1.ai) acc=9.715 ms-2
aii)1. 0.376% 2. 0.676%
b)
Use s = 1/2 at2, a = 2s/t2,
% error in a = % error in s + 2(% error in t)= 0.376 + 2(0.676)= 0.017%
g = 9.715 + 0.2 = 9.7 +0.2 ms-2
c) 1 . Reaction time when stopping the stopwatch and starting the stopwatch.
2. The determination of the exact point where the ball reaches the distance h .
2. b) i) direction of electric field X to C
ii) F = qE where E is read off the graph at x = 2 cm
F= 1.6 x 10-19 x (2.4 x 103) = 3.84 x 10-16 N
ci)pd = 96V Use Fd=qV, V = Fd ,q= Ed which is area under the graph= 2.4x103 (4x10-2)
ii) This pd is an underestimate as the force at x=2cm is below the average force over 4 cm. Or the actual area is larger than the estimated area , hence an underestimate.
3b) Three conditons for observable interference pattern : 1. the waves must meet. 2. the amplitudes are approx the same 3. the sources are coherent
ii)1. number of directions is 7
2. draw a minimum line in between two maxima line.
4. a) out of page
Bqv = m v2/r
B(2e)r = mv where mv = p
p = 2Ber
ii) Ke = 1/2 mv2=p2/2m
hence Ek = (2Ber)2/2m
= 2(Ber)2/m
c) The speed is greater , the radius of curvature is larger . (straighter)
5. a) electrons from n-type diffuse to the left side the p -type
b) As the electron diffuse to the p-type material at the junction, immobile ions exist in the n and p type create an internal electric field that prevent further diffusion of charges. This region is depleted of mobile charge carrier and hence the formation of depletion region.
c) Positive terminal of battery to the n type material will increase the width of the depletion region. Negative terminal is to the ptype material.
6. a)X is proton
b) Sum (N + He) - Sum of (O+X) = m . Convert m to energy using E = mc2
ii) The reaction can occur if the KE of the reactant is larger than the KE of the product by 1.9x 10-13J.
iii) The oxygen and nucleus X move apart and hence the conservation of momentum is valid as the helium nucleus is moving with high velocity. The KE of the helium nucleus is larger than the KE of the oxygen and X by the difference suggest in ii)
Showing posts with label past papers_2010. Show all posts
Showing posts with label past papers_2010. Show all posts
Wednesday, October 13, 2010
Saturday, October 9, 2010
H1_07_P1
Some commonly asked questions are:
4. Steepest gradient is highest speed .
5. The cyclist down the slope has weight component , hence acc downwards giving positive gradient. Along the horizontal road, friction is opp to his motion , hence acc is negative giving negative gradient.
11 The only vector triangle that has the 3 arrows forming a closed triangle and flowing in the same direction is A and D . D does not show W to be vertical ,hence A.
18. Particle B is at zero displacement. The particle to the right is displaced(-) towards B. The particle to the left is displaced(+) towards B also. Hence B is jammed in between the neighbouring particles. Hence it is compressed,
19. Summation of P and Q at time zero is Y. Shift P and Q a little towards each other , the resultant is Z . Hence the sequence is YZYX YZYX ....
25. Positive voltage shows presence of diode in forward biased. Negative voltage shows presence of ohmic device. Hence parallel connection of diode and resistor.
4. Steepest gradient is highest speed .
5. The cyclist down the slope has weight component , hence acc downwards giving positive gradient. Along the horizontal road, friction is opp to his motion , hence acc is negative giving negative gradient.
11 The only vector triangle that has the 3 arrows forming a closed triangle and flowing in the same direction is A and D . D does not show W to be vertical ,hence A.
18. Particle B is at zero displacement. The particle to the right is displaced(-) towards B. The particle to the left is displaced(+) towards B also. Hence B is jammed in between the neighbouring particles. Hence it is compressed,
19. Summation of P and Q at time zero is Y. Shift P and Q a little towards each other , the resultant is Z . Hence the sequence is YZYX YZYX ....
25. Positive voltage shows presence of diode in forward biased. Negative voltage shows presence of ohmic device. Hence parallel connection of diode and resistor.
Thursday, October 7, 2010
IGSE_6_Oct_2010
1. No resultant force . weight component = frictional force
b) Weight comp greater than friction ,
ii) resultant force = ma
iii) 24 N
ci) resultant force = 38 N , a = 3.2 ms-2
ii) v = 0.8 ms-1
d) acceleration
2. a)200g and 100g at 4 cm and 8 cm respectively
b) Summation of force is zero and summation of torque is zero
c) Moment = 0.2 x 9.81x 4x10-2= 0.078 Nm
d) Reaction force = 0.5 x 9.81= 4.9 N
3ai) hdg = 735 000Pa
ii) 8.35 x 105 Pa
b) 1.625 x 106 Pa
c) density of fresh water less than density of salt water
4a) Random path or zig zag path
b) air molecules hit the dust in all direction. Just as likely to be up or down.
c) random movement smaller
5ai) funnel no longer giving heat to ice.
ii) inside the large piece could well be below freezing point.
or better contact between heater and ice
b) mass of beaker and water
ci) Pt = mc@
c=4.9 x 103 Jkg-1K-1
ii) heat lost/gained or impurities in water
7a) show spreading of waves . keep to same wavelength
b) 8 Hz
c) the same
8a) change ac to dc
b) 86400C
c) 12 J of energy delivered for every coulomb of charge
di) series ii) total power - 16W , 1.33 A iii) 57600J
9 a) pump water to a higher level of shortage
b) less heat loss as current is low when transmitted at high voltage.
c) I2R
d) 34900
e) I = 25A
b) Weight comp greater than friction ,
ii) resultant force = ma
iii) 24 N
ci) resultant force = 38 N , a = 3.2 ms-2
ii) v = 0.8 ms-1
d) acceleration
2. a)200g and 100g at 4 cm and 8 cm respectively
b) Summation of force is zero and summation of torque is zero
c) Moment = 0.2 x 9.81x 4x10-2= 0.078 Nm
d) Reaction force = 0.5 x 9.81= 4.9 N
3ai) hdg = 735 000Pa
ii) 8.35 x 105 Pa
b) 1.625 x 106 Pa
c) density of fresh water less than density of salt water
4a) Random path or zig zag path
b) air molecules hit the dust in all direction. Just as likely to be up or down.
c) random movement smaller
5ai) funnel no longer giving heat to ice.
ii) inside the large piece could well be below freezing point.
or better contact between heater and ice
b) mass of beaker and water
ci) Pt = mc@
c=4.9 x 103 Jkg-1K-1
ii) heat lost/gained or impurities in water
7a) show spreading of waves . keep to same wavelength
b) 8 Hz
c) the same
8a) change ac to dc
b) 86400C
c) 12 J of energy delivered for every coulomb of charge
di) series ii) total power - 16W , 1.33 A iii) 57600J
9 a) pump water to a higher level of shortage
b) less heat loss as current is low when transmitted at high voltage.
c) I2R
d) 34900
e) I = 25A
Wednesday, October 6, 2010
H1_07_p2
1a) The rate of change of momentum of a body is directly proportional to the net external force acting on the body and takes place in the direction of the force.
b) The tension is not zero and act opp to W , hence resultant force is less than mg , acc is less than g
ii) Write two equation for the bucket A and B , solve to obtain expression for acc . Note the direction positive is down for bucket A and up for bucket B .
iii) Sub the mass in the expression to obtain a = 1.6 ms-2 . Use equation of motion to obtain time taken to travel 4 m. time = 2.2s
2i) resistance is infinite from 0 to 0.6 V. Resistance is negligible from 0.6 V onwards.
iii) A curve of decreasing gradient and starting from origin.
b) R = pL/A but V=AL , subs to obtain R=pL2/V
ii) Using the formula above, fractional change in resistance is equal to twice the fractional change in length. The fractional change in length = 0.052 (5.2%) . The fractional change in resistance = 0.104 .The new length is 1.104 of the initial length
b) Draw two lines one directly from source to X , the other to mirror and then to X. ii) At X, constructive interference occur implies phase diff = 0 . Which also means other angles like 360 deg and 720 deg.
c) Use the young double slit formula to calc wavelength.The fringe separation is read from the fig. 3.3 as 0.45 mm , wavelength is 6x10-7 m.
ii) if one of the slit is blocked , the intensity arriving will be 1/4 (amplitude A) of the maximum intensity at constructive interferece (2A)
4 a) Minimum frequency of EM radiation that can emit photoelctrons from the material.
b) As the pd becomes less negative, some of the more energetic photoelectrons will be able to reach the collector and constitute the photoelectric current.
c) eVs = 1/2 mv2
v = 8.6 x 105 ms-1
d) At the intensity is halved , the photo-current will be halved but the KE remain the same hence cut off voltage the same. Indicate which is original and which is the answer.
5a)one kg yields 45x 106 J, hence by proportionality 28 kg will yield 1260 MJ
b) Power = VI and power = energy / time , hence time = energy / VI= 8.1 Hour.
c) 20 kg yields 2.8 MJ hence 1260 MJ needs 9000 kg .
cii) mass of car is about 1000kg m , the mass of the batteries are a few times greater than the mass of the car. Hence mor energy is required to carry the batteries in the car.
d) At constant speed , force of engine = Drag force. Work done by engine = 0.25 ( 1260 MJ ) Work done = F d where F = 580 N , d= 543 km
e) Power = kv3
power=k(0.8v)3
dividing first and second equation
Power /new power = 2, new power is half of the old power .
