Final Mock Examination Physics I

Time allowed : 3 hours.

1. When told to open this question and answer book, check that all the questions are there. Look for the

words ‘END OF PAPER’ after the last question.

2. ANSWER ALL QUESTIONS. Write your answer in the spaces provided in the question book. In

calculations you are advised to show all the main steps in your working.

3. Assume: Velocity of light in air = 3 x 108 ms-1

Acceleration due to gravity = 10 ms-2

Question No. 1 2 3 4 5 6 7 8 9 10 TotalMarks 9 16 11 16 9 12 12 9 9 17 120

1. An object of mass M = 0.4 kg rests on a smooth table. A bullet of mass m = 0.025 kg with speed v0 = 450 m/s is fired horizontally towards the object. The bullet penetrates the object and emerges from it. It is found that the object falls to the ground and lands at a point 30 m from the table.

(a) (i) What is the time taken for the object to reach the ground? (2 marks)S = 1/2 a t2

10 = 1/2 x 10 x t2 (1)

t = 1.41 s (1)

(ii) Hence determine the horizontal component of the velocity of the object when it reaches the ground. (2 marks)

s = vt

30 = v x 1.41 (1)

V = 21.3 m/s (1)

(b) By using the principle of conservation of linear momentum, determine the velocity of the bullet when it just emerges from the object. (2 marks)

450 x 0.025 = v x 0.025 + 21.3 x 0.4 (1)

v = 109.2 m/s (1)

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(c) Use the result of part (b) to compute the value of R, which is the distance of the bullet from the table when it reaches the ground. (1 mark) R = 109.2 x 1.41 (0.5)

= 154.4 m (0.5)

(d) What fraction of the kinetic energy is lost during the collision between the bullet and the object? (2 marks)Loss = 1 – [ (0.5 x 0.4 x 21.32 + 0.5 x 0.025 x 109.22)/(0.5 x 0.025 x 4502)](1)

= 90.5 % (1)

2 (a) (i) The period of a pendulum is 2.2s. Its bob has a maximum speed of 0.46m/s and a mass of 65 g. Calculate the maximum kinetic energy of the bob and the amplitude of the oscillation (2 marks)Max K.E. = 1/2 x 0.065 x 0.46 2 (1)

= 36.88 mJ (1)

A = max velocity

A 2 / 2.2 = 0.46 A = 0.16m

(ii) The figure (a) below shows how the kinetic energy W of the pendulum bob varies with time t. Mark the axes with appropriate scales. (2 marks)

figure (a)(iii) Sketch two more lines on the same graph, one to represent the variation of potential energy

with time and a second line to represent the total mechanical energy stored in the system. Label these line “P.E” and “System energy” respectively. (2 marks)

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System energy

Potential energy

2.2s 1.1s

3.44mJ

Energy / J

Time / s

(b) A space shuttle crew is sent to service the Hubble Space Telescope H which is in a circular oibit 6.0105 m above the earth’s surface. The crew succeed in moving the space shuttle S into the same orbit as H and its thrust rocket are shut down. The telescope is positioned a few kilometres in front as shown in following figure. Let G be the gravitational constant and ME the mass of the earth.(Given : radius of the earth = 6.4 106 m)

(a) What is the apparent weight of an astronaut of mass 75 kg inside the shuttle? (1 marks) 0 N

(b) (i) Calculate the value of the gravitational field strength in the orbit. (3 marks)Field strength = -GM / r2 = - GM / R e

2 x Re2 / r2 (1)

= 10 x ( 6.4 x 106 / (6.4x 106 + 6 x 105 ) 2 (1)

= 8.36 N/kg (1)

(ii) Calculate the speed and period of the shuttle in the orbit. (4 marks)

v2 / r =- 8.36 (1)

v = 7649 m/s (1)

T = 2r / v (1)

= 5750s (1)

(iii) Show that the total mechanical energy of the shuttle is proportional to 1/r, where r is the radius of its orbit. (2 marks)

mv2 / r = GMm / r2 (.5)

K.E. = 1/2 mv2 = GMm / 2r (.5)

P.E. = - GMm / r (.5)

Total Energy = - GMm / 2r (.5)

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3.

Figure (a)An audio source S is placed somewhere between an observer O and a reflecting plate R, both lie on the same straight line perpendicular to the reflecting plate. Suppose the speed of sound in air is c and a note of frequency fo is emitted from the source S.(a) The source is moving with speed v (v < c) towards the reflecting plate R.

