 10.11 = 12.11 seconds

5

25 km/h

vCT = vCR + vRT

North

vTR = 60 km/h north

vCR = 85 km/h north vRT = 60 km/h south

vCT = 25 km/h north

u = 4.0, s = −150, a = −10, t = ?

6 5.9 s

s = u t + ½ a t2

−150 = 4 t − 5 t2

solve for t using the quadratic formula gives t = 5.89 or −5.09

OR solve for v first (−54.92 m s−1) then for t (5.892 s)

Reproduced under licence from STAV Publishing 2007


Q Answer Solution

7 6.26 m s−1

for Janette's vertical motion:

u = U sin 80°, v = 0 at the top, a = −10, s = 1.9

v2 = u2 + 2 a s

0 = (U sin 80°)2 + 2 × −10 × 1.9 = U2 (sin 80°)2 − 38

U2 = 2)80(sin38

°= 39.18

U = 6.2595

8 1.34 m

for Janette's vertical motion:

u = 6.26 sin 80° = 6.16, v = 0 at the top, a = −10, s = 1.9, t = ?

v = u + a t

0 = 6.16 − 10 t

t = 6.16 ÷ 10 = 0.616 s

then use this in the horizontal motion:

u = 6.26 cos 80° = 1.09, a = 0, t = 0.616 halfway, s = ?

s = u t + ½ a t2

s = 1.09 × 0.616 + 0

s = 0.67 m

∴ total horizontal distance = 2 × 0.67 = 1.34 m

9

1.24 m s−1

accept

1.2 – 1.3 m s−1

area under a force-time graph = impulse or Δp = m Δv

area = approximately 31 squares × area of one square

area = 31 × (50 × 1) = 1550 kg m s−1

1550 = m Δv

Δv = 1550 ÷ 1250 = 1.24 m s−1

10 730 J

e.p.e stored in catapault = area under the graph

area = triangle + trapezium

= (½ × 0.6 × 600) + ½ (600 + 1600) × 0.5

= 180 + 550 = 730 J



Q Answer Solution

7.64 m s−1

maximum possible speed is achieved if all e.p.e. is converted to k.e.

730 = ½ m v2

730 = ½ × 25 × v2 11

v2 = 730 ÷ 12.5 = 58.4

v = √(58.4) = 7.64 m s−1

12 The maximum horizontal distance (i.e. the range) is achieved from a launch angle of 45° to the horizontal (or to the vertical).

13 3.6 × 104 N

The tyres apply a sideways force equal in magnitude to the centripetal force

F = r

mv 2

= 61

451100 2× = 36516 N

14

Since the car is in uniform circular motion, the net force exerted by the road on the tyres will be towards the centre of the circle. Therefore the force exerted on the road by the tyres is outwards from the centre of the circle.

The change in kinetic energy equals the loss of g.p.e.

15

6.9 × 108 J

accept

6.6 – 7.2 × 108 J

Loss of g.p.e. = area under the graph × mass of the satellite

= approximately 25 squares × (200 × 103 × 0.5) × 275

= 6.9 × 108 J

The point of action of the force is variable.

16 0.96 m s−1

m1 u1 + m2 u2 = (m1 + m2) v

52 × 1.5 + 60 × 0.5 = (52 + 60) v

78 + 30 = 112 v

v = 108 ÷ 112 = 0.964



AREA 2 – ELECTRONICS AND PHOTONICS

Q Answer Solution

1

50 Ω

5.45 Ω

I: series resistors add together 20 + 30 = 50

II: this is a parallel combination

Rparallel = (20−1 + 10−1 + 30−1)−1 = 5.45

2 9 V Vout = 12 × ⎟⎠⎞

⎜⎝⎛

+18618 = 9 V

3 203 J

V = 0.5 V, I = 150 mA = 0.15 A, t = 45 min = 45 × 60 = 2700 s

E = V I t = 0.5 × 0.15 × 2700

= 202.5

4

405

C or coulombs

q = I t = 0.15 × 2700

= 405

5 2500 Ω

At 20°C the graph gives a resistance of 5 kΩ. Can use kΩ on the top and bottom of the voltage divider formula.

Vout = inVRR

R×⎟⎟⎠

⎞⎜⎜⎝

⎛+ 21

2

8 = 125

5

1

×⎟⎟⎠

⎞⎜⎜⎝

⎛+R

8 = 5

60

1 +R

8 R1 + 40 = 60

R1 = 20 ÷ 8 = 2.5 kΩ

6 D

Zero biased means the graph passes through (0,0).

Non-inverting means the graph has a positive gradient.

Graph D has both of these characteristics.

7 The capacitor filters (or blocks or decouples) the DC component of the input voltage allowing the small AC component to pass through.

8 1 mA

If Vout = 4.5 V, VRc = 6 − 4.5 = 1.5 V

IC = C

Rc

RV

= 3105.15.1×

= 1 × 10−3 A = 1 mA



Q Answer Solution

Vbe = 0.7 V

∴ VRb = 6 − 0.7 = 5.3 V

Ib will be 1/600th of Ic

Ib = 1 mA ÷ 600 = 1.67 × 10−6 A 3.2 × 106 Ω 9

Rb = 61067.13.5

−×=

b

Rb

IV

= 3,180,000 = 3.18 × 106

10 B, C, D

a thermistor's resistance varies depending on temperature ∴ not A

an LED emits light ∴ not E

a diode controls current flow ∴ not F

LDRs, photodiodes & phototransistors all detect light ∴ B, C, D



Detailed study 1 – Einstein's special relativity

Q Answer Solution

4 × 108 m s−1

the speed of each spacecraft is added in order to get the relative speed

2 × 108 + 2 × 108 = 4 × 1081

The observable speed of an object cannot exceed the speed of light 3 × 108 m s−1.