6. Summation of force is zero and summation of torque is zero.
i) At equilibrium , draw vector triangle that has arrows flowing from one to the next and ending at the starting point.
ii) Solve by sine rule the ratio T1/T2
Since tensions only depends on the angle , the weight will not change the ratio.
c) Taking moments about O, Wd1=Fd2 Where d2 = 42 cm and d1 = 18.3 cm.
F = 77 N
di) Pressure increases linearly with depth of seawater.
ii) Draw F1 shorter at the top and F2 longer at the base.
iii) Resultant = F2 -F1
= hpg A =1640N
2 Upthrust = weight
1640 = mg
m= 167 kg
3. Average density = mass /vol
= 167 /120 x0.19 = 733 kgm-3
iv) The grad is steepest at the earliest time 0.45s. Gradient is the speed.
7. a) he amount of charge which passes through a given cross section of a conductor is 1 second when a constant current of 1 ampere is flowing.
bi) Q = Ne
N = Q/e= 2.0 x 1020
bii) Q = IT
T= Q/I = 32/1.5 x 104=2.1 x 10-3 s
ci) The energy converted by a source from other forms of energy into electrical energy when a unit charge passes through it.
cii) R eff = 0.57 R
2 V= IR
I = E/0.57
R=1.57 E/R
di) V= IR2
V=(E/R1+R2)R2
diii) As the temp of the thermistor increases , its resistance decreases. Hence potential difference across thermistor decreases.This lead to corresponding increase in pd across 100 ohm as the two resistors are connected series.
ei) I = 2.0 A
ii) pd across R = 3V
r= 1.5 ohm.
8a) vertically downwards,
ii) the mass will move in parabolic path towards the ground.
bi) horizontally leftwards
bii) decelerate to a rest and then accelerated along the same line of motion in the opp direction.
c) the molecule will experience a clockwise rotation about its centre , with no translational movement .
di) see def.
dii) draw to scale : Ans 53 mT
e) force F = BIL
torque = Fxd= BIab = BIA Since B and I are constants , torque is directly proportional to area.
fi) r = 8.1 x 10-3 m
fii) F/L = BI= 3.3 x 10-3 Nm-1
b) The tension is not zero and act opp to W , hence resultant force is less than mg , acc is less than g
ii) Write two equation for the bucket A and B , solve to obtain expression for acc . Note the direction positive is down for bucket A and up for bucket B .
iii) Sub the mass in the expression to obtain a = 1.6 ms-2 . Use equation of motion to obtain time taken to travel 4 m. time = 2.2s
2i) resistance is infinite from 0 to 0.6 V. Resistance is negligible from 0.6 V onwards.
iii) A curve of decreasing gradient and starting from origin.
b) R = pL/A but V=AL , subs to obtain R=pL2/V
ii) Using the formula above, fractional change in resistance is equal to twice the fractional change in length. The fractional change in length = 0.052 (5.2%) . The fractional change in resistance = 0.104 .The new length is 1.104 of the initial length
b) Draw two lines one directly from source to X , the other to mirror and then to X. ii) At X, constructive interference occur implies phase diff = 0 . Which also means other angles like 360 deg and 720 deg.
c) Use the young double slit formula to calc wavelength.The fringe separation is read from the fig. 3.3 as 0.45 mm , wavelength is 6x10-7 m.
ii) if one of the slit is blocked , the intensity arriving will be 1/4 (amplitude A) of the maximum intensity at constructive interferece (2A)
4 a) Minimum frequency of EM radiation that can emit photoelctrons from the material.
b) As the pd becomes less negative, some of the more energetic photoelectrons will be able to reach the collector and constitute the photoelectric current.
c) eVs = 1/2 mv2
v = 8.6 x 105 ms-1
d) At the intensity is halved , the photo-current will be halved but the KE remain the same hence cut off voltage the same. Indicate which is original and which is the answer.
5a)one kg yields 45x 106 J, hence by proportionality 28 kg will yield 1260 MJ
b) Power = VI and power = energy / time , hence time = energy / VI= 8.1 Hour.
c) 20 kg yields 2.8 MJ hence 1260 MJ needs 9000 kg .
cii) mass of car is about 1000kg m , the mass of the batteries are a few times greater than the mass of the car. Hence mor energy is required to carry the batteries in the car.
d) At constant speed , force of engine = Drag force. Work done by engine = 0.25 ( 1260 MJ ) Work done = F d where F = 580 N , d= 543 km
e) Power = kv3
power=k(0.8v)3
dividing first and second equation
Power /new power = 2, new power is half of the old power .
6. Summation of force is zero and summation of torque is zero.
i) At equilibrium , draw vector triangle that has arrows flowing from one to the next and ending at the starting point.
ii) Solve by sine rule the ratio T1/T2
Since tensions only depends on the angle , the weight will not change the ratio.
c) Taking moments about O, Wd1=Fd2 Where d2 = 42 cm and d1 = 18.3 cm.
F = 77 N
di) Pressure increases linearly with depth of seawater.
ii) Draw F1 shorter at the top and F2 longer at the base.
iii) Resultant = F2 -F1
= hpg A =1640N
2 Upthrust = weight
1640 = mg
m= 167 kg
3. Average density = mass /vol
= 167 /120 x0.19 = 733 kgm-3
iv) The grad is steepest at the earliest time 0.45s. Gradient is the speed.
7. a) he amount of charge which passes through a given cross section of a conductor is 1 second when a constant current of 1 ampere is flowing.
bi) Q = Ne
N = Q/e= 2.0 x 1020
bii) Q = IT
T= Q/I = 32/1.5 x 104=2.1 x 10-3 s
ci) The energy converted by a source from other forms of energy into electrical energy when a unit charge passes through it.
cii) R eff = 0.57 R
2 V= IR
I = E/0.57
R=1.57 E/R
di) V= IR2
V=(E/R1+R2)R2
diii) As the temp of the thermistor increases , its resistance decreases. Hence potential difference across thermistor decreases.This lead to corresponding increase in pd across 100 ohm as the two resistors are connected series.
ei) I = 2.0 A
ii) pd across R = 3V
r= 1.5 ohm.
8a) vertically downwards,
ii) the mass will move in parabolic path towards the ground.
bi) horizontally leftwards
bii) decelerate to a rest and then accelerated along the same line of motion in the opp direction.
c) the molecule will experience a clockwise rotation about its centre , with no translational movement .
di) see def.
dii) draw to scale : Ans 53 mT
e) force F = BIL
torque = Fxd= BIab = BIA Since B and I are constants , torque is directly proportional to area.
fi) r = 8.1 x 10-3 m
fii) F/L = BI= 3.3 x 10-3 Nm-1
Friday, October 1, 2010
H2_09_p1
1. Use calc to divide. Cannot be whole number.
2. Make a good set of a few hundred eV
3. Deduce the gradient at 3parts
4. Decide to take up as positive. Hence a is negative . After 2s, stone on the way down . Choose negative velocity
5. Find using component method . Vertical component . And horizontal comp separately. Using pythagoaros thm.
6. Centrifugal is new term used in exam. It refers a force opp to centripetal force .
7. Elastic . Use velocity of separation = velocity op approach.
8 Take moment about a point. Not vertical forces and horizontal forces.
9. Has mass and is negatively charged .
10 Upthrust is weight of water displaced. Consider only outer vol of boat. The inner vol is to find weight of oil.the increase in weight due to load = increase in weight of water displaced.
11. Draw 2 forces : W and Fe. If the charge move in dir of W ; decrease in gpe. Move opp to W increase in gpe. So it works for epe the same way. Check if it works for you
12 Drag force= driving force =200 N
Driving force = drag force + weight component .hence new power can be found
13. Drag force =driving force of boat. Power= force x v = kv3. Use power ~v3
14 The resultant force is friction
15 The velocity = rw where T = 24 h
16 F = G 2MM/9x2 = 2/9GMM/x2
17 GMm/r=3.2MJ . Total energy at 2r = 0.5 GMm/2r =1/4GMm/r=3.2 /4 MJ
18 Change the n cycles to radians
19. Graph shows SHM. Hence U shape graph for energy.
20. Two hydrogen atoms in one molecule. Hence twice the number of molecules
21 Suddenly expand imply q=o, expansion means W is negative. U is negative
22 At L1 + x 1/4wavelength. At L2 + x = 3/4 wavelength. Solve for wavelength.
23 The distance bet P and Q read on the x axis is less than a quarter . One quarter of one wavelength is 45 degrees. So less than one quarter is 30 degrees
24 Use young double slit equation . The satellite moving with speed v travel a distance x in a time T = 1/f. So x = v/ f
25 Point 3 arrows to show E due to the charges at the 3corners at the fourth corner.
26 Power = VI .total power = EI , E =( P+p)/I
27 The ratio is the R value. Change in resistance is the difference in the ratio
28 In series resistor is NR. But in parallel is NR/N= R
29 The effective value across voltmeter is 1500 ohm . Voltmeter reading = (12/3500)1500 = 5.14 V
30 UseLHR to determined charge . The current direction and the charge motion is the same , hence positive charge. Smaller radius means decreasing speed
31The two currents are opposite in direction as the flux density at O decrease. Now you can consider point P , if the current in X and Y are in opposite direction it decreases. At Q, the B due to X and Y are in the same direction , so increase.
