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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:

$math jc$

$f math$

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H2 Physics Answers

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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 HOLID AYS INTENSIVE REVISION

A-LEVEL H2 PHYSICS JUNE HOLIDAYS INTENSIVE REVISION

JUNIOR COLLEGE / A-LEVEL TUITION:

$math jc$

$f math$

_______________________________________________________________________

______________________________________________________________

By EX-MOE TEACHERS & EXPERIENCED TUTORS

@ BLK 644, BUKIT BATOK CENTRAL, #01-68. S(650644).

CALL 65694897 OR SMS 98530744 OR 97860411.

H2 Physics Answers

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JUNIOR COLLEGE / A-LEVEL TUITION:

$math jc$

$f math$

_______________________________________________________________________

______________________________________________________________

By EX-MOE TEACHERS & EXPERIENCED TUTORS

@ BLK 644, BUKIT BATOK CENTRAL, #01-68. S(650644).

CALL 65694897 OR WHATAPPS 88765498 OR 97860411.

H2 Physics Answers

H2 PHYSICS HANDS-ON PRACTICAL CRASH COURSE

H2 PHYSICS PRACTICAL

H2 CHEMISTRY PRACTICAL

H2 BIOLOGY PRACTICAL

HOW TO BOOK A LAB SESSION:

(3) Register Online by clicking below:

(4) Pay Registration fee of $30 via Paynow or funds transfer.

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

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