Download Physics_1996_Paper_I_+_ANS

1.
In Figure 1, the image of an illuminated letter ‘J’ formed by
a lens is caught by a translucent screen.
( a ) ( I ) What kind of lens is being used ? Explain your answer.
( 2 marks )
( II ) Sketch the shape of the image seen by the observer. ( 2 marks )
( b ) Figure 2 shows the relation between the image distance v and
the object distance u.
( I ) Find the magnification of the image when u=18cm. ( 2 marks )
( II ) Find the value of u when u=v, Hence find the focal length of the
lens. ( 3 marks )
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( c ) The illuminated letter is now placed closer to the lens . In Figure
3, AB represents the letter and the path of a ray from A through the lens
is shown.
Figure 3
( I ) In Figure 3, draw the path of a ray from A which passes through the
optical center of the lens and construct the image of AB ( 3marks )
( II ) State an application of the lens in which an image like that shown
in Figure 3 is formed . ( 1 mark )
( III ) Comment on the following statement:
After adjusting the position of the screen, the image formed in Figure
3 can still be caught. ( 2 marks )
3. (a) A boy 1.5 m tall stands a few metres in front of a plane mirror
AB which is hung on a vertical wall. The boy’s eyes are 1.4 m above
the ground . He can see all of himself in the mirror. In Figure 5,
PQ represents the boy and E denotes his eyes.
i.
State three properties of the boy’s image as formed by the mirror. (2 marks)
(ii) In Figure 5,draw
1.
the image of the boy formed by the mirror,
(2) the paths of two rays, one from P and one From Q ,to
show how the rays reach his eyes. (4 marks)
iii.
Using (ii) , or otherwise ,find the minimum length of the mirror AB for the
boy to see all of himself. (2 marks)
(iv ) If the boy moves a few steps towards the mirror and the length
of the mirror is equal to that found in (iii),can the boy still see
all of himself in the mirror? In Figure 6,draw a ray diagram to
illustrate your answer. (3 marks)
b.
There are two types of rear-view driving mirrors.
(i) Which type of mirror (plane , concave and
convex)cannot be used as a driving mirror? Give a
reason . (2 marks)
i.
State the advantage of each type of driving mirror. (2 marks)
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2.Susan takes part in a 100 m race at an athletic meet. She starts at
time t=0 and accelerates at a uniform rate of 1.6 m s-2 for 5 s. she then
maintains a uniform speed afterwards and reaches the finishing line at
t=1.5s.
a. Find
i.
ii.
the speed of Susan at t=5s,
Susan’s average speed for the whole journey.
(3 marks0
b. On graph paper, draw the graph of Susan’s speed against time from t=0 to
15s.
State the physical meaning of the area under the graph.
(5 marks)
c. Susan’s mass is 45 kg . Find the resultant force acting on Susan in the race
i.
ii.
during the first 5 s;
after t=5s.
Mary also takes part in the same race. She first accelerates at a uniform
rate of 1.5 m s-2 for 6 s and then maintains a uniform speed afterward.
Explain whether Susan or Mary will reach the finishing line first.
(4 marks)
3.
Two metal balls p and Q are suspend by light inextensible strings.
Ball P is pulled to a point A which is at a height h above its initial
position B and is then released. (See Figure 4.) After colliding
at B, the two balls move away in opposite directions. (See Figure
5)
a. Draw a diagram to show all the forces acting on P when it swings
from A to B. Label the forces.
(2 marks)
b. Describe the energy changes in the balls, from the moment P is
released until the balls swing up to their maximum heights after the
collision.
marks)
C.The mass of P is 0.3 kg and its speeds immediately before and
after the collision are found to be 1.0 m s-1 and 0.5 m
s-1respectively.
(i)Find h.
ii.
Find the average force acting on P during the collision, assuming that the time
of contact is 0.02 s.
(2 marks)
iii.
Consider the following set of date :
Mass/kg
Velocity before
P
Q
0.3
0.75
0
0
-1
collision/ m s
(towards the right)
Velocity after
collision/ m s-1
0.5
0.6
(towards the left) (towards the right)
1.Show that the above set of date obeys the law of conservation of
momentum.
2.Explain why the above set of date is impossible.
( 5 marks)
4. (a)
A string is fixed at one end to a wall and a vibrator generates a
stationary wave on the string . The distance between the vibrator
and the wall is 0.6 m . Figure 6 shows the string at the instant
of maximum displacement.
i.
ii.
iii.
Find the wavelength of the stationary wave. (1 mark)
Describe the motion of particles P and Q at this instant.(2 marks)
Copy Figure 6 into your answer book and mark in the figure positions
of the nodes (labeled as N) and antinodes (labeled as A) . (2 marks)
(i v) Figure 7 shows the displacement-time graph of particle
P.
Copy Figure 7 into your answer book and in the same
figure ,sketch the displacement-time graph of particle Q .
(2 marks)
iv.
The vibrating also sets the neighbouring air into vibration . State two
differences between the waves on the string and those in air. (2 marks)
(b)Figure 8 shows the traces on a CRO of three notes produced by different
musical instruments. (Note: The settings of the CRO remain unchanged.)
