H2 PHYSICS PRACTICAL (20% weightage)
1 ai) The horizontal component of the velocity is unchanged.
ii) Along the vertical direction, the object’s velocity increases linearly from zero. This is because a resultant downward force acts on the object to cause it to accelerate. Thus velocity increases with t by the relationship of v(y) = gt.
bi) Horizontal component of the velocity decreases with time due to air resistance acting opposite to the object’s direction of motion.
ii) Vertical component of the object’s velocity increases from 0 with time at a slower rate due to air resistance. Thus the downward acceleration is less than g.
c) Path with air resistance will fall closer to the building.
2 ai) The angle which is subtended at the centre of a circle by an arc length equal to the radius of the circle.
ii) The product of 2π radian and the frequency.
bi) Ek = 0 , Ep = mgh = 4.7 x 10^-3 J
ii) 0.557 Hz
3 a) Increase in internal energy of a system is the sum of work done on the system and the heat supplied to the system.
bi) 4.98 x 10^-2 mol
ii) 1. -226 J
2. 0 J
iii) work done on gas: -226 , heat supplied to gas: -884
increase in internal energy: 344, 540, -884
4 a) Resultant displacement at a point due to two or more waves is the vector sum of the displacements due to those waves acting individually.
bi) 57.3º
ii) 0.49 : 1
5 a) Emax = h (c/f) – hf
bi) 4.35 x 10^-7
ii) Gradient of graph = hc
h = 6.667 x 10^34 J
c) Draw a line parallel to the original line but cutting the x-axis at a larger value.
6 a) A region of space set up by a mass, a charged particle or magnet, and surrounding the body, a field which can exert a force on another body that is not physical.
ii) The statement did not specify the type of field and state of particle. Hence the field may be gravitational field, and the force is due to the mass of the particle. The field may also be magnetic field. The magnetic force on the particle is only zero when it is in motion and not moving parallel to the magnetic field.
bi) Any point within a charged sphere will have no electric field.
ii) 1. A = 4.0cm B = 2.0cm
2. The spheres have opposite signs. Since E is always >0, indicating that the electric fields from A and B are in the same direction.
iii) 1. 3.00 x 10^-15 J
2. a= 1.24 x 10^13 m/s²
3. From 4 to 18.5 cm, the electric field is predomniantly contributed by sphere A and decreasing with distance from A. Acceleration is +3eE/m
The acceleration of the nucleus decreases from a max value at 4cm to a minimum value at 18.5cm
From 18.5 to 28 cm, the minimum E starts to increase again, indicating that the E-field from 18.5cm onwards is predominantly from B. The E field’s direction is the same, and E >0, the lithium nucleus acceleration increases again.
7 ai) The ratio of the potential difference across the wire to the current flowing through the wire.
ii) The resistivity of a wire is a property of the material. It is related to the resistance by the equation R = pl/A
bi) 4.9Ω
ii) 118 turns
c) B = 0.72μ (NI/r)
di) 8.54 x 10^-24
ii) 4.58 x 10^-2
e) Vertical component of the electron’s velocity is perpendicular to the B field and hence it is subjected to a centripetal force. Horizontal component is parallel to the B field and remains constant as there is no magnetic force accelerating the horizontal velocity. Hence the resultant path is helical.
8 ai) A spontaneous and random process in which an unstable nucleus changes into a different nuclide, emitting radiation.
ii) It is unpredictable in both when and which particular nuclide will decay.
iii) Not triggered by external factors.
bi) 2.54 x 10^-10
ii) 4.25kg
ci) 156 W
di) The half life of plutonium is much longer than polonium. For a space flight lasting several years, the original quantity of polonium will have decayed to less than half in a year. Since plutonium has a long half life, over seven years, more than half would still be remaining.
ii) Plutonium has a higher power density as strontium has a much smaller mass and energy release.