32.The magnetic flux increase when iron rod enters the solenoid. To oppose the increase in flux the current dip momentarily. When the iron rod leaves , the opp occurs.
33 The magnetic flux is the same but not the magnetic flux density . flux linkage is increased with more turns N.
34The graph is of sine form and period T=2.5ms
35 mean power = 0.5 peak power
36same gradient and lower threshold freq.
40 Use A=0.693(N/half life). Calc (n/half life) and choose the maximum value.
2. Make a good set of a few hundred eV
3. Deduce the gradient at 3parts
4. Decide to take up as positive. Hence a is negative . After 2s, stone on the way down . Choose negative velocity
5. Find using component method . Vertical component . And horizontal comp separately. Using pythagoaros thm.
6. Centrifugal is new term used in exam. It refers a force opp to centripetal force .
7. Elastic . Use velocity of separation = velocity op approach.
8 Take moment about a point. Not vertical forces and horizontal forces.
9. Has mass and is negatively charged .
10 Upthrust is weight of water displaced. Consider only outer vol of boat. The inner vol is to find weight of oil.the increase in weight due to load = increase in weight of water displaced.
11. Draw 2 forces : W and Fe. If the charge move in dir of W ; decrease in gpe. Move opp to W increase in gpe. So it works for epe the same way. Check if it works for you
12 Drag force= driving force =200 N
Driving force = drag force + weight component .hence new power can be found
13. Drag force =driving force of boat. Power= force x v = kv3. Use power ~v3
14 The resultant force is friction
15 The velocity = rw where T = 24 h
16 F = G 2MM/9x2 = 2/9GMM/x2
17 GMm/r=3.2MJ . Total energy at 2r = 0.5 GMm/2r =1/4GMm/r=3.2 /4 MJ
18 Change the n cycles to radians
19. Graph shows SHM. Hence U shape graph for energy.
20. Two hydrogen atoms in one molecule. Hence twice the number of molecules
21 Suddenly expand imply q=o, expansion means W is negative. U is negative
22 At L1 + x 1/4wavelength. At L2 + x = 3/4 wavelength. Solve for wavelength.
23 The distance bet P and Q read on the x axis is less than a quarter . One quarter of one wavelength is 45 degrees. So less than one quarter is 30 degrees
24 Use young double slit equation . The satellite moving with speed v travel a distance x in a time T = 1/f. So x = v/ f
25 Point 3 arrows to show E due to the charges at the 3corners at the fourth corner.
26 Power = VI .total power = EI , E =( P+p)/I
27 The ratio is the R value. Change in resistance is the difference in the ratio
28 In series resistor is NR. But in parallel is NR/N= R
29 The effective value across voltmeter is 1500 ohm . Voltmeter reading = (12/3500)1500 = 5.14 V
30 UseLHR to determined charge . The current direction and the charge motion is the same , hence positive charge. Smaller radius means decreasing speed
31The two currents are opposite in direction as the flux density at O decrease. Now you can consider point P , if the current in X and Y are in opposite direction it decreases. At Q, the B due to X and Y are in the same direction , so increase.
32.The magnetic flux increase when iron rod enters the solenoid. To oppose the increase in flux the current dip momentarily. When the iron rod leaves , the opp occurs.
33 The magnetic flux is the same but not the magnetic flux density . flux linkage is increased with more turns N.
34The graph is of sine form and period T=2.5ms
35 mean power = 0.5 peak power
36same gradient and lower threshold freq.
40 Use A=0.693(N/half life). Calc (n/half life) and choose the maximum value.
Tuesday, September 28, 2010
H2_08_paper3
1. To show inverse proportionality plot p vs 1/V
ii) A straight line passing through the origin is obtained.
bi) Find the area under the graph for ever 1s interval . To plot the graph of W vs d , The cummulative area is the work done at each d on the x axis . A curve is obtained. After (1m, W = 5J) , (2m , W= 17.5J) , (3,37.5) and (4, 47.5)
c) the grad of the garph is -2
ii) hence g is proportional to r-2 . Then g is inversely proportional to r2.
2. ai) Magnetic flux linkage of the coil changes as it rotates. By Faraday's law emf is generated.
aii) The two factors are speed of rotation and the No of turns or B
iii) The flux linkage of the coil depends on the angle @ between the coil and field . This angle changes sinuisoidally with time . Hence emf is sinuisoidal.
bi) peak input emf = 72 x 1.414 where sq root of 2 = 1.414
peak input emf= 102 V
bii)rms output voltage = pd across the resistor
= 20x 72 V = 1440 v
iii) rms current = rms voltage / R
= 1440/160 =9 A
iv) mean power = I2R = 92(160) = 1.3 x 104 W
v) VI input = VI output
I input = 1.3 x 104/72 = 180 A
3. a)Wavefronts of stationary waves do not advance. Energy is trapped within the waves.
b) Diffraction is the spreading of waves when the waves pass through a slit or an obstacle.
c) Two waves have constant phase relationship are said to be coherent.
d) The oscillation of the wave is in one direction in a plane normal to the direction of transfer of energy.
4. The internal energy is the sum of ke and pe all all molecules of the gas .
ii) The increase in internal energy is the sum of the heat supplied to the gas and the work done on the gas.
b) i) Work done by the gas is the area under the graph .
W= p(change in vol) = 1.5 J
ii) A-B : 0 / 4.2 / 4.2
B-C : -8.5 / 0 / -8.5
C-A : 5.8 / -1.5 / 4.3
5a) force acting on per unit positive charge placed at that point.
ii) Work done = F d but F = qE
W = qEd.
iii) V = W/q , sub W = qEd
V = Ed
bi) The number of electrons per s = N/t
I = Ne/t , N /t = I/e =5.4 x 1016 s-1
qV = 1/2 mv2
v2 = 2eV /m
v=1.45 x 10 8 ms-1
iii) power = N( ke) of electrons
P= N/t ( KE) = 5.2 x 105 W
or Power = IV = 8.6 mA ( 60kV) = 516 W
c) power supplied by electrons= rate of heat removed by coolant.
516 W = mc@/t , m/t = 516 /c@= 4.9 x 10-3 kgs-1
d) The lines appear to radiate from the centre. So the charge is concentrated at the centre of the sphere.
ii) E = kQ/r2= 5.4 x103 NC-1
iii) The Va and Vb and Vc are to be calculated separately . potential difference is the same . Hence Va - Vb = Vb - Vc.
Cancel Q and k . we have (1/0.4 -1/0.5 ) = (1/0.5 - 1/r)
r = 0.67 m
6a) freq is no of cycles per s but the angular freq has units in radians per s. Therefore angular freq is w= (2x3.142xf) rad s-1
bi) loss in gpe = mgh = 0.4x 9.81x 0.2=0.785J
ii) epe = 1/2 kx2 = 1/2 x19.62x(0.2)2= 0.392 J
c) loss in gpe is not equal to gain in epe as energy is loss as heat due to the stretching of the spring.
d) T-mg = ke - mg = 19.62(0.4) - mg = 3.93 N
(Note ; Important to draw a free body diag at the bottom of the motion. The positive is taken to be up (towards the eqm) T-mg = ma
ii) a = w2Xo
w2 = a/xo = F/mxo= 3.93 /0.4x0.2.
w= 7.0 rads-1
iii) vo=wxo = 7.0x 0.2 = 1.4 ms-1
e) lowest point : 0/1.57/0/1.57
equilm point : 0.78/0.39 /0.39/1.57
highest point: 1.57/ 0/0/1.57
7. This expt involves alpha particle and gold foil. The particles travel without deflection through the atom (not nucleus and not between the atoms) . The nucleus is small in comparision with the atom. Use the correct words for description.
b) x= 138, y=40
ii) E released = (235.0439+1.0087-137.9603-97.9197)x 1.66x10-27xc2= 2.6x 10-11J
iii) The remaining energy is released as gamma photon.
iv) Com: Initial p = final p
0=m1v1 +m2v2 where m1 is zirconium and m2 is tellerium , hence ratio of their speed
v1/v2=m2/m1=138/98
=1.4
2. The ratio of the ke = 1/2 m1(v1)2 divide by 1/2 m2(v2)2
ratio = m1/m2(1.4)2 = 98/138(1.4)2= 1.4
v) ke of zirconium = (1.4 /2.4 ) (2.3 x 10-11) =1.34 x 10-11J
v2 = 2x1.34x 10-11/98u
v = 1.3 x 10 7 ms-1
vi) Assumption one : The initial momentum assumed zero as the neutron speed is negligible.
Assumption 2: No other external forces are acting and hence conservation momentum is true.
ii) A straight line passing through the origin is obtained.
bi) Find the area under the graph for ever 1s interval . To plot the graph of W vs d , The cummulative area is the work done at each d on the x axis . A curve is obtained. After (1m, W = 5J) , (2m , W= 17.5J) , (3,37.5) and (4, 47.5)
c) the grad of the garph is -2
ii) hence g is proportional to r-2 . Then g is inversely proportional to r2.