use the relative speeds equation

2 2.8 × 108 m s−1 V =

21cuvvu

+

+ =

16

16

8

1091041

104

××

+

× = 2.77 × 108 m s−1

3 45 m

γ =

2

2

1

1

cv

−

= ( )( )28

28

1031021

1

×

×−

= 1.3416

l = γol = 61 ÷ 1.3416 = 45.468

4 B T = γ To where γ =

2

2

1

1

cv

−

∴ answer B is correct

5 5.4 years γ = 1.3416 T = γ To = 1.3416 × 4 = 5.3664

6 8 × 10−16 J to convert eV to J, multiply by 1.6 × 10−19

5.0 keV = 5000 × 1.6 × 10−19 = 8 × 10−16 J

7 4.2 × 107 m s−1 KE = ½ m v 2 v = 31

16

101.910822

−

−

×××

=mKE = 4.193 × 107

8 1.01

(γ − 1) mo c 2 = KE γ = 12 +cmKE

o

γ = ( )2831

16

103101.9108

×××

×−

−

+ 1 = 0.009768 + 1 = 1.009768 = 1.01



9

total mass-energy = γ mo c 2 OR KE + mo c 2

γ =

2

2

1

1

cv

−

=

28

28

)103()105.1(1

1

××

−

= 1.1547 9.5 × 10−14 J

= 1.1547 × 9.1 × 10−31 × (3 × 108)2 = 9.457 × 10−14

10 1.05 × 10−30 kg m = mo γ

= 9.1 × 10−31 × 1.1547 = 1.05 × 10−30



Detailed study 2 – Investigating materials and their use in structures

Q Answer Solution

1 A brittle material fails just after its elastic limit. A ductile material has an elastic and a plastic region. OR A ductile material undergoes permanent deformation and a brittle material does not. OR A ductile material deforms plastically before fracture. A brittle material has little or no plastic deformation before fracture.

2 2.0 × 103 J

E = area under a stress-strain graph E = ½ base × height = ½ ε σ

= ½ × 2.3 × 10−4 × 1.7 × 107 = 1955 J Assumption: The elastic limit has not been exceeded OR The material has remained in the elastic region OR All energy is stored as EPE.

3 4.3 mm Y =

lAlFΔ×

×= 0

εσ 46 103.1104.0

5.1150×××

×=

××

=Δ −YAlFl

= 0.0043 m = 0.0043 × 103 mm = 4.3 mm

4 Reinforced concrete is an example of a composite material (many others possible).

Its application is in house slabs (or footpaths, driveways etc).

It is superior to non reinforced concrete because it combines the compressive strength of concrete with the tensile strength of the steel reinforcing rods.

5 4.9 × 104 N

tan θ = 5.25.1 θ = tan−1 ⎟

⎠⎞

⎜⎝⎛

5.25.1 = 31°

sin 31° = CF

102500× FC = °31sin

25000 = 48540 N

6

or

The rod is in compression so the forces acting on it push in towards the centre. The arrow therefore pushes up towards the centre of the rod. (Since the rod is light, its weight force can be ignored). If the weight is not ignored, the arrow pushes up along the rod and also up slightly.

7 B

To determine the force on an individual member of a structure, imagine what would happen if the member was replaced with a string or rope. Strings can only hold tension forces. If the road bed from A-B was replaced with a string, it would hold, so the road bed must be in tension. ∴ Answer B

8 4.7 × 104 N

Take torques about the right hand end of the drawbridge:

60,000 × 2.6 = T sin 40° × 5.2

T = 2.540sin6.260000

×°×

= 46672 N



Q Answer Solution

9 4.7 × 104 N The tension force in the rope must remain constant.

10

The equation in question 8 becomes:

Weight of knight and horse × distance from RH end + 60,000 × 2.6 = T sin 40° × 5.2 As the knight and his horse step onto the drawbridge, the value of T must increase initially. As the knight passes along the drawbridge, T will gradually reduce again, returning to 4.7 × 104 N when the knight is across the drawbridge.



Detailed study 3 – Further electronics

Q Answer Solution

1 34 V Vpeak-peak = 2 × √2 × Vrms = 2√2 × 12 = 33.9 V

2 20 2012240

==S

P

NN

3

OR

A bridge rectifier will give full-wave rectification either all above or all below the centre.

4



Q Answer Solution

5

A larger capacitor will increase the discharge time resulting in less ripple. Any graph with less ripple is okay.

The voltage at X-Y is a varying D.C. voltage. This variation is visible on a CRO but not on a digital voltmeter. 6

7 B τ = RC R = τ ÷ C = 6 ÷ (470 × 10−6) = 12766 Ω ≈ 13 kΩ

8 10 mV Ripple voltage is measured from a peak to a trough. The graph gives a value of 10 mV for this measurement.

The voltage regulator keeps the output voltage constant despite variations in the input voltage as long as the input voltage is above the voltage of the regulator i.e. ~ 6.3 V. 9

The heat sink dissipates the heat generated when high currents flow through the voltage regulator. 10

The zener diode is connected in reverse bias ∴ take the negative voltage reading where the graph goes vertical, indicating conduction. This value is ≈ 6.3 ± 0.2 V

11 6.3 V or −6.3 V

To use a zener diode as a voltage regulator it needs to be connected in reverse bias (i.e. "pointing" towards the positive terminal) and the output voltage must be taken across the zener. This is shown in diagram D.

12 D

IC or integrated circuit 13


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