CENTRALE COMMISSIE VOORTENTAMEN NATUURKUNDE
PHYSICS EXAM
Date
: Friday 28 April 2017
Time
: 13.30 tot 16.30
Number of questions
:5
Number of answer sheets
: 1 (for question 1)
Each question must be answered on a separate sheet (because each question
is marked by different examiners).
State your name on every sheet you hand in.
Do not write in pencil and do not use Tipp-Ex or any similar product.
Answers without argumentation will not be deemed correct.
Additional data can be found in the BINAS science data reference book (4th or
5th edition).
The standardized scores are:
Question 1
Question 2
Question 3
Question 4
Question 5
: 18 points
: 12 points
: 15 points
: 8 points
: 16 points
The marks are calculated as follows: mark = score/69 x 9 + 1
Information concerning the procedure and progress of the proces of marking
see: www.ccvx.nl > verloop van de correctie
QUESTION 1 - pilot emergency training
An ejector seat is used by fighter pilots who need to evacuate their plane in
an emergency situation. There is a specific installation to train pilots for this
eventuality. This installation consists of long, vertical rails along which a pilot
in a seat can be launched vertically upwards. This is schematically shown in
Figure 1.
The force propelling the pilot and the chair upwards only operates for the first
1.81 s immediately after launch. After that time, the only active forces are
friction and, of course, gravity.
Figure 2 shows the v-t graph for the first 6.00 seconds after launch. An
enlarged version of the v-t graph is provided on the answer sheet.
2p
a. Calculate the maximum speed reached by the pilot.
3p
b. Calculate the height reached by the pilot. (If you were unable to
calculate this part, assume a value of 85 m for the remainder of this
assignment. Please note that is value is incorrect.)
3p
c. Calculate the maximum acceleration during the upwards launch.
The pilot and the chair together have a mass of 96 kg.
3p
d. Calculate how much work the force with which the pilot is launched exerts on the
pilot and chair if there was no friction.
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In reality, there is friction and part of the work is converted into heat.
At t = 4.26 s, the graph shows a slight kink.
3p
e. Explain this kink.
When coming down, assume that the pilot descends at a constant speed until he hits the
ground.
The v-t graph has only been drawn up to t = 6.00 s but at that time the pilot has not
reached the ground.
4p
f. Using the v-t graph, calculate whether the pilot hits the ground before or after
t = 8.00 s.
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QUESTION 2 - resonance
A string with a certain lengthl is strung between
two fixed points, P and Q. At point P, there is a
source that can be used to make the string vibrate.
Next to the string, there is a hollow pipe, closed at
one end, with a depth (d) of 25.0 cm. The scene is
represented as a diagram in the figure.
The speed of sound in air is 343 m/s.
The air in the cylinder starts resonating.
3p
a. Calculate the cylinder’s lowest and second
lowest resonant frequency.
(If you have been unable to work out this
section, please assume a value of 400 Hz
as the lowest frequency and 1200 Hz as the
second lowest resonant frequency).
The vibrations in the string are caused by a source with an adjustable frequency. Both
ends of the string are attached to a fixed point. The reproduction rate of the waves in the
string is 823 m/s.
3p
b. What should the length lof the string be to ensure that resonance occurs at the
lowest frequency in section a.?
4p
c. Calculate the shortest length the string can have be to ensure that both resonances
under section a. can occur.
2p
d. If resonance occurs initially and the temperature falls subsequently, this resonance
will cease. Explain why.
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QUESTION 3 - hot air gun
A hot air gun is a device in which a fan moves an air flow
over a filament between points K, L, M, N and O. The hot
air produced can be used to remove old coats of paint
from a door you want to repaint, for example.
The wiring diagram is shown below. The hot air gun is
connected to an effective alternating current of 220 V.
Some time after switching on, the device enters its “ready to use” mode. At this time, a
current of 750 mA runs through the fan, and a current of 8.50 A runs through the filament
between K, L, M, N and O.