(i) If f1 is the apparent frequency heard by the observer O for the wave comes directly from the source S ,write down an expression for f1, in term of fo , c and v. (1 mark)

(ii) If f2 is the apparent frequency heard by the observer O for the wave reflected from plate R, write the expression of f2, in term of fo , c and v. (1 mark)

(b) Suppose S is moving at speed v = 2 ms-1, emitting a note of frequency 676 Hz. and speed of sound is 340 ms-1.(i) How many beats per second will be observed for the two notes received by

the observer ? (2 marks)

beat frequency = f2 – f1 = -

= 680 – 672 = 8 beat per second

(ii) The reflecting plate R then starts to move and at a particular speed vR The beats is double to that in part(b)(i). (5 marks)(I) What is the apparent frequency received by the reflector?

The apparent frequency received by the reflector

(1)

(II) Find vR.

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S RO

vfo

Frequency heard by O after reflection

(1)

if beat frequency is doubled

- = 16 (1)

VR = 2 (1)

(III) In which direction is it moving ?Direction =Left (1)

Direction = right if vR = 6 m/s

(c) Mention two applications of Doppler effects (2 marks)

Observation of red shift leading the theory of expanding Universe (1)

Car speed trap (1)

4. The diagram shows a switched-range voltmeter

figure (a)

(i) The voltmeter is set to the 10V range. Find the output voltage from the op-amp when the input is 10V. Your are not required to derive any formula you use. (2 marks) Vout = Vin ( - 3.0k /10k ) (1)

= - 3 V (1)

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selectorVoltage

(ii) The indicating meter has a resistance of 150. It needs a current of 2.0 mA for full scale deflection. Calculate the value of R1 if the meter is to indicate full scale deflection when the voltmeter input is 10V. (2 marks) -3V = 2.0 mA x (R +150) (1)

R = 1350 (1)

(iii) Calculate the feedback resistances R2 and R3 for the 1V and 100mV ranges. (3 marks)

- 3V = - R2 / 10 K x 1 V (.5)

R2 = 30 k (1)

- 3V = - R3 / 10 K x 100m V (.5)

R2 = 300 k (1)

(b)

Figure (b) Figure (c)

(i) The circuit in Figure (b) shows an NPN silicon transistor and two resistors connected to a 6 V d.c. power supply. The current gain of the transistor is 100.

Calculate the value of the base current, stating clearly any assumptions you make. (2 marks) Assume VBE = 0.6 V (1)

IB = (6-0.6) / 10k

= 5.4 x 10-4 A (1)

Estimate the potential difference between the collector and the emitter ? Give reasoning for your estimation. (3 mark) IC = 100 x 5.4 x 10-4 = 0.054 (1)

VCE = 6V - 0.054A x 1k < 0 (1)

The transistor is saturated and VCE =0. (1)

(ii) The same transistor is now used in the circuit in Figure c. Inputs 1 and 2 can be connected to either +6 V d.c. or to 0 V. If an input is connected to +6 V it is said to be HIGH and if connected to 0 V it is said to be LOW.

The output of the circuit is HIGH when the voltmeter reading is approximately 6 V and LOW when the reading is close to 0V. Complete the following table for the circuit shown in Figure (c) : (2 marks)

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1k

Input 1 Input 2 OutputLOW LOW HighLOW HIGH LowHIGH LOW LowHIGH HIGH Low

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(iii) A student is now given two light-dependent resistors (LDRs) and a small d.c. buzzer. The buzzer can be connected to the output of the circuit in Figure (c) and emits a loud noise when the output is high but no noise when it is low.

The student is asked to add the two LDRs to the circuit in Figure (c) to produce an alarm system. This system should emit a warning sound when the length of an object passing along a conveyor belt is greater than the distance l between the two LDRs (see Figure (d) below).

Figure (d)

Each LDR is mounted in a cardboard tube and is illumined by a light beam falling directly onto its surface. W

Complete the circuit diagram in Figure (e) below showing how the two LDRs should be connected to perform the necessary task. (2 marks)

Figure (e)Complete this circuitby adding two LDRs

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5. (a) The diagram shows a beam of protons of mass m and charge q traveling at speed v and a beam of alpha particles traveling at the same speed v. They both enter a region where a uniform magnetic field B acts perpendicular to their direction of travel.

figure (a)(i) Write down an expression for the force required to accelerate the protons in a circular path of r p.

(1 mark) mv2 / rp (1)

(ii)Write down an expression for the magnetic force acting on the protons. (1 mark)Bqv (1)

(iii) Use the above expressions show that (1 mark)

mv2 / rp = Bqv (.5)

(.5)

(iv) Calculate the value r:rp where r is the radius of the path of the alpha particles. (1 mark) r:rp = (q/m) p : (q/m) (.5)

= 2 (.5)

(v) Hence, on the figure (a) above, draw the path of the protons and the path of the alpha particles after they enter the magnetic field. (2 marks)

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(b)

A student uses a flame probe to investigate the variations in potential in the region around a positively charged sphere. The probe, in the form of a small gas flame at the point of a needle, is connected to an electroscope calibrated to measure potentials.