A-LEVEL H2 MATH JUNE HOLIDAYS INTENSIVE REVISION
A-LEVEL H2 PHYSICS JUNE HOLIDAYS INTENSIVE REVISION
JUNIOR COLLEGE / A-LEVEL TUITION:
_______________________________________________________________________
______________________________________________________________
By EX-MOE TEACHERS & EXPERIENCED TUTORS
@ BLK 644, BUKIT BATOK CENTRAL, #01-68. S(650644).
CALL 65694897 OR SMS 98530744 OR 97860411.
Suggested Answers to 2014 A Level H2 Physics 9646 PAPER 1 and 2
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H2 PHYSICS HANDS-ON PRACTICAL CRASH COURSE
H2 CHEMISTRY HANDS-ON PRACTICAL CRASH COURSE
H2 BIOLOGY HANDS-ON PRACTICAL CRASH COURSE
Practical training:
H2 PHYSICS PRACTICAL
H2 CHEMISTRY PRACTICAL
H2 BIOLOGY PRACTICAL
1) C 2) A 3) A 4) D 5) D
6) C 7) C 8) D 9) C 10) D
11) B 12) B 13) B 14) C 15) A
16) B 17) A 18) C 19) A 20) D
21) D 22) C 23) B 24) D 25) A
26) D 27) D 28) B 29) A 30) D
31) D 32) C 33) D 34) B 35) D
36) C 37) A 38) D 39) A 40) C
1 ai) Y = 60.0 N
ii) θ = 13°
b) For S to be in equilibrium, Rx = Ry = 0. If weight of S and angle of rope A is maintained, then Rx = X cos 30° > 0
2 a) P-type doping introduces more holes to the semiconductor, increasing the number of charge carriers. This increases the semiconductor’s ability to conduct electric current.
b) When forward biased, the holes in the p-type will be attracted to the negative terminal of the battery while electrons of the n-type will be attracted towards the positive terminal. A large current flows through and depletion region is decreased.
When battery is reversed, the holes in p-type are attracted to negative terminaal and elecrtons of the n-type are attracted to the positive terminal. Width of depletion region increases, junction voltage increases. No current flow.
3 a) As R increases from 0, V of qx decreases, V of xq increases. Since the terminal pd is V(wq) = V(wx) + V(xq) = 5.0 + V(xq) , V(wq) increases with R.
bi) 4.84V
ii) 96.8%
ci) 1.2V
ii) 0.531m
iii) When J is moved towards Q, L will be more than 0.531m. Hence the V in pj will be greater than in pk. Therefore the pd across K and J is greater than 0, which means there is a current through the ammeter now.
4 ai) After time, the coil would have turned an angle about the axis PQ. The magnetic flux linkage is 120BA cos (2π/T)t.
According to Faraday’s law, the induced emf is directly proportional to the rate of change of magnetic flux linkage through the coil, hence E is a sinusoidal function. The negative sign comes from Lenz’s law which states that induced emf opposes the change producing it.
ii) 1. 0.170V
- 25.0 Hz
b) 6.94 x 10^-3 T
5 a) The work done per unit mass by an external agent in bringing a mass from infinity to that point.
bi) 1.19 x 10^27 kg
ii) -1.36 c 10^31 J
c) 1.41 x 10^7 m/s
d) Both Gravitational and electric field strength is related to their respective potentials by the negative of their potential gradient.
Gravitational potential is always < 0 whereas electric potential can be either positive or negative.
6 a) Rate of change of A decreases with increasing time. At the start, the system M and the disc moves very quickly. There is alot of resistive force from the fluid for the disc to overcome and thus A decreases at a high rate. As M and the disc moves slower, there is lesser resistive force and thus a slower energy loss, causing A to decrease at a lower rate.
bi) 5.00, 1.8, -4.0
ii) Plot the point ( 5.00 , -4.0 ), Draw the best fit line.
iii) -0.310 kg
iv) Equation 1 represents a straight line graph of ln A against 1/m with a negative gradient and a non-zero y-intercept.
v) 1. 0.124 kg/s
- 8.62 x 10^-2
c) 5.59 s
d) since t is directly proportional to m, as m increases, t also increases.
- The lamp, detector and material are all placed in a black box.