2. ai) Magnetic flux linkage of the coil changes as it rotates. By Faraday's law emf is generated.
aii) The two factors are speed of rotation and the No of turns or B
iii) The flux linkage of the coil depends on the angle @ between the coil and field . This angle changes sinuisoidally with time . Hence emf is sinuisoidal.
bi) peak input emf = 72 x 1.414 where sq root of 2 = 1.414
peak input emf= 102 V
bii)rms output voltage = pd across the resistor
= 20x 72 V = 1440 v
iii) rms current = rms voltage / R
= 1440/160 =9 A
iv) mean power = I2R = 92(160) = 1.3 x 104 W
v) VI input = VI output
I input = 1.3 x 104/72 = 180 A
3. a)Wavefronts of stationary waves do not advance. Energy is trapped within the waves.
b) Diffraction is the spreading of waves when the waves pass through a slit or an obstacle.
c) Two waves have constant phase relationship are said to be coherent.
d) The oscillation of the wave is in one direction in a plane normal to the direction of transfer of energy.
4. The internal energy is the sum of ke and pe all all molecules of the gas .
ii) The increase in internal energy is the sum of the heat supplied to the gas and the work done on the gas.
b) i) Work done by the gas is the area under the graph .
W= p(change in vol) = 1.5 J
ii) A-B : 0 / 4.2 / 4.2
B-C : -8.5 / 0 / -8.5
C-A : 5.8 / -1.5 / 4.3
5a) force acting on per unit positive charge placed at that point.
ii) Work done = F d but F = qE
W = qEd.
iii) V = W/q , sub W = qEd
V = Ed
bi) The number of electrons per s = N/t
I = Ne/t , N /t = I/e =5.4 x 1016 s-1
qV = 1/2 mv2
v2 = 2eV /m
v=1.45 x 10 8 ms-1
iii) power = N( ke) of electrons
P= N/t ( KE) = 5.2 x 105 W
or Power = IV = 8.6 mA ( 60kV) = 516 W
c) power supplied by electrons= rate of heat removed by coolant.
516 W = mc@/t , m/t = 516 /c@= 4.9 x 10-3 kgs-1
d) The lines appear to radiate from the centre. So the charge is concentrated at the centre of the sphere.
ii) E = kQ/r2= 5.4 x103 NC-1
iii) The Va and Vb and Vc are to be calculated separately . potential difference is the same . Hence Va - Vb = Vb - Vc.
Cancel Q and k . we have (1/0.4 -1/0.5 ) = (1/0.5 - 1/r)
r = 0.67 m
6a) freq is no of cycles per s but the angular freq has units in radians per s. Therefore angular freq is w= (2x3.142xf) rad s-1
bi) loss in gpe = mgh = 0.4x 9.81x 0.2=0.785J
ii) epe = 1/2 kx2 = 1/2 x19.62x(0.2)2= 0.392 J
c) loss in gpe is not equal to gain in epe as energy is loss as heat due to the stretching of the spring.
d) T-mg = ke - mg = 19.62(0.4) - mg = 3.93 N
(Note ; Important to draw a free body diag at the bottom of the motion. The positive is taken to be up (towards the eqm) T-mg = ma
ii) a = w2Xo
w2 = a/xo = F/mxo= 3.93 /0.4x0.2.
w= 7.0 rads-1
iii) vo=wxo = 7.0x 0.2 = 1.4 ms-1
e) lowest point : 0/1.57/0/1.57
equilm point : 0.78/0.39 /0.39/1.57
highest point: 1.57/ 0/0/1.57
7. This expt involves alpha particle and gold foil. The particles travel without deflection through the atom (not nucleus and not between the atoms) . The nucleus is small in comparision with the atom. Use the correct words for description.
b) x= 138, y=40
ii) E released = (235.0439+1.0087-137.9603-97.9197)x 1.66x10-27xc2= 2.6x 10-11J
iii) The remaining energy is released as gamma photon.
iv) Com: Initial p = final p
0=m1v1 +m2v2 where m1 is zirconium and m2 is tellerium , hence ratio of their speed
v1/v2=m2/m1=138/98
=1.4
2. The ratio of the ke = 1/2 m1(v1)2 divide by 1/2 m2(v2)2
ratio = m1/m2(1.4)2 = 98/138(1.4)2= 1.4
v) ke of zirconium = (1.4 /2.4 ) (2.3 x 10-11) =1.34 x 10-11J
v2 = 2x1.34x 10-11/98u
v = 1.3 x 10 7 ms-1
vi) Assumption one : The initial momentum assumed zero as the neutron speed is negligible.
Assumption 2: No other external forces are acting and hence conservation momentum is true.
H2_08_paper2
1. vertical component can be found using the vertical height and initial velocity is zero. Ans : 25ms-1
ii) The component 34 sin @= 25 , hence angle @ = 47.3
b) The is rate of change of momentum .
F= m(change in vel) /t = 0.13 ( 34 - 2 ) / 0.95 = 4.4 N
c) using only energy consideration, account for the loss of ke of the stone. The water gain ke and pe .
2. The straight line passing through the origin shows that acc is proportional to displacement.
The negative grad shows that the dir of acc is opp the the displacement .
bi) When the sand lose contact with the plate, acc= g . The Weight is the resultant force that gives rise to acc a . mg = ma , hence a = g
ii) Read from graph the x when g = 9.81 ms-2. x = 3.8 mm
3. a) Equate centripetal acc to gravitational acceleration
Obtain the expression for the speed. subs the speed to otain the kinetic energy of the satellite .
11) gpe/ke = GMm/r divide by 1/2(GMm/r) = -2
b) Plot the graph of KE by using the Gpe value and dividing it by 2
ii) Read off the two values of GPE at the two distances. Half the GPE to obtain KE . Find the speed separately. Deduct to find change in speed.
4. a) current = power/ V
I = 5.62/ 4.5=1.25 A
ii) pd= IR , R = V/I = 4.5 /1.25 = 3.6 ohm
b) E = 4.5 + 1.25 (r) and E= 1.6(2.03 + r) equate the two equations.
r = 3.6 ohm
5. When temperature increase , electrons in valence band have sufficient energy to overcome band gap and are raised up to conduction band. This create increase in number of electrons in the CB and holes in the VB. More charge carriers contribute to the decrease in resistance.
6. a) Nuclei that emits particles like alpha , beta and gamma radiation is said to be radioactive.
bi) The decay constant is the fration of the nuclei that decay per unit time.
ii) decay constant = A/N = 1.26 x 105/ N =7.8 x 10-10 s-1
N can be obtained by dividing mass with 90u.
c) Activity would change considerably during the time of measurement if decay constant were large.
7. a) gradient indicates the rate. The rate of increase of radius is higher at the start and becomes lower at a later time.
ii) The volume of the room = 12x5x3= 180 m3. The radius would be 3.5 m . Check the time taken for the fire to fill the room would be 1 ms . Hence the fire ball spread the whole room too quickly and hence would be hazardous.
b) read off graph
d)i) the constant c may type of liquid or volatility of the liquid.
ii) Other situation like explosion of petrol tank in cars.
ii) The component 34 sin @= 25 , hence angle @ = 47.3
b) The is rate of change of momentum .
F= m(change in vel) /t = 0.13 ( 34 - 2 ) / 0.95 = 4.4 N
c) using only energy consideration, account for the loss of ke of the stone. The water gain ke and pe .
2. The straight line passing through the origin shows that acc is proportional to displacement.
The negative grad shows that the dir of acc is opp the the displacement .
bi) When the sand lose contact with the plate, acc= g . The Weight is the resultant force that gives rise to acc a . mg = ma , hence a = g
ii) Read from graph the x when g = 9.81 ms-2. x = 3.8 mm
3. a) Equate centripetal acc to gravitational acceleration
Obtain the expression for the speed. subs the speed to otain the kinetic energy of the satellite .
11) gpe/ke = GMm/r divide by 1/2(GMm/r) = -2
b) Plot the graph of KE by using the Gpe value and dividing it by 2
ii) Read off the two values of GPE at the two distances. Half the GPE to obtain KE . Find the speed separately. Deduct to find change in speed.
4. a) current = power/ V
I = 5.62/ 4.5=1.25 A
ii) pd= IR , R = V/I = 4.5 /1.25 = 3.6 ohm
b) E = 4.5 + 1.25 (r) and E= 1.6(2.03 + r) equate the two equations.
r = 3.6 ohm
5. When temperature increase , electrons in valence band have sufficient energy to overcome band gap and are raised up to conduction band. This create increase in number of electrons in the CB and holes in the VB. More charge carriers contribute to the decrease in resistance.
6. a) Nuclei that emits particles like alpha , beta and gamma radiation is said to be radioactive.
bi) The decay constant is the fration of the nuclei that decay per unit time.
ii) decay constant = A/N = 1.26 x 105/ N =7.8 x 10-10 s-1
N can be obtained by dividing mass with 90u.
c) Activity would change considerably during the time of measurement if decay constant were large.
7. a) gradient indicates the rate. The rate of increase of radius is higher at the start and becomes lower at a later time.
ii) The volume of the room = 12x5x3= 180 m3. The radius would be 3.5 m . Check the time taken for the fire to fill the room would be 1 ms . Hence the fire ball spread the whole room too quickly and hence would be hazardous.
b) read off graph
d)i) the constant c may type of liquid or volatility of the liquid.
ii) Other situation like explosion of petrol tank in cars.