(i)Which of the notes is produced by a tuning fork?
(1 mark)
(ii)Compare the pitch and loudness of the three notes. Explain
briefly.
(5 marks)
5. A student sets up a circuit as shown in Figure 9.
(a) Name the decices C and D.
(b)
Step
Switch A
Switch B
Device D
1
Closed
Closed
OFF
2
Closed
Open
3
Open
Closed
4
Open
Open
ON
The steps shown above are performed to investigate how the circuit works.
When both switches A and B are closed , device D dose not emit light(OFF) .
When switches are open, D emits light(ON).
*(1) Explain why device D is OFF when both switches are closed. (5 marks)
ii.
State whether D is ON or OFF in steps 2 and 3. (2 marks)
(iii) What kind of logic gate does the circuit represent if switches A
and B are taken as the inputs and device D as the output ? (Note : A closed
switch denotes high input and device D ON denotes high output.) (1 mark)
(I v) What is the function of the resistor R in the circuit? (1 mark )
(c) In Figure 10 , draw wires to connect the terminals of the devices
according to the circuit in Figure 9. (4 marks)
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6.
A factory produces detergent contained in plastic bottles.
The following method is used to monitor the amount of
detergent contained in each bottle : a radioactive source
is placed on one side of the conveyor belt at the level to
which the detergent is expected to fill and a detector is
placed at the same level on the other side .(See Figure 11)
a. Which type of radioactive source (α, β or γ) should be
used?
Explain briefly why the other two types are not suitable.
(3 marks)
(b) Suggest one suitable detector for the above system. (1
mark)
*(c)Explain how the monitoring system can detect bottles of
detergent that have not been filled up to the required level. ( 4
marks)
d. Two sources emitting the suitable type of radiation of half-lives 10 minutes
and 5 years are available.
i.
ii.
Explain what is meant by the half-life of a radioactive source . (2 marks)
Which source should be used ? Explained briefly.(3 marks)
(e). State two safety precautions that the factory workers should
take when handling radioactive sources. (2 marks)
7. (a) When the magnet moves toward the solenoid, a current passes
from X through G to Y(OR from X to Y )When the magnet is inside
the solenoid, there is no current. When the magnet moves away from
the solenoid, a current passes from the solenoid , a current passes
from Y through G to X (OR from Y to X). (4M)
(b) (i) The plane of the coil is parallel to the magnetic field
at points Q and S.
(ii) The peak voltage of the output of the generator =1×50=50mV
(OR 0.05V)
The frequency of the output of the
generator=1/T=1/4×20×10-3Hz=12.5Hz
(iii)(1) If the speed of rotation p of the coil is increased , the
peak voltage and frequency of the output of the generator increase.
(2) If the coil is would on a soft-iron core, the peak voltage
increases while the frequency remains unchanged.
(iv) The two other practice means of driving generators are:
*Wind
*Running water (OR hydroelectric power).
*Sea wave
*Tidal power (11M)
CE Physics (English) 1996 Paper 1 第 1 題 答案
1. (a) (i) A convex (OR converging lens)is being because
*a real image is formed
*the image can be caught by a screen
(ii) (4M)
(b) (i) When u=18cm, v=36cm
Magnification of the image when u = 18cm
= v/u=36/18=2
(ii) From Figure 2, u=24cm when u=v
Focal length of the lens = 24/2
=12cm
(5M)
(c) (i)
(ii) An application of the lens is being used
*as a magnifying glass
*to make a telescope/a microscope
*to correct for long sight
(iii)After adjusting the position of the screen, the image
formed in Figure 3 cannot be caught because
the image formed is virtual . Therefore, the statement is wrong.
(6M)
2. (a) (i) Speed of Susan at t=5s=u+at
=0+1.6×5=8ms-1
(ii) Susan’s average speed for the whole journey
=Total distance/time = 100/15
=6.7 ms-1
(3M)
b.
c.The area under the graph represents the distance
Traveled by Susan. (5M)
d.(i) The resultant force acting on Susan in the race during the
first 5s
=ma
=45×1.6
=72N
(ii) The resultant force acting on Susan in the race after
t=5s=0N (3M)
b. Marking critera:
-2M awarded for using an appropriate method;
-1A awarded for the accuracy of the computations
involved;
-1M awarded for a correct conclusion.
2. (a) (i) Speed of Susan at t=5s=u+at
=0+1.6×5=8ms-1
(ii) Susan’s average speed for the whole journey
=Total distance/time = 100/15
=6.7 ms-1
(3M)
b.
c.The area under the graph represents the distance
Traveled by Susan. (5M)
d.(i) The resultant force acting on Susan in the race during the
first 5s
=ma
=45×1.6
=72N
(ii) The resultant force acting on Susan in the race after
t=5s=0N (3M)
b. Marking critera:
-2M awarded for using an appropriate method;
-1A awarded for the accuracy of the computations
involved;
-1M awarded for a correct conclusion.