After the lid is put in place, the intensity meter should read zero. The holes which the wires are should be kept as small as possible.
The lamp itself is isolated in a compartment and light can only enter via a converging lens. Distance between light source and material d is 1m. Leave a small gap of 1mm between the detector and material.
- Set up the experiment.
-
Switch on the lamp and measure the intensity.
-
Switch off the lamp, put the slab in and measure I through the slab.
-
Use a pair of vernier calipers to measure t.
-
Repeat steps by using 2 slabs taped together, then 3 slabs, and so on, till a total of 6 results is obtained.
-
Record readings in a table.
-
Plot a graph of ln I vs t.
Reliability
Ensure minimum spread of light waves by adding a converging lens or using a laser source.
Safety
If laser source is used, keep flat reflecting materials away and avoid looking directly at laser source.
A-LEVEL H2 MATH JUNE HOLIDAYS INTENSIVE REVISION
A-LEVEL H2 PHYSICS JUNE HOLIDAYS INTENSIVE REVISION
JUNIOR COLLEGE / A-LEVEL TUITION:
_______________________________________________________________________
______________________________________________________________
By EX-MOE TEACHERS & EXPERIENCED TUTORS
@ BLK 644, BUKIT BATOK CENTRAL, #01-68. S(650644).
CALL 65694897 OR SMS 98530744 OR 97860411.
Suggested Answers to 2015 A Level H2 Physics 9646 Paper 3
H2 PHYSICS HANDS-ON PRACTICAL CRASH COURSE (Sat 7/9, 10am – 3pm)
H2 PHYSICS PRACTICAL
H2 CHEMISTRY PRACTICAL
H2 BIOLOGY PRACTICAL
HOW TO BOOK A LAB SESSION:
(1) Decide which Program or Lab sessions you need or most suitable for, by visiting the webpages of the practical subjects you are interested in (if you can’t find the labs you need, message us at 88765498).
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Suggested Answers to 2015 A Level H2 Physics 9646 Paper 3
1a) 1. Net external force acting on the system is zero
- Net torque on the system about any point is zero
bi) Equilibrium system. Ball is falling at constant speed, resultant force is zero.
ii) Not equilibrium. There is resultant force in the form of centripetal force on the satellite due to the earth. There is a constant change in direction therefore there is change in velocity.
2a) 6.33 x 10^7 m
b)Bright fringes get dimmer, dark fringes get brighter.
c)For interference pattern to be seen, a should be much smaller than d. When this condition is not met, fringes cannot be seen. When a becomes too large, the central maximun diffraction bands will not be wide enough to produce inteference pattern.
3a) 14.0ºC
b) With an increase in temperature, the resistance of the thermistor drops. This leads to a drop in the circuit’s total resistance which increases the current through the thermistor, increasing the heating on the thermistor and further lowering the resistance. When the resistance is too low, the whole process will continue and result in a thermal runway, resulting in the thermistor permanently damaged by heat.
4a) Tesla is a unit of magnetic flux density when a uniform magnetic field normal to a wire carrying a current of 1A produces a force per unit length of 1N/m
bi) Force acting on the particle is always normal to the velocity of the particle.
ii) 2.38 x 10^-21 Ns
c) *Line is downwards with a smaller curvature.
5a) A photon is a packet of energy of electromagnetic radiation
bi) 3.84 x 10^-19 J
ii) 8.36 x 10^-25 Ns
c) 52.4º
6ai) Rate of change of velocity
ii) Since the radius of earth is 6400km, several km above the earth’s surface would not affect the acceleration of free fall.
bi) Surface area of the ball is too small or distance traveled by the ball is too short.
ii) 1. 0.299 s
- 0.0362 s
c) 9.8%
di) The height to fall has now been decreased, thus it takes less time for the ball to travel to the point.
ii) The ball will be affected by the magnetic force exerted onto it, thus reducing the net downward force acting on the ball. As a result, the time taken will be longer.
e) Acceleration will decrease till it reaches zero. When air resistance is equal to the weight of the ball, the resultant force acting on the ball will be zero. Hence it has reached terminal velocity and acceleration is zero.