Monday, September 27, 2010
H2_09_paper2
1. Vectors have direction but scalar does not have.
acceleration is the rate of change of velocity.
c) Pythagoras theorem = c2= a2+b2. To find change in velocity = 25.5 sm-1
ii) acc= change in vel / time taken = 5.8 ms-1
iii) Arrow pointing to the centre of circle
iv) force perpendicular to direction of motion. No work done by this force and hence no change in ke.
2. vertical component = 5.94 ms-1
b) av force =( change in m v /t) = 24.1 N .Direction : leftwards.
c) inelastic as ke is lost.
d) the time taken to fall to the ground is the same but the horizontal comp is reduced and hence the range is smaller/nearer to the wall.
3. The mass is not uniformly distributed. More mass on the leftside.
b) Net force = 0
net moment about any point =0
c) T1 = 59380 N, T2 = 5630 N
4. diffraction: Light waves spread out after passing through each slit of the grating.
coherence : Light waves passing through each slit have a constant phase difference.
superposition: When the two waves from each slits meet , the resultant displacement is the sum of their individual displacement at that point. The displacement is eletric field.
bi) 40.5 deg
bii) 31.7 deg
iii) Part of 2nd order overlap with the 3rd order.
5.a) F= 57.5 N
The shaded area represent the work done in bringing p1 from infinity to a point 2 x 10-15 away from p2.
c) When fusion occurs, nuclear energy (binding energy) released due to loss in mass of both particle is converted to ke of the product particle.
6 a) The change in flux linkage = NBA = 0.019 Wb
b) There is change in flux , by Faradays law there is an induced current in the coil.
ii) the emf = 0.019 / 2.0 = 4.75 mV
7. USe the values of P1 and V1 to compare with P2 V2 . Subs values to show that T2 is greater than T1.
b) If the graph is linear , the equation is valid , the gradient = & and the lg c is the intercept.
ci) 5.41
cii) & =1.45
c = 12.3 Pa -4.35
d) T= 520 K
e ) The inital temp is greater than the surrounding temp. If it expands rapidly , change in Q = 0 . Big fie 7.4 the temp decreases rapidly.
acceleration is the rate of change of velocity.
c) Pythagoras theorem = c2= a2+b2. To find change in velocity = 25.5 sm-1
ii) acc= change in vel / time taken = 5.8 ms-1
iii) Arrow pointing to the centre of circle
iv) force perpendicular to direction of motion. No work done by this force and hence no change in ke.
2. vertical component = 5.94 ms-1
b) av force =( change in m v /t) = 24.1 N .Direction : leftwards.
c) inelastic as ke is lost.
d) the time taken to fall to the ground is the same but the horizontal comp is reduced and hence the range is smaller/nearer to the wall.
3. The mass is not uniformly distributed. More mass on the leftside.
b) Net force = 0
net moment about any point =0
c) T1 = 59380 N, T2 = 5630 N
4. diffraction: Light waves spread out after passing through each slit of the grating.
coherence : Light waves passing through each slit have a constant phase difference.
superposition: When the two waves from each slits meet , the resultant displacement is the sum of their individual displacement at that point. The displacement is eletric field.
bi) 40.5 deg
bii) 31.7 deg
iii) Part of 2nd order overlap with the 3rd order.
5.a) F= 57.5 N
The shaded area represent the work done in bringing p1 from infinity to a point 2 x 10-15 away from p2.
c) When fusion occurs, nuclear energy (binding energy) released due to loss in mass of both particle is converted to ke of the product particle.
6 a) The change in flux linkage = NBA = 0.019 Wb
b) There is change in flux , by Faradays law there is an induced current in the coil.
ii) the emf = 0.019 / 2.0 = 4.75 mV
7. USe the values of P1 and V1 to compare with P2 V2 . Subs values to show that T2 is greater than T1.
b) If the graph is linear , the equation is valid , the gradient = & and the lg c is the intercept.
ci) 5.41
cii) & =1.45
c = 12.3 Pa -4.35
d) T= 520 K
e ) The inital temp is greater than the surrounding temp. If it expands rapidly , change in Q = 0 . Big fie 7.4 the temp decreases rapidly.
Thursday, September 23, 2010
H2_08_paper1
1 ADDCD
6 CCBDB
11 BCDBA
16 DCDBC
21 DADAD
26 BCCBB
31BBCDB
36BDDCC
6. The total momentum = 0 . final vel = 0 . To find the force on each , consider either the 6 kg or the 10 kg alone. The change in p = 30 . The force = 30/0.2s= 150 N
8. Find the centre of mass of the woooden and rubber handle. The ratio of masses
Mw(0.9)= Mr(1.6) using summation of moment. density = mass / vol , hence dr/dw= (mr/mw) ( vw/vr) but v = AL.
The ration dw/dr = (0.9/1.6) ( Lw/Lr) = (0.9/1.6) (4/1) = 2.25
10. The work done depends on opposing force which is dependant upon the induced current. Hence F = BIL where I= E/r = BLv/r. Hence force F is directly proportional to v . Hence workdone is directly proportional to v.
18. gradient of graph = 1/nrT. Hence if amount of gas and temperature is doubled .
grad = 1/4 (nrT) . hence 1/4 of original gradient.
21. K = hc = 1.99 x 10 -25 Jm . converting the unit to eV and nm .
The K = 1.99 x 10-25 J(1 eV/1.6x-19J)=1.24x10-6 eVm= 1.24 x 10-6 eV(109nm/1m)m = 1.24x103 eVnm. Now it is properly converted to eVnm . Note : the conversion to unit that you want involves multiplying by unit . The bracketed ratio = 1
23 If the maximum order is 5 , so the total image is 5 + 5 +1 =11
25. Work done on the charge must be a positive value. Hence q(higher V- lower V) = q (V1-V2) .
29. The pd drop across the thermistor is 3V , the pd drop across the R (at the base is also 3V. Hence the voltmeter is 0V.
The pd drop across thermistor (5R) is now 5V , the pd drop across the R(at the base) is still 3V . Hence the pd across the voltmeter is 2V.
32. Resolve B to the components . Bsin@ is perpendicular to area . Hence flux linkage= BAnsin@. When rotated to zero degrees , flux linkage = 0 . hence the change in flux linkage = BAn sin @
35. If velocity is increased , the de broy wavelength of the electron is decreased. Using dsin@ = n(wavelenth) , the angle @ is smaller , so diameter of circle is decreased.
39. The activity 532 per min must deduct 24 per min and then divide by 4 due to 2 half lives. The final answer add 24 per min.
6 CCBDB
11 BCDBA
16 DCDBC
21 DADAD
26 BCCBB
31BBCDB
36BDDCC
6. The total momentum = 0 . final vel = 0 . To find the force on each , consider either the 6 kg or the 10 kg alone. The change in p = 30 . The force = 30/0.2s= 150 N
8. Find the centre of mass of the woooden and rubber handle. The ratio of masses
Mw(0.9)= Mr(1.6) using summation of moment. density = mass / vol , hence dr/dw= (mr/mw) ( vw/vr) but v = AL.
The ration dw/dr = (0.9/1.6) ( Lw/Lr) = (0.9/1.6) (4/1) = 2.25
10. The work done depends on opposing force which is dependant upon the induced current. Hence F = BIL where I= E/r = BLv/r. Hence force F is directly proportional to v . Hence workdone is directly proportional to v.
18. gradient of graph = 1/nrT. Hence if amount of gas and temperature is doubled .
grad = 1/4 (nrT) . hence 1/4 of original gradient.
21. K = hc = 1.99 x 10 -25 Jm . converting the unit to eV and nm .
The K = 1.99 x 10-25 J(1 eV/1.6x-19J)=1.24x10-6 eVm= 1.24 x 10-6 eV(109nm/1m)m = 1.24x103 eVnm. Now it is properly converted to eVnm . Note : the conversion to unit that you want involves multiplying by unit . The bracketed ratio = 1
23 If the maximum order is 5 , so the total image is 5 + 5 +1 =11
25. Work done on the charge must be a positive value. Hence q(higher V- lower V) = q (V1-V2) .
29. The pd drop across the thermistor is 3V , the pd drop across the R (at the base is also 3V. Hence the voltmeter is 0V.
The pd drop across thermistor (5R) is now 5V , the pd drop across the R(at the base) is still 3V . Hence the pd across the voltmeter is 2V.
32. Resolve B to the components . Bsin@ is perpendicular to area . Hence flux linkage= BAnsin@. When rotated to zero degrees , flux linkage = 0 . hence the change in flux linkage = BAn sin @
35. If velocity is increased , the de broy wavelength of the electron is decreased. Using dsin@ = n(wavelenth) , the angle @ is smaller , so diameter of circle is decreased.
39. The activity 532 per min must deduct 24 per min and then divide by 4 due to 2 half lives. The final answer add 24 per min.
Wednesday, September 22, 2010
Tuesday, September 21, 2010
H2_09_lesson2
Section B
5ai) force per unit nass
ii)The def is force per unit mass , so F/m = GMm/R2 divide by m gives g = GM/R2.
b) The density = 2.52 x 107 kgm-3
bii) The density is greater in the centre as the outer layers compresses the inner layers.
ci) g = 1.2 x 1012 Nkg-1 cii) acc = 1.59 x 107 ms-2
iii) comparing this values , a is less than g , the normal force is greater than 0 , particle will not leave.
d) The accelerating protons has ke and when it approaches the star and undergo decceleration and emit x rays. The decrease in KE is converted to Xray photon.