2p
a. Calculate the electrical resistance of the
filament (KLMNO) is this situation.
3p
b. Calculate the overall power consumed by the
device in the “ready to use” mode.
Immediately after switching on, the magnitude of the
current in the filament is greater than in the “ready to
use” mode.
2p
c. Explain why.
At a certain moment in time, a short circuit occurs
between points M and O of the filament. The length
between M and O equals 1/2 of the total length of KLMNO.
4p
d. Calculate the value of the current intensity immediately after the short circuit:
1) through the filament part KLM
2) through the filament part MNO
3) through the fan motor
In China, mains voltage is 220 V. The Chinese manufacturer of the hot air gun intends to
make the device suitable for use in Europe, where mains voltage is 230 V. This can be
done by placing an additional electrical resistor in the device.
1p e1.
Draw a wiring diagram that includes this additional resistor.
3p e2. Calculate the value of the additional resistance.
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QUESTION 4 - radioactivity
Two different decay mechanisms are known for the radioactive isotope 137Cs.
See the figure below.
Six percent of the 137Cs nuclei decay directly to the stable 137Ba, throwing out an electron
in the process. This electron emission releases 1.18 MeV in the form of kinetic energy of
the electron.
2p
a. Provide the equation for this radioactive decay.
Ninety-four percent of the 137Cs nuclei decay after an intermediate step.
These caesium nuclei first decay to an excited state of barium nuclei, with the emission of
an electron.
This releases 0.51 MeV in the form of kinetic energy. This excited state of barium nuclei is
referred to as 137mBa.
From this excited state, the barium nucleus then decays to a ground state, with the
emission of ã-radiation (gamma radiation). See the figure again.
Einstein has discovered that mass and energy can be converted into one another
according to the famous formula:
with:
E the energy,
m the mass and
c the speed of light.
3p
b. Calculate the mass difference between a 137Cs nucleus and a 137Ba nucleus by
using the data above.
3p
c. Calculate the frequency of the ã-radiation produced during the decay from 137mBa to
137
Ba.
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QUESTION 5 - the electron
The ratio between the mass and charge of an electron can be calculated experimentally
using a setup of which a simplified version is shown in Figure 1.
The electrons released at the cathode are accelerated between the cathode and anode.
The electrons pass through the hole in the anode then enter a uniform electric field
between two plates, M and N, with a velocity parallel to these plates. The strength of the
electric field is 2.1·104 V m–1 . The distance between the plates is 1.8 cm. The whole setup
is in vacuum.
2p
a. Calculate the potential difference between plates M and N.
3p
b. Demonstrate using a calculation that the effect of gravity on the electrons is
negligible compared to the force experienced by the electrons due to the electric
field between the plates.
The deflection of the electron beam between the plates can be avoided by also creating a
uniform magnetic field in the area between the plates. The effect of gravity on the
electrons is negligible.
In the figure 1, a point P is shown on the trajectory of the electrons.
4p
c. Explain using a drawing on the answer sheet which direction the magnetic field has
at point P.
The speed of the electrons travelling straight through is 3.1·107 m s–1.
3p
d. Calculate the magnitude of the magnetic induction.
Continued on the next page
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Next, the magnetic field is switched off. This means the electron beam will be deflected in
the electric field between the plates. This is schematically shown in Figure 2.
The length of the plates is 3.5 cm.
From this deflection, it can be calculated that the electrons will have a speed with a
vertical component of 4,2·106 m s–1 once they have passed through the electric field.
4p
e. Calculate the resulting value for the ratio between the mass and charge of an
electron. Hint: first, calculate the vertical acceleration achieved by the electrons in
the electric field. Keep in mind that in the time an electron moves in a horizontal
direction from right to left between the plates, the electron gets the aforementioned
vertical component speed of 4,2·106 m s–1.
END
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ANSWERSHEET - QUESTION 1
Naam : ...........................................................