Download VCE Physics Exam 2 2003 Solutions

VCE Physics Exam 2 2003 Solutions
AREA 1: Motion
Q1 Distance travelled = area under speed-time
graph
 10 105  115m .
Q2 Acceleration = gradient =
v
30
 10ms  2 .
=
t
3
Q3 Total distance = 115 + 300 + 210 = 625m
625
 52ms 1 .
Average speed 
12
Q4 80kmh1 
80
ms 1 .
3.6
1
1
80 
Car B: s  u  v t , 400   0 
t , t  36
2
2
3.6 
80
 36  800m
Car A: s  ut , d 
3.6
Q5 Both cars have the same speed 80kmh1 at tY .
Distance covered by car A is twice the distance
covered by car B at tY . C.
Q8 Constant velocity, Fnet  0 .

Fd   5000  3  2000  0 , Fd  11000N .
Q9 ma  Fnet ,
100.0  10 3   a   4.6  10 4   11000 ,
a  0.35ms 2 .
Q10
2000N
Consider net force on the last carriage,
ma  Fnet , 20.0  10 3   0.20  T   2000 ,
T  6.0  10 3 N .
Q11 For uniform circular motion, direction of the
net force is towards the centre of the path. C.
Q12
Normal reaction of the road on the car
Q6 Kinetic energy = elastic potential energy
1 2 1 2
mv  kx ,
2
2
m
200
x
v 
 2.0  0.050m
k
3.2 105
Q7 Initial momentum  200  2.0  400kgms1 .
Final momentum  200  2.0  400kgms1 .
T
Force of friction between the
tyres and the road
Weight of the car
Q13 G.
Consider the dodgemcar and the earth as an isolated
system.
Change in momentum of the dodgem car =
Final momentum – initial momentum =

400  400  800kgms1 .
Therefore the change in momentum of the earth
must be  800kgms1 , so that the change in
momentum of the isolated system is zero, i.e. the
total momentum of the system remains constant.
Q14 The wheels spin in clockwise direction and
push the road backwards at the points of contact due
to friction between the tyres and the road surface.
According to Newton’s third law, the road pushes
back on the wheels in the forward direction.This
accelerates the car forward if the force is greater
than the total resistance to motion.
AREA 2: Gravity
AREA 3: Structures and materials
3
2
r 3 GM
2r

,T
.
2
2
T
4
GM
r  0.4  10 6  3.4  10 6  3.8  10 6 m
Q1 Kepler’s third law:
T

2  3.8  10 6

3
2
6.67  10 11  6.4  10 23
 7124s  2.0h
Q2 Find the area under the graph from 3 200 000m
to 1 200 000m. This area gives the decrease in
gravitational potential energy that is transformed to
kinetic energy (increase) of the spacecraft. Thus the
total kinetic energy of the spacecraft is the sum of
the initial kinetic energy (at 3200km) and the
increase in kinetic energy (area under graph). Then
1
use mv 2  E K to determine its speed at 1200km.
2
Q1 Bricks and stones are weak in tension but very
strong in compression. In older structures, e.g.
bridges, arches made with stones and bricks were
employed. Such a structure is always under
compression, due to its own weight and the load it
supports, which keeps the stones or bricks in
position together.
Q2 Since pin P is in equilibrium, net force on it is
zero. The upper walkway exerts a downward force
of 5000N on it, therefore the rod exerts an upward
force of 5000N on it.
Q3 B, for the same reason as in Q2.
Q4
T
v 2 GM
GM
Q3 a  g ,
,
 2 , v
r
r
r
16sin50o
6.67  10 11  2.0  10 30
v
 4.5  10 3 ms 1
12
6.5  10
50o
O
700g
Q4 Incorrect. If Kiera was correct, any one on
Earth would experience apparent weightlessness
because both the person and the Earth would be
accelerating towards the Sun at the same rate.
However we do not experience apparent
weightlessness on Earth because there is force of
gravity between the Earth and us. Similarly, there is
force of gravity on Quaoar but the effect is small
because the mass of Quaoar is much smaller than
the mass of the Earth.
16m
12m
10000g
If the walkway is in equilibrium, the net torque
equals zero about point O.

T  16 sin 50 o   10700 g  12  0 , T  1.0  10 5 N .
The cable will hold.
1.5  10 5
0.01
 0.0004
,
2
25
r
1.5  10 5
 3.4  1011  0.0004 ,
  Y ,
2
r
r  0.0187m , d  0.037m
Q5  
Q6
F1
F2
4.5m
O
Q3 Kinetic energy gained by electron  qV ,
2qV
1
mv 2  qV , v 
,
2
m
50.0  10 3 g
6.75m
300  10 3 g
The slab is in equilibrium, net torque about O
equals zero.

F2  13.5  50.0  10 3 g  4.5  300  10 3 g  6.75  0
 F2  1.5  10 6 N
Net force equals zero.

F1   1.5  10 6   50.0  10 3 g   300  10 3 g  0 ,
F1  1.9  10 6 N .
Q7 Cast iron fractures in the elastic region.
Q8 Energy absorbed = area under stress-strain
graph  volume of material.
1
E  1.5  10 3 150  10 6   0.10  0.05  10
2
 5.6  10 3 J
AREA 4: Light and matter
Q1 3.75  2.10  1.65eV , A.

hc


Q4 de Broglie wavelength of softball
h 6.63  10 34
 

 1.1  10 34 m .
mv
0.20  30
Width of bat w  0.1m .
Wave nature of moving softball is insignificant

because the ratio  1 .
w
Q5 Vertical axis intercept at –2v.
Work function  qV  1.6  10 19  2  3.2  10 19 J
OR  2eV
Q6 Max E K  hf  W
 4.14  10 15 1.93  1016  2  78eV

Q2 Photon energy  2.10eV , E 

2 1.6  10 19 2500 
 2.96  10 7 ms 1
 31
9.1  10
v
2.96
 100% 
 100%  10% .
c
3.0
v
,
hc 4.14  10 15  3.0  10 8

 5.9  10 7 m
E
2.10