(i) The electroscope reading is a measure of the potential at the point where the probe is situated. The experiment has to be performed with the charged sphere remote from the floor and neighboring walls. Explain briefly why. (1 mark)

Avoid induced charge to affect the measurement

(ii)

Figure (c)The student measures the potential at points A and B (see Figure (c)), which are 1 cm apart. He finds that the potentials at A and B are 450 V and 400 V respectively. Give an estimate of the electric field in the region between A and B. In what direction does it act ? (2 marks) Field = 450 – 400 / 0.01 (1)

5000 N/C (.5)

From A to B (.5)

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6 (a)

Figure (a)Figure a shows an aluminum plate with a current of 50 mA passing through it.(i) Calculate the average drift velocity of the conducting electrons, given that there are

1029 conducting electrons per m3 of aluminum and the charge of an electron is 1.6 10-19 C. (2 marks) I = nAqv

50 m = 1029 x 0.1 x10-3 x 0.02 x 1.6 x 10-19 x v (1)

v = 1.6x 10-6 m/s (1)

(ii) A uniform magnetic field of 1.5 T is now applied normally downwards to the plate and covers whole surface area. Mark on Figure (a) the direction of the force experienced by each electron and calculate its magnitude. (3 marks)

F = Bqv = 1.5 x 1.6 x 10-19 x 1.6x 10-6 (1)

= 3.8 x 10-25 N (1)

(b) An experiment is being set up to demonstrate the Hall voltage and it is decided to use a germanium slice. The circuit diagram for the experiment is shown in Figure (b) below

Figure (b)(i) Explain why a germanium slice was chosen in this experiment instead of an aluminum

plate. (2 marks)Lower carrier concentration (1)

Which results in larger Hall voltage observed. (1)

(ii) After switch S has been closed, a small p.d. is found to exist between X and Y even in the absence of a magnetic field. Explain why this is so. How would you arrange X and Y to be at the same potential ? (2 marks) X and Y are not opposite (1)

Adjust X and Y until V = 0 or adjust R (1)

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(d) A uniform magnetic field of 0.2 T is now applied acting perpendicularly downwards into the plane of the paper, covering the whole surface of the slice. If the reading of the milliammeter is 1 mA, estimate the Hall voltage that exists across the slice.

(the thickness of the slice = 0.1 mm,number of charge-carriers per unit volume for germanium = 1020 m-3,the charge on each carrier = 1.610-19 C.) (2 marks)

Bqv = qV /d

V = 0.125 V

(e) Mention one practical application of the Hall effect. (1 mark)Hall probe to measure magnetic field (1)

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7. (a) State Faraday’s law of electromagnetic induction. ( 1 mark)The induced emf in a conductor is directly proportional to the rate of change at

Which the conductor cut through the magnetic field lines ( or rate of change of magnetic flux )

(b) An exhibit at a science museum consists of three apparently identical vertical tubes, T1, T2 and T3, each about 2 m long. With the tubes are three apparently identical small cylinders, one to each tube.

figure aWhen the cylinders are dropped down the tubes those in T1, T2 and T3 , arrange the time required for the cylinders to reach the bottom in ascending order. Explain briefly. (4 marks)T1 = T2 < T3 (1)

Induced emf and hence current flow in T3. The induced magnetic field oppose the

Motion of the bar magnetic and hence acceleration less than g (1)

The cylinder in T2 is unmagnetized and hence no induced emf. (1)

The tube T3 is plastic and no induced current can flow (1)

(c)

figure (b) figure (c)A d.c. source of 200 V supplies power to a circuit containing a motor with armature resistance 5 , a switch S, a rheostat R and an ammeter A, all connected in series. After the switch S is closed, the reading on the ammeter changes with time as shown by the curve in Figure (c).(a) Calculate the resistance of the rheostat R. (2 marks)

V=IR (1)

R = 15 (1)

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(b) After the motor has reached constant speed, calculate (i) the back e.m.f. of the motor, (2 marks)

V = E + IR (1)

200 = E + 1 x 20 => E = 180 V (1)

(ii) the mechanical power output of the motor, and (2 marks)Power = EI (1)

= 180W (1)

(iii) the power dissipated as heat in the armature. (1 mark)I2R = 5W (1)