7ai) 1. 24.04 V
- 60.5 Hz
ii) There is a heating effect regardless of direction of the current. It depends on the root mean square current instead of the average current. The heating effect is caused by power loss.
bi) The amount of thermal energy required to raise the temperature of 1 unit mass of the substance by 1ºC
ii) This indicates the power loss from the system to the surrounding.
iii) 4.22 x 10^3 J/kg K
c) *same gradient, y-intercept should be more negative.
di) The increase in internal energy of a system is equal to the sum of the thermal energy supplied to the system and the amount of work done on the system
ii)If specific heat capacity is measured at a constant volume, no work is done on system. If specific heat capacity is measured at constant pressure, there will be a change in volume. Hence, there is work done on the gas.
8ai) An object undergoing oscillatory motion with its acceleration directly proportional to its displacement from its equilibrium position. The acceleration is always towards the equilibrium position.
ii) 1. Acceleration is always opposite in direction of displacement. Hence it is oscillating.
- Acceleration is not directly proportional to its displacement. Hence, oscillations are not simple harmonic.
bi) 1. A wave in which the particles oscillates parallel to the direction of propagation.
- The speed of the wave means the distance traveled per unit time by a wave front.
ii) 1. 1.20 x 10^-3
- 1.20 x 10^-3
-
3.20 x 10^-3 m/s
-
2.71 x 10^-31 J
iii) Sound waves moves relatively fast in gas but the gas molecules oscillate back and forth about their original positions with relatively small speeds.
ci) Energy of the wave originates from the vibrational motion of the gas molecules.
ii) It is caused by the pressure imbalance from the disturbance of the gas’ uniform density.
A-LEVEL H2 MATH JUNE HOLIDAYS IN TENSIVE REVISION
A-LEVEL H2 PHYSICS JUNE HOLIDAYS INTENSIVE REVISION
JUNIOR COLLEGE / A-LEVEL TUITION:
_______________________________________________________________________
______________________________________________________________
By EX-MOE TEACHERS & EXPERIENCED TUTORS
@ BLK 644, BUKIT BATOK CENTRAL, #01-68. S(650644).
CALL 65694897 OR WHATAPPS 88765498 OR 97860411.
Suggested Answers to 2015 A Level H2 Physics 9646 PAPER 1 and 2
H2 PHYSICS HANDS-ON PRACTICAL CRASH COURSE
H2 PHYSICS PRACTICAL
H2 CHEMISTRY PRACTICAL
H2 BIOLOGY PRACTICAL
HOW TO BOOK A LAB SESSION:
(1) Decide which Program or Lab sessions you need or most suitable for, by visiting the webpages of the practical subjects you are interested in (if you can’t find the labs you need, message us at 88765498).
(2) Whatsapp or Message our staff at 88765498 with your Name, Private or School Candidate, A or O level, Subject or Lab Name (e.g Lab PP2), Date and Time of Lab. (Our staff will then guide you on how to register and make payment. If you are not sure about the lab sessions, just state your Name and the Subjects and we will get back to you)
(3) Register Online by clicking below:
(4) Pay Registration fee of $30 via Paynow or funds transfer.
(5) Make the required payment for each lab session at least 5 days before the date of the lab session. (You may also pay for several sessions at one go to ensure that you will have a place in future slots)
Physics Paper 1 MCQ
1) A 2) D 3) C 4) A 5) C
6) A 7) D 8) C 9) A 10) C
11) B 12) A 13) D 14) A 15) A
16) A 17) C 18) A 19) C 20) B
21) C 22) D 23) B 24) B 25) A
26) B 27) B 28) B 29) D 30) B
31) B 32) D 33) B 34) C 35) C
36) C 37) C 38) A 39) D 40) B
Physics Paper 2
1. a) EPE = 0.662J
bi) The system has only GPE before the 300g mass is released. After releasing, part of the GPE is converted to KE and to EPE. At the lowest point, once the mass comes to a stop, all of the GPE will have been converted to EPE.
