2011 J2 H2 9646
2011 J2 H2 Physics 9646/01 H2 Physics Paper 1 Suggested Solutions (Term 4) 1.
Ans: A Using
,
,
Graph plotted is
vs
=> accurate but not precise 2.
Ans: C
Vf
v
Only option is C 3.
Ans: C
-vi
2011 J2 H2 9646 4.
Ans: D Acceleration initially is downwards. Air resistance increases as speed of object increases. Air resistance will decrease till 0.
5.
Ans: A
Assuming object falls from rest. Speed of object when it reaches gate 1, (using ) Speed of object when it reaches gate 2,
-
(1)=(2)
-----(1)
-
----- (2)
-
6.
Ans: B By conservation of momentum, - ( elastic collision, using relative speed) Solving for , -
2011 J2 H2 9646
7.
Ans: B Consider forces on 6.0 kg
54 – 6 – F = 6a Consider forces on 2.0 kg
F = 2a Solving for F: F = 12 N Resultant force on 6.0 kg, R = 54 – 6.0 – 12 = 36 N 8.
Ans: A Upthrust is small as density of air is small. So arrow of U must be shorter than R. droplet in equilibrium, Sum of length of arrows upwards = sum of length of arrows down
9.
Ans: B Taking moments about X, ! "
10.
Ans: B See lecture notes.
2011 J2 H2 9646
11.
Ans: D # =>
12.
# $ %$ =># & Ans: D
% (rest are constants)
' ( r increases =>a increases ) )(' 13.
' constant
Ans: D String will pull ball inwards to provide centripetal force. So ball will pull string outwards. Pull of pole on string will be opposite to the pull of ball on string.
14.
Ans: B Total energy of satellite
*+ ,-+
,. - (
15.
Ans: B Neutral point must be close to the moon as it has smallest mass. The point cannot be A due to the presence of the Sun.
16.
Ans: D Damping => amplitude falls & frequency of which resonance takes place decreases. Cannot be A as amplitude must be smaller than undamped system at all frequencies.
17.
Ans: B Total energy
' / 0
-
1
2011 J2 H2 9646
18.
Ans: C See lecture notes.
19.
Ans: C For thermal equilibrium, Heat lost by sphere P = Heat gained by sphere Q .- 2 3- 3 .4 2$3 34 % 3- 3 .4 5 $ .4 5.- % $3 34 % .=> 3- -3 5$3- 34 %
20.
Ans: D Using
6 and
0(
=> ( 6 $ % 6 6
$ %
6 ! 21
Ans: C The reflected wave cannot be the case in option A and option B. Since it is a closed end, the stationary wave must have a node at the barrier hence only the reflected wave in option C is possible. We observe that the incident wave has undergone a change in phase of 180o to produce the reflected wave. We can apply the procedure below to obtain the reflected wave (1) Reflect the incident wave about the barrier. (2) Next, reflect the wave about the horizontal axis. The above two procedures, which give a phase change of 180o, is illustrtaed below.
2011 J2 H2 9646
The incident wave undergoing a phase change of 180o to give the reflected wave is only applicable when we examine (1) a string attached to a fixed end, (2) electromagnetic waves, (3) the displacement of air particles in a sound wave. 22
Ans: A
x= 23
λd a
1 hence graph A a
Ans: A
θ = sin−1
24
x∝
nλ d
where n = 1,2,3...
Ans: B All the squares are in static equilibrium but Square A has a net clockwise moment about it centre, Square C has a net anti-clockwise moment about its centre, Square D has a net clockwise moment about its centre. There for only square B is in static and rotational equilibrium
25
Ans: A Options C and D are not possible as the electric field lines are evenly spaced. Option B is not possible as the charges are evenly distributed on surface of the sphere.
2011 J2 H2 9646
26
Ans: B R E R+r as variable resistance R decreases, terminal p.d. decreases. terminal p.d. =
Power dissipated in cell =
E R+r
2
r=
r
(R + r )
2
E2
as variable resistance R increases, Power dissipated in cell increases.