8 (a)direction of ball

Figure (a)(i) Figure (a) shows the streamlines around a tennis ball when it is projected in a straight

line through still air. If, apart from the forward motion, it is also spinning about an axis, through its centre, perpendicular to the plane of paper in an anti-clockwise direction according to figure (b).

forward motion

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Figure (b)Describe, with reasons, the subsequent motion of the ball. (4 marks)Ball rotates and drags surrounding air (1)

Air on LHS flow faster than that in RHS (1)

Pressure on LHS less (1)

Ball deflected towards the left (1)

(b) (i) One end of an open tube is put vertically into water. By blowing strongly across the open end, water can be drawn up the tube. Suppose a few centimetres of the tube is above the water surface. What should be the air velocity at the open end for water in the tube to rise up by 1 cm? Explain your working. (Surface tension effects may be ignored.)( Density of air = 1.29 kg m-3,

Density of water = 1 000 kg m-3,Acceleration due to gravity = 10 m s-2 ) (3 marks)

Pressure at upper end reduced due to air flow (1)

1/2 v 2 = dgh (1)

V = 12.5 m/s (1)

(ii) Mention two daily applications making use of the principle described in (b)(i).(2 mark)

Carburetors in car

aerosol spray

9. A circular wire ring is dipped into soap solution and held vertically. When viewed in a dark room with monochromatic light of wavelength 6.5 10-7 m reflected normally from the film, a series of interference fringes are seen. The pattern of fringes at a particular instant is shown is Figure (a) (The refractive index of soap solution = 1.33)

Figure (a)

(a) What colour are the fringes ? ( 1 mark)Red or orange

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(b) Why is there dark area at the top of the ring ? (2 marks)Thickness less than /4n (1)

No constructive interference occurs (1)

(c) What is the thickness of the film at point A ? (3 marks) 2nd order destructive interference (1)

2t = 2 / n (1)

T = 4.9 x 10-7 m (1)

(d) As time goes on and if the film drains downwards and does not break, the fringe pattern changes from that shown in Figure (a) describe the changes you would expect to see (explanation are not required). (3 marks)Dark area become larger (1)

No. of fringes increases (1)

Fringes are more closely spaced towards the bottom (1)

10. (a) It is thought that an extremely short-lived radioactive isotope , which decays by -emission, has a half-life of 200s. After a series of decays the element is formed from the original isotope. There are no decays.

(i) Deduce the value of A (2 marks) No. of decay = (110-104) / 2 =3 (1)

A = 269 – 4x3 = 257 (1)

(ii) Calculate the decay constant of (2 marks) T(1/2) = ln2 / (1)

= 3466 s-1 (1)

(iii) The number of nuclei of in a sample of mass 0.54 g is 1.2 x 1015. Determine the activity of 0.54 g of . (2 marks) A = N (1)

=- 4.15x1018 decay / second (1)

(iv) Why is the above calculated value for the activity is only an approximate? (1 mark)Radioactivity is random in nature (1)

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(b) In a fusion reaction, two nuclei of the isotope of hydrogen (called deuterium) fuse together producing another isotope of helium and a neutron.

( Mass of the deuterium nucleus = 2.015 uMass of the helium nucleus =3.017 uMass of neutron = 1.009 uAtomic mass unit 1u = 1.66x10-27 kg. )

(i) Write an equation for the above fusion reaction. (1 mark)

(ii) Explain how and why the mass of a helium atom differs from the sum of the masses of its constituent particles. (2 marks)Energy is needed to separate the particles (1)

And the energy is transformed into mass (1)

(iii) Calculate the energy released by fusion of 1 kg of deuterium. (3 marks)In fusing two deuterium nuclei to form a helium nucleus and a neutron, the

decrease in mass is (2 Í 2.015) – (1.009 + 3.017) = 0.004 u (1)

Using E = mc2, the energy released is 0.004 Í (1.66 Í 10-27) Í (3 Í 108)2 (1)

no. of pairs of deuterium atoms = 1/(2 Í 2.015 Í 1.66 Í 10-27)

Total energy released is = 8.93 Í 1013 J

(iv) If 50% of the energy released in part (c) were used to produce 1 MW of electricity continuously in a station, for how many days would the station be able to function?(2 mark)Day = 8.9 Í 1013 Í 50% / 1 Í 106 / (24 Í60 Í 60)

= 517 days

(v) The energy released in each fusion reaction is smaller than that in each fission reaction. Statetwo reasons why it would still be advantageous if an operational fusion reactor could be developed. (2 mark)

Start material readily available / cheap

Fission product is radioactive and have long half life

Energy released per kg of material is greater in fusion

End of paper

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