ii) v = 1.21 m/s
iii) 0.40m
2a) Select 2 points from the graph: (6.0, 1.25) and (12.0, 2.50)
(V – 12.0)/(I – 2.50) = (12.0 – 6.0) / (2.50 – 1.25)
V – 12.0 = (6.0/1.25) (I – 2.50)
V = 4.81I
V is proportional to I
bi) 4.8Ω
ii) 1.3 A , 1.6A
iii) 6.35 V
3a) Incident photons must have energy greater than or equal to the work function of M to emit one electron. Energy of photon is related to the frequency of the electromagnetic radiation by E = hf
bi) Free electrons at the surface are emitted with maximum KE but electrons of the outer shell require more energy to escape thus they will be emitted with less KE. A minimum potential difference is present to stop the photoelectrons with the maximum KE from reaching the collector in order to reduce the current to zero.
ii)It is limited by rate of emission of photoelectrons, which is dependent on the intensity.
c) 9.45 x 10^14 Hz
d) Starts from -2.2v and follows the same shape of the original. The value of y-intercept is doubled due to the doubled intensity.
4a) 0.712 m/s^2
bi) Since the 2 moons move in circular orbits around Jupiter’s center of mass, thus the centripetal force is also the gravitational force due to Jupiter, which acts towards Jupiter’s center of mass.
ii) 2.33
5ai) 4.80 x 10^-14 N
ii) -2.30 x 10^-16 N
b) Electrostatic force on A due to B and B due to A are internal forces. They have the same magnitude but different in direction. Thus the resultant force action on the molecule is zero.
c) 1.6628 x 10^-25 Nm
6a) 128, 54
b) 1510 s
7ai) s = -3.0
ii) -3.170 , -0.693
iii) Plot the point and draw the best fit line.
iv) 3.10
v) There is a linear relationship between ln(y1/m) and ln(l/m) with a gradient of 3.1 and with a y- intercept of -1.02.
vi) 3
b) k = 2.70 x 10^-10
ci) Acceleration is directly proportional to the displacement. The negative sign indicates that the acceleration is opposite to the displacement and is always directed to the position of zero displacement.
ii) 0.560
8.
Apparatus.
Filament lamp, power supply, beaker, thermometer, Stirrer, Stopwatch, Voltmeter, Ammeter, Rheostat, Switch, Weighing machine, Waterproof tape.
Draw a diagram with a filament lamp immersed in water, with the thermometer and stirrer inside the beaker. Draw the circuit of the lamp with connecting ammeter and voltmeter and the power source.
Procedure.
- Measure mass of water in beaker, m.
-
Insulate the beaker.
-
Connect filament lamp with circuit, wrap the metal cap with the waterproof tape.
-
Set up apparatus as shown.
-
Record initial temperature.
-
Close the circuit and start the stopwatch.
-
When the temperature hits 70ºC, stop the stopwatch and record the readings of the voltmeter and ammeter.
-
Repeat steps 5 to 7 after replacing the hot water in the beaker with the same amount of water at room temperature. Obtain different potential differences by changing the resistance of the rheostat.
Working
- 1 – (Thermal energy/electrical energy output)
-
Relationship between I and t is given by η = aV^b , where a and b are constants. The equation can be written as lg η = b lg V + lg a
-
Plot a graph of lg η against lg V and obtain a and b
Accuracy
- Minimise heat loss to surroundings by using the lid and lagging.
-
Conduct preliminary experiments to decide on the suitable values of resistance and voltage supply.
-
Ensure experiment is not conducted over long periods of time so that efficiency of the lamp is not affected by the thermal energy released.
Safety
- Handle with dry hands to prevent electrocution.
-
Ensure the circuit is not in contact with the water.
-
Handle the beaker with care when replacing the water.
A-LEVEL H2 MATH JUNE HOLIDAYS INTENSIVE REVISION
A-LEVEL H2 PHYSICS JUNE HOLIDAYS INTENSIVE REVISION
JUNIOR COLLEGE / A-LEVEL TUITION:
_______________________________________________________________________
______________________________________________________________