. (a)
(2M)
(b) Marking criteria
1A for the energy change when P swing downward,
1A for the energy change during the collision.
1A for the energy change when the balls swing upward,
1C for effective communication.
When P is released from A to B, its potential energy is converted to kinetic
energy.
When P collides with Q some of its (kinetic) energy is converted to
(kinetic) energy of Q (and some may be lost as internal energy)When P and
Q swing upward after the collision ,their kinetic energies are converted
back into potential energies.
(4M)
(c) (i) By conservation of energy,
mv2/2=mgh
h= v2/2g=12/2×10=0.05m(or 5cm)
(ii) Average force acting on P during the
collision=m(v-u)/t=0.3[0.5-(-1)]/0.02=22.5N
(iii) (1)Momentum before the collision =0.3(1.0)+0.75(0)=0.3kg m s
Momentum after the collision=0.3(-0.5)+0.75(0.6)=0.3 kg m s
–1
–1
∴The above set of date obeys the law of conservation of momentum.
Alternative solution
Change in momentum of P
=0.3(-0.5 –1)
–1
=-0.45kg m s
Change in momentum of Q
=0.75(0.6-0)
=0.45kg m s
–1
Total change of momentum=0
∴The above set of date obeys the
law of conservation of momentum.
(2) Total kinetic energies before the collision =(0.3)(1.0)2/2=0.15J
Total kinetic energies after the collision=(0.3)(0.5)2/2+(0.75)(0.6)2/2
=0.1725J(or～0.17J)
Since the total kinetic energies after the collision increase (OR As no
extra energy can be introduced during the impact),the above set of
4. (a) (i) The wavelength of the stationary have=0.4m
(ii) Both part ∴les P and Q
(iii)
(iv)
v.
Any two of the following :
Waves in string Waves in air
*Stationary wave Travelling wave
*Transverse wave Longitudinal wave
*They have
different
wavelengths OR
speeds.
(9M)
(b) (i) Note Z is produced by a tuning fork.
(ii) The pitches of the three notes are identical because
*the frequencies of the three notes are the same.
*the pitch increases with the frequency of the note.
The loudness of note X is larger than that of Y, which is larger than of
Z because
*the amplitude of note X is larger than that of Y, which is larger than
of Z.
*the loudness increases with the amplitude of the note. (6M)
5. (a) Device C is a relay/a reed relay/an electromagnetic relay
Device D is a light emitting diode (LED)
(2M)
(b) (i) Marking criteria
1A for current passing through the relay
the coil,
1A for closing the relay,
2A for explaining why D is off,
1C for effective communication.
6. (a) A β source should be used .
An α source is not suitable because a particles cannot pass through the
container/detergent.
A γ source is not suitable because it penetrating power is too strong.(OR
completely passes through/cannot be absorbed by the
container/detergent).
(3M)
(b) A Geiger-Muller (G-M) tube is a suitable director for the above system.
(1M)
(c)
Marking criteria
2A for describing how the monitoring system
works,
1A for providing an explanation,
1C for effective communication.
If a bottle which has not been filled up to the required level passes the
radio active source, the counter will record a much higher reading (count
rate) than that when an acceptable bottle passes the radio active, it is
because there is no detergent absorbing the B radiation and hence more
B particles are absorbed by the counter. (4M)
(d) (i) The half-life of a radioactive source is the time taken for
*half of the nuclei present in the source (OR in any given sample ) to
decay .
*the activity of the source (OR any given sample) to fall to half its
initial value.
(ii) the source with half-life 5 years should be used
Because * The source will decay slowly (OR The activity of the source will
be very stable) and can be used for a long time.
The source with half-life 10 minutes will decay rapidly and the registered
count rate is unstable even when no bottles are present. (5M)
(e) The factory working should take the following two precantion when
handling radioactive source:
*Wearing film badges or other detecting detecting devices.
*Working behind lead-glass windows.
*Handling radioactive sources using special forceps/remote-controlled
robots.
*Wearing protective coveralls.
*Radioactive sources should not be pointed towards anybody or held
near anyone’s eye.
*Radioactive sources should be stored in lead castles/returned to the
storage box after use.
*Working should wash their hand with soap after handling radioactive
sources.(2M)
7. (a) When the magnet moves toward the solenoid, a current passes from
X through G to Y(OR from X to Y )When the magnet is inside the solenoid,
there is no current. When the magnet moves away from the solenoid, a
current passes from the solenoid , a current passes from Y through G to
X (OR from Y to X). (4M)
(b) (i) The plane of the coil is parallel to the magnetic field at points
Q and S.
(ii) The peak voltage of the output of the generator =1×50=50mV (OR 0.05V)
The frequency of the output of the generator=1/T=1/4×20×10-3Hz=12.5Hz
(iii)(1) If the speed of rotation p of the coil is increased , the peak
voltage and frequency of the output of the generator increase.
(2) If the coil is would on a soft-iron core, the peak voltage increases
while the frequency remains unchanged.
(iv) The two other practice means of driving generators are:
*Wind
*Running water (OR hydroelectric power).
*Sea wave
*Tidal power (11M)