6. The gradient of the graph is not constant hence the net force not constant . Since Weight is constant , the air resistance veries with speed.
b) gradient of the graph at velocity = 0 gives the magnitude of free fall as no drag force act at v = 0.
c) count the area under the graph. height = 19.2 m
cii) Energy lost = KEi - mgh)/KEi = 0.4
d) i) acc of the ball is gradient of the graph at v = 10 ms-1.
ii) Fd = ma - mg
Fd = 0.35(13-9.81) = 1.12N
e) The area below the x axis is the same above the x axis. Height of rise and fall is the same. Since the gradient will be less slopping, the area is found to extend for a longer time. Gradient of graph decrease when the object is moving downwards.
7.b) Charge = It = 72C
ii) E = 108 J iii) E = 104 J iv) R= 6 ohm
ci) new R = 6.8 ohm by taking the emf to be 3V and internal resistance = 0.5 ohm cii) The current increase leads to higher temp, hence resistance increase.
d The resistance of thrmistor and fixed resistor is too large compared to the internal resistance.
ii)1. p.d = (3/4000 + 2000) 4000= 2.0 V
2 pd = 3/2000 + 1800)x 1800 = 1.4 V
iii) for the p.d to be 1.2 V , R = 2700ohm
The same 2700 ohm will give a pd = 1.8 V and not 2.4 V as desired.hence no fixed resistor can acheived the desired range of 1.2 V to 2.4 V .Note that part iii) the pd is across the fixed resistor but part ii) the pd is across the thermistor.
5ai) force per unit nass
ii)The def is force per unit mass , so F/m = GMm/R2 divide by m gives g = GM/R2.
b) The density = 2.52 x 107 kgm-3
bii) The density is greater in the centre as the outer layers compresses the inner layers.
ci) g = 1.2 x 1012 Nkg-1 cii) acc = 1.59 x 107 ms-2
iii) comparing this values , a is less than g , the normal force is greater than 0 , particle will not leave.
d) The accelerating protons has ke and when it approaches the star and undergo decceleration and emit x rays. The decrease in KE is converted to Xray photon.
6. The gradient of the graph is not constant hence the net force not constant . Since Weight is constant , the air resistance veries with speed.
b) gradient of the graph at velocity = 0 gives the magnitude of free fall as no drag force act at v = 0.
c) count the area under the graph. height = 19.2 m
cii) Energy lost = KEi - mgh)/KEi = 0.4
d) i) acc of the ball is gradient of the graph at v = 10 ms-1.
ii) Fd = ma - mg
Fd = 0.35(13-9.81) = 1.12N
e) The area below the x axis is the same above the x axis. Height of rise and fall is the same. Since the gradient will be less slopping, the area is found to extend for a longer time. Gradient of graph decrease when the object is moving downwards.
7.b) Charge = It = 72C
ii) E = 108 J iii) E = 104 J iv) R= 6 ohm
ci) new R = 6.8 ohm by taking the emf to be 3V and internal resistance = 0.5 ohm cii) The current increase leads to higher temp, hence resistance increase.
d The resistance of thrmistor and fixed resistor is too large compared to the internal resistance.
ii)1. p.d = (3/4000 + 2000) 4000= 2.0 V
2 pd = 3/2000 + 1800)x 1800 = 1.4 V
iii) for the p.d to be 1.2 V , R = 2700ohm
The same 2700 ohm will give a pd = 1.8 V and not 2.4 V as desired.hence no fixed resistor can acheived the desired range of 1.2 V to 2.4 V .Note that part iii) the pd is across the fixed resistor but part ii) the pd is across the thermistor.
H2_09_lesson1
1. acc directly proportional to displacement . dir of acc is opp to the displacement.
b1i) w = 3.9 x 103 rads-1
i2) a = 3.2 x 103 ms-2
ii)plot the values a = 3.2x 10-3 and x = -0.21 mm , straight line with negative gradient and passing origin.
c) If it is , cone will resonate when driving freq is close to nat freq , causing damage to the cone.
2. ai) R= 1.8 ohm aii) number = 361 iii) 3000 turns per m
b) force per meter of 1 Nm-1 acting on conductor carrying a current of 1 A placed perpendicular to the field.
ii) B = 9.42 x 10-3 T
c) i) 3.46 x 107 ms-1 ii) 2.0 x 107ms-1
d) magnetic force acting perpendicular to velocity. Hence this force provides the centripetal force . Equate magnetic force = centripetal force
the formula can be shown.
e) The radius r= 0.012 ( taking the vertical component of v)
It will hit the sides of the wall as the diameter of the circle is 2.4 cm which is larger than the radius of th solenoid of 1.4 cm .
3. p= 2.8 x10-14 Ns, b) V = 26.1 V
4a) The number of nuclei is incorrect as the radioactive nuclei is reducing in number . The number of nuclei does not reduce as daughter nuclei is the product of the decay. Hence the nuclei number is the same.
b) Use decay constant = A/N , where N = m /60u. The unit of m must be changed to kg.
The half life = 0.693/decay constant.
half life= 5.27 yrs.
b1i) w = 3.9 x 103 rads-1
i2) a = 3.2 x 103 ms-2
ii)plot the values a = 3.2x 10-3 and x = -0.21 mm , straight line with negative gradient and passing origin.
c) If it is , cone will resonate when driving freq is close to nat freq , causing damage to the cone.
2. ai) R= 1.8 ohm aii) number = 361 iii) 3000 turns per m
b) force per meter of 1 Nm-1 acting on conductor carrying a current of 1 A placed perpendicular to the field.
ii) B = 9.42 x 10-3 T
c) i) 3.46 x 107 ms-1 ii) 2.0 x 107ms-1
d) magnetic force acting perpendicular to velocity. Hence this force provides the centripetal force . Equate magnetic force = centripetal force
the formula can be shown.
e) The radius r= 0.012 ( taking the vertical component of v)
It will hit the sides of the wall as the diameter of the circle is 2.4 cm which is larger than the radius of th solenoid of 1.4 cm .
3. p= 2.8 x10-14 Ns, b) V = 26.1 V
4a) The number of nuclei is incorrect as the radioactive nuclei is reducing in number . The number of nuclei does not reduce as daughter nuclei is the product of the decay. Hence the nuclei number is the same.
b) Use decay constant = A/N , where N = m /60u. The unit of m must be changed to kg.
The half life = 0.693/decay constant.
half life= 5.27 yrs.
HI_09
1. Give best answer by estimating: 20km x 15 km = 3.0 x 108 m2
b) acc= 20ms-1 in 5s= 4ms-2
c) Power = Fxv
Estimate car mass =1000kg
acc = 4ms-2
v - 80kmh-1
2. T= 3.5 ms
b) 0.87m
c) v = 0.87/3.5x10-3 =250 ms-1
4b)The two forces are not action and reaction forces as they act on the book . They are also not of the same nature. One is gravitational and the other is contact force.
c) The total force is zero. Hence Sum of the forces is zero .The change in momentum of of A + the change in momentum of B = 0.There fore the total momentum is constant . This is principal of conservation of momentum.
5i) when two waves meet the resultant displacement at a point is the sum of the displacement of the individual displacements of the two waves.
ii) The diffraction is the bending of waves as it passes an aperture or obstacle.
iii) Coherence is the condition when two waves has a constant phase difference.
b)i) The wavefront AB reaches the slit A and also at slit B .
ii) Mark out the crosses on the screen YZ . You have 3 marked C including at O . In between is marked D . You can find 2 marked D.
Label C-C and marked it x
vi) wavelength = ax/D
v) The equation is only approx as the length L is not large enough .
7. The time taken using formula s= 0.5 at2 = 2.89 s
aii) The actual time is 2.9 s . The air resistance is negligible.
bi) The extension of the rope = 73 - 41 = 32 m
ii) The elastic potential energy = 0.5 at2 or ( area under the graph = area of traingle)=5.4 x 104 J
ci) complete the table :
GPE : 5.4 x 104 // 2.4 x 104J // 0J
EPE : 0 //0 // 5.4 x 104 J
KE : 0 // 3.0 x 104 J // 0
ii) Use graph to find extenstion when Tension = weight of man , extension = 7m
distance fallen when he has Ke max = 41 + 7 m = 48 m
iii)KE max = total energy - epe - gpe
KE max = 5.4 x 104 - 2600J - 1.84 x 104 J=3.3 x 104 J
To draw graph , plot the points with values calculated
GPE: Draw a straight line sloping downwards to 0 at 73 m
KE: Draw a straight line with postive gradient until 41 m . From 41m to 73 m , the graph curve up to peak at 48 m and 0 at 73 m
EPE: The graph starts from 0 at 41 m and U shape increase to 73 m.
8. b) Parallel to each other hence , i) 400 ohm ii) 100W iii) 23 ohm
c) lamp: 200// 0.50// 400// 100.
tele: 200// 1.2// 167// 240.
cooker: 200// 6.0 // 28.6 // 1400
d) Resistance on ammeter and internal resistance is taken to be negligible.
e) The lamp must have smaller cross sectional area than the cooker as the resistance is much higher. The lamp resistance is 14 X higher than the cooker , so the cross sectional area must be 14 times smaller . Using a longer length of wire may not fit the lamp . So the length should not be increased but change the wire to a thinner one. Use R=pL/A.
9. E=hc/~ = 1.63 x 10-18 J ,
ii) The current will increase when the intensity is increased.
iii) The faster photoelectrons have higher kinetic energy can overcome the electric force to reach the anode .
ii) eV = 0.5 mv2
v = 1.16 x 106 ms-1
iii) work function = hf - KE= 1.63 x 10-18 -6.08 x 10-19 J= 1.0 x 10-18 J.
c) The reduction in intensity does not change the photon energy and therefore the ke of the electron remain the same. The rate of photon will decrease and hence will decrease the photocurrent and not the KE max.
d) The transition from level 2 to level 1 will give the same energy as the 122 nm uv radiation.
The data analysis:
6. I choose W = 3N , and looking vertically upwards I recorded the values of d for 10 different L values in the table. The taking the log , I plotted the graph suggested , the value of n is found to be 3.00
b) acc= 20ms-1 in 5s= 4ms-2
c) Power = Fxv
Estimate car mass =1000kg
acc = 4ms-2
v - 80kmh-1
2. T= 3.5 ms
b) 0.87m
c) v = 0.87/3.5x10-3 =250 ms-1
4b)The two forces are not action and reaction forces as they act on the book . They are also not of the same nature. One is gravitational and the other is contact force.
c) The total force is zero. Hence Sum of the forces is zero .The change in momentum of of A + the change in momentum of B = 0.There fore the total momentum is constant . This is principal of conservation of momentum.
5i) when two waves meet the resultant displacement at a point is the sum of the displacement of the individual displacements of the two waves.
ii) The diffraction is the bending of waves as it passes an aperture or obstacle.
iii) Coherence is the condition when two waves has a constant phase difference.
b)i) The wavefront AB reaches the slit A and also at slit B .
ii) Mark out the crosses on the screen YZ . You have 3 marked C including at O . In between is marked D . You can find 2 marked D.
Label C-C and marked it x
vi) wavelength = ax/D
v) The equation is only approx as the length L is not large enough .
7. The time taken using formula s= 0.5 at2 = 2.89 s
aii) The actual time is 2.9 s . The air resistance is negligible.
bi) The extension of the rope = 73 - 41 = 32 m
ii) The elastic potential energy = 0.5 at2 or ( area under the graph = area of traingle)=5.4 x 104 J
ci) complete the table :
GPE : 5.4 x 104 // 2.4 x 104J // 0J
EPE : 0 //0 // 5.4 x 104 J
KE : 0 // 3.0 x 104 J // 0
ii) Use graph to find extenstion when Tension = weight of man , extension = 7m
distance fallen when he has Ke max = 41 + 7 m = 48 m
iii)KE max = total energy - epe - gpe
KE max = 5.4 x 104 - 2600J - 1.84 x 104 J=3.3 x 104 J
To draw graph , plot the points with values calculated
GPE: Draw a straight line sloping downwards to 0 at 73 m
KE: Draw a straight line with postive gradient until 41 m . From 41m to 73 m , the graph curve up to peak at 48 m and 0 at 73 m
EPE: The graph starts from 0 at 41 m and U shape increase to 73 m.
8. b) Parallel to each other hence , i) 400 ohm ii) 100W iii) 23 ohm
c) lamp: 200// 0.50// 400// 100.
tele: 200// 1.2// 167// 240.
cooker: 200// 6.0 // 28.6 // 1400
d) Resistance on ammeter and internal resistance is taken to be negligible.
e) The lamp must have smaller cross sectional area than the cooker as the resistance is much higher. The lamp resistance is 14 X higher than the cooker , so the cross sectional area must be 14 times smaller . Using a longer length of wire may not fit the lamp . So the length should not be increased but change the wire to a thinner one. Use R=pL/A.
9. E=hc/~ = 1.63 x 10-18 J ,
ii) The current will increase when the intensity is increased.
iii) The faster photoelectrons have higher kinetic energy can overcome the electric force to reach the anode .
ii) eV = 0.5 mv2
v = 1.16 x 106 ms-1
iii) work function = hf - KE= 1.63 x 10-18 -6.08 x 10-19 J= 1.0 x 10-18 J.
c) The reduction in intensity does not change the photon energy and therefore the ke of the electron remain the same. The rate of photon will decrease and hence will decrease the photocurrent and not the KE max.
d) The transition from level 2 to level 1 will give the same energy as the 122 nm uv radiation.
The data analysis:
6. I choose W = 3N , and looking vertically upwards I recorded the values of d for 10 different L values in the table. The taking the log , I plotted the graph suggested , the value of n is found to be 3.00
Wednesday, September 15, 2010
H1_N09_P1
1 CDDCB
6 CCADC
11BDABC
15CBCAA
20CBADB
25DAABB
Check your answers and any queries will be answered next week during tutorial
Questions with difficulty:
4. Starting from rest hence x= 0.5 at2. After next interval means that the time is 2t. The value of x' = 4at2 which is 4x. During this second time interval the distance travelled is 4x-x= 3x.
9. The work done by the force F= elastic potential energy + heat energy. Hence it is not elastic potential energy unless no heat is produced.
12 . Power = Fv , P = kv3.
Writing ration of the two cases P2 can be found.
12. Work done by the external pressure = p ext(change in V). Work done by the gas = Net work done = (change in pressure)xchange in vol.
17. L = n1(24) = n2 (30) . Hence the n1=5 and n2 = 4. Length = 120 m
24. The broken lamp has 240 V reading and the unbroken one has 0 reading. use the rule that pd = IR. If I = 0 , pd = 0 eventhough the lamp is not broken.
6 CCADC
11BDABC
15CBCAA
20CBADB
25DAABB
Check your answers and any queries will be answered next week during tutorial
Questions with difficulty:
4. Starting from rest hence x= 0.5 at2. After next interval means that the time is 2t. The value of x' = 4at2 which is 4x. During this second time interval the distance travelled is 4x-x= 3x.
9. The work done by the force F= elastic potential energy + heat energy. Hence it is not elastic potential energy unless no heat is produced.
12 . Power = Fv , P = kv3.
Writing ration of the two cases P2 can be found.
12. Work done by the external pressure = p ext(change in V). Work done by the gas = Net work done = (change in pressure)xchange in vol.
17. L = n1(24) = n2 (30) . Hence the n1=5 and n2 = 4. Length = 120 m
24. The broken lamp has 240 V reading and the unbroken one has 0 reading. use the rule that pd = IR. If I = 0 , pd = 0 eventhough the lamp is not broken.
Sunday, September 12, 2010
Paper3_Prelim2010YJC
1.a) resultant of all the weights acting on the car by the Earth
resultant of the normal and the fricitonal forces.
b) Zx = 11600 N,Zy = 11772 N Z = 16.5 kN
Angle = 45.4
2. Use work done = force x distance moved where force = mg if the mass move at constant speed. Hence work done is change in PE = mgh
ii) AT larger distances away from the Earth mgh can be used provided the g is the value at that location and the distance moved is kept small.
b) The total heat = mL + mc@ = 774 J WHERE @ = 68 deg. and L is latent heat of evaporation.
c) The work done against atm is large as the intermolecular distances is larger from gas to liq . as compared to solid to liq.
3. Do not omit free body diagram. R+mg=mv2/r
where R=0 , v = sq root of rg
vmin = 13.3 ms-1 and entry speed = 29.8 ms-1 and R-mg = mv2/r where R = mg +mv2/r
= 14.8kN.
Take note that the final ke +mgh + Wfriciton= initial KE
b) The magnitude of force at the bottom is greater than the weight as the resultant force is towards the centre.
4. when the freq is below the threshold freq, no photoelectrons are emitted no matter how high the intensity.
Reason : Photon energy is insufficent to overcome work function and hence no emission .Increasing the intensity only increases the number of photons per sec but the photon energy remains the same. hence particulate nature of light.
The uncertainty in time is longer and hence the uncertainty in E is decreased. Hence the energy level is defined.
e) At higher energy the barrier width shorter hence transmission coefficient greater giving shorter average waiting time before tunnelling therefore shorter half-life
5. As temp increases the number charge carrier increases in SC and hence better conductivity. The resistance wire has free electrons whose path are obstructed by lattice vibration and hence greater resistivity.
b) The energy band gap is small and hence electron overcome band gap easily and moved across to Conduction band. The hole left behind at the valence band and electron in conduction band contribute to conduction of charges.
c) The electron diffuse across to the p region creating immobile ions and free of mobile charge carrier.
d) forward biased circuit diagram
6
resultant of the normal and the fricitonal forces.
b) Zx = 11600 N,Zy = 11772 N Z = 16.5 kN
Angle = 45.4
2. Use work done = force x distance moved where force = mg if the mass move at constant speed. Hence work done is change in PE = mgh
ii) AT larger distances away from the Earth mgh can be used provided the g is the value at that location and the distance moved is kept small.
b) The total heat = mL + mc@ = 774 J WHERE @ = 68 deg. and L is latent heat of evaporation.
c) The work done against atm is large as the intermolecular distances is larger from gas to liq . as compared to solid to liq.
3. Do not omit free body diagram. R+mg=mv2/r
where R=0 , v = sq root of rg
vmin = 13.3 ms-1 and entry speed = 29.8 ms-1 and R-mg = mv2/r where R = mg +mv2/r
= 14.8kN.
Take note that the final ke +mgh + Wfriciton= initial KE
b) The magnitude of force at the bottom is greater than the weight as the resultant force is towards the centre.
4. when the freq is below the threshold freq, no photoelectrons are emitted no matter how high the intensity.
Reason : Photon energy is insufficent to overcome work function and hence no emission .Increasing the intensity only increases the number of photons per sec but the photon energy remains the same. hence particulate nature of light.
The uncertainty in time is longer and hence the uncertainty in E is decreased. Hence the energy level is defined.
e) At higher energy the barrier width shorter hence transmission coefficient greater giving shorter average waiting time before tunnelling therefore shorter half-life
5. As temp increases the number charge carrier increases in SC and hence better conductivity. The resistance wire has free electrons whose path are obstructed by lattice vibration and hence greater resistivity.
b) The energy band gap is small and hence electron overcome band gap easily and moved across to Conduction band. The hole left behind at the valence band and electron in conduction band contribute to conduction of charges.
c) The electron diffuse across to the p region creating immobile ions and free of mobile charge carrier.
d) forward biased circuit diagram
6
H1_N08_P2
1a No external force acting on the system
b The force is opposite and hence the gradient is positive and negative
ii) The initial momentum is equal to final momentum = 36KNs
iii) The force is change in momentum divide by time taken F= 2.7 kN
iv) The KE is not conserved as intial is 11.4 kJ and final is 10.8kJ. Hence change of 6kJ. The collision is inelastic.
3a Force multiply by perpendicular distance
b) The three forces are Wbricks , W beam and Rx and Ry at the hinge.
ii) The tension increases as the W bricks is moved towards the end P as the moment increase. To remain balance T must remain the same
iii) the Tension = 2880N
4 Energy level is the energy state an electron can occupy
b) A: Emission line spectra
B: line absorption spectrum.
A :cooler gas excited by the white light will emit photon of specific wavelength
B: White light is absorped by cooler gas has some dark lines missing over bright backgd .
ii) 1 The freq f= 6.2 x 1014Hz 2 E2-E1 = 4.10 x 10-19 - hf = 3.0 x 10-19. f= 4.52 x 1014 and wavelength = 6.65 x10-7m
5. change MeV to Joule
b) Photon E = hf.
c) the loss in ke = 550 (photon energy) divide by energy of proton
d) Use photoelectric equation max KE = hf - work function
e) No of electron = 3 power of 10 = 59049
f) I = Q/t = 550e /time = 4.1 x 10-7
7. ii) R = 3 ohm using P = V2/R
2. length of wire is 9.5 cm
3. I = E/R+r
where R = parallel of 1.5 and 6 ohm .
3 I = 9.8 A
4 When a large current is drawn from the battery , the pd across the battery is E - Ir which is much smaller . Hence the lamps will be less bright as pd across the lamps is reduced.
5. 2 % of 48 W divide by photon energy =
rate of photon emitted .N = 2.7 x 1018 s-1
8 Force is perpendicular to velocity hence the force in the dir of vel is zero . The direction of motion changes but the speed remain constant,
9 qV = 0.5 mv2
v2 = 2eV/m where V = 2.8 kV
v = 3.1 x 107 ms-1
ii) 1 a = QE/m = 3.5 x 1015 ms-2
2. L = v/t, t = 3.87 x10-9 s
3. W = Fs where s = 0.5 at2
W = Fs= 3.2 x 10-15 x 0.5 (3.51 x 1015( x( 3.9 10 -19 ) =0.263 J
4. v = 1.07 x 104 ms-1
5 angle = 23.7 deg.
iii Proton deflect below the horizonal and the velocity and angle of deflection will be smaller than that of electron as the acc is small
b The force is opposite and hence the gradient is positive and negative
ii) The initial momentum is equal to final momentum = 36KNs
iii) The force is change in momentum divide by time taken F= 2.7 kN
iv) The KE is not conserved as intial is 11.4 kJ and final is 10.8kJ. Hence change of 6kJ. The collision is inelastic.
3a Force multiply by perpendicular distance
b) The three forces are Wbricks , W beam and Rx and Ry at the hinge.
ii) The tension increases as the W bricks is moved towards the end P as the moment increase. To remain balance T must remain the same
iii) the Tension = 2880N
4 Energy level is the energy state an electron can occupy
b) A: Emission line spectra
B: line absorption spectrum.
A :cooler gas excited by the white light will emit photon of specific wavelength
B: White light is absorped by cooler gas has some dark lines missing over bright backgd .
ii) 1 The freq f= 6.2 x 1014Hz 2 E2-E1 = 4.10 x 10-19 - hf = 3.0 x 10-19. f= 4.52 x 1014 and wavelength = 6.65 x10-7m
5. change MeV to Joule
b) Photon E = hf.
c) the loss in ke = 550 (photon energy) divide by energy of proton
d) Use photoelectric equation max KE = hf - work function
e) No of electron = 3 power of 10 = 59049
f) I = Q/t = 550e /time = 4.1 x 10-7
7. ii) R = 3 ohm using P = V2/R
2. length of wire is 9.5 cm
3. I = E/R+r
where R = parallel of 1.5 and 6 ohm .
3 I = 9.8 A
4 When a large current is drawn from the battery , the pd across the battery is E - Ir which is much smaller . Hence the lamps will be less bright as pd across the lamps is reduced.
5. 2 % of 48 W divide by photon energy =
rate of photon emitted .N = 2.7 x 1018 s-1
8 Force is perpendicular to velocity hence the force in the dir of vel is zero . The direction of motion changes but the speed remain constant,
9 qV = 0.5 mv2
v2 = 2eV/m where V = 2.8 kV
v = 3.1 x 107 ms-1
ii) 1 a = QE/m = 3.5 x 1015 ms-2
2. L = v/t, t = 3.87 x10-9 s
3. W = Fs where s = 0.5 at2
W = Fs= 3.2 x 10-15 x 0.5 (3.51 x 1015( x( 3.9 10 -19 ) =0.263 J
4. v = 1.07 x 104 ms-1
5 angle = 23.7 deg.
iii Proton deflect below the horizonal and the velocity and angle of deflection will be smaller than that of electron as the acc is small
Wednesday, September 1, 2010
Paper1_Preliminary 2010
1 - 11 B 21 D 31 C
2 B 12 B 22 D 32 D
3 B 13 B 23 B 33 D
4 D 14 D 24 A 34 C
5 A 15 B 25 A 35 B
6 D 16 B 26 C 36 D
7 A 17 C 27 C 37 C
8 C 18 D 28 A 38 D
9 A 19 C 29 C 39 A
10 C 20 B 30 A 40 A
Total upon 39
2 B 12 B 22 D 32 D
3 B 13 B 23 B 33 D
4 D 14 D 24 A 34 C
5 A 15 B 25 A 35 B
6 D 16 B 26 C 36 D
7 A 17 C 27 C 37 C
8 C 18 D 28 A 38 D
9 A 19 C 29 C 39 A
10 C 20 B 30 A 40 A
Total upon 39
Tuesday, August 24, 2010
N06/II/4 cont
The graph is used to read off the temperature at 1500 ohm.
c) Potential at A is 3V . Potential at B is also 3V when the voltmeter read zero. That means the resistance of thermistor is equal to 1200 ohm. Hence read off from graph its temperature.
ii) If the voltmeter reads 1.2 V , the potential at B could be 1.2 higher or lower than A. Hence the resistance of thermistor is either greater than 1200 ohm or less than 1200 ohm . Hence two different temperature.
2. To calc the lower temperature, by looking at the graph it must be having a greater resistance. Hence greater than 1200 ohm.
The working is done mathematically now,
current through thermistor = current through 1200ohm (belowB) No current from voltmeter.
1.2V higher than 3V must be 4.2 V across the thermistor.
6V - 4.2 V = 1.8 V across the 1200 ohm
the equation of same current :
4.2 / Rt = 1.8/1200
Rt= 2800 ohm
Read off graph temp = 282K
The principle of this is potential divider . The current is the same down one line , so p.d is proportional to resistance
c) Potential at A is 3V . Potential at B is also 3V when the voltmeter read zero. That means the resistance of thermistor is equal to 1200 ohm. Hence read off from graph its temperature.
ii) If the voltmeter reads 1.2 V , the potential at B could be 1.2 higher or lower than A. Hence the resistance of thermistor is either greater than 1200 ohm or less than 1200 ohm . Hence two different temperature.
2. To calc the lower temperature, by looking at the graph it must be having a greater resistance. Hence greater than 1200 ohm.
The working is done mathematically now,
current through thermistor = current through 1200ohm (belowB) No current from voltmeter.
1.2V higher than 3V must be 4.2 V across the thermistor.
6V - 4.2 V = 1.8 V across the 1200 ohm
the equation of same current :
4.2 / Rt = 1.8/1200
Rt= 2800 ohm
Read off graph temp = 282K
The principle of this is potential divider . The current is the same down one line , so p.d is proportional to resistance
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