27
Ans: C E = 6.0 .......... (1) 1.2 + r E = 5.0 .......... (2) 1.6 + r Solving (1) and (2) simultaneously, r = 0.8 Ω, E = 12 V
28
Ans: C
R=
29
V 6 = = 1.71 Ω I 3.5
Ans: D
I = 0.3 A, VX = 1.0 V, VR = 1.5 V ∴ emf = VX + VR = 2.5 V When connected in parallel to cell, VX = 2.5 V I X = 0.5 A VR = 2.5 V
IR = 0.5 A
Itotal = IR + I X = 1.0 A
2011 J2 H2 9646
30
Ans: A When R = 0 7, all the current will pass through it, bypassing the 3.0 7 and 6.0 7 resistor i.e. a short circuit. Hence the effective resistance is: −1 1 1 + Reff = 10 + 40 10 = 18 Ω ∞
1 1 1 + + ∞ 3 6 = 20 Ω
R eff =
31
−1
+ 10 +
1 1 + 40 10
−1
Ans: C
τ = F ×d = N.B.I.L × d = 20 ( 0.83 )( 4.5 )( 0.17 ) × 0.11 = 1.4 Nm 32
Ans: C The magnetic flux (not flux linkage) is the same throughout coils A, B, C and D since the no. of field lines passing through all four coils must be the same. When the current through coil B peaks in one direction, the magnetic field in coil C will have the largest magnitude (as compared to the other three coils) since it has the smallest cross sectional area (recall: B =
φ
A
When the current through coil B peaks in the opposite direction, the magnetic field in coil C will again have the largest magnitude but will now be pointing in the opposite direction. Thus the magnetic field strength in coil C exhibits the greatest variation.
33
Ans: B 1 = 40 Hz 25 × 10 −3 V 110 Vrms = o = = 77.8 V 2 2
f =
2011 J2 H2 9646
34
Ans: A There is a lower probability of finding the particle to the right of the barrier thus the amplitude of the wavefunction to the right of the barrier is smaller. The transmitted wave however has the same energy as the incident wave thus the wavelength remains the same. Hence option A.
35
Ans: D
hf = φ + KEmax 36
Ans: D
5 4
1
2
3
Photons released from transitions 1, 2 and 3 are highly energetic and have a high frequency i.e. they should appear on the right of the spectrum. Hence options A and C are not possible. The difference in energy of the photons released in transitions 3 and 4 is larger than the difference in energy of the photons released in transitions 4 and 5. Hence we expect the spectral lines for transitions 4 and 5 to be more closely spaced as compared to the lines for transition 3 and 4. Hence option D and not B.
37
Ans: A Refer to Page 11-12 of your Lasers & Semiconductors notes.
2011 J2 H2 9646
38
Ans: B Red light has a wavelength of about 600 nm.
Ered =
hc
λred
≈ 2.07 eV
The transition with a ∆E that is about 2 eV is E6 → E5 .
39
40
Ans: C 108 46
Pd has 46 protons and 108 − 46 = 62 neutrons
109 47
Ag has 47 protons and 109 − 47 = 62 neutrons
Ans: C First, we note that the no. of tin-124 nuclei is equal to the no. of antimony-124 nuclei that has decayed.
Hence a ratio of
No. of tin-124 nuclei =6 No. of antimony-124 nuclei
6 1 of the original no. of antimony-124 nuclei has decayed leaving of the 7 7 N 1 original no. of antimony-124 nuclei yet to decay i.e. = . N0 7 means that
1 the original number of antimony-124 nuclei. 2 2 1 1 In 2 half-lives, we are left with = the original number of antimony-124 nuclei. 2 4
In 1 half-life, we are left with
In 3 half-lives, we are left with
Since we are left with
1 2
3
=
1 the original number of antimony-124 nuclei. 8
1 of the original number of nuclei, the time period is between 2 7
and 3 half-lives i.e. between 120 and 180 days.
2011 J2 H2 9646
2011 J2 H2 Physics 9646/02 H2 Physics Paper 2 Suggested Solutions (Term 4) 1
(a)
Newton’s second law of motion states that the rate of change of momentum of an object is proportional to the resultant force acting on it and the change takes place in the direction of the force.
(b)(i)
Resultant force = ma Resistive force, F = resultant force = ma = 750 × 4.8 = 3600 N (b)(ii) Work done by friction = |Change in KE | F × d = |0 - ½ mv2| 3600 × d = 234375 J d = 65.1 m (b)(iii) As the car is brought to rest, the car tyres experiences a resistive force F acting against the direction of motion of the car due to friction by the road. By Newton’s third law, an equal but opposite force then acts on the road by the car tyres in the same direction as the motion of the car. (c)(i)
Originally, when the car travels on the level road, Work done by friction = decrease in kinetic energy only. When the car travels down the slope, it undergoes a decrease in potential energy. Work done by friction = decrease in kinetic energy + decrease in potential energy. More work needs to be done by friction compared to before. Since friction acting on the car is the same, the distance travelled by the car increases.
(c)(ii) resultant F along the slope = 3600 – mgsin8 = 3600 – 750 × 9.81 × sin 10o = 2322.38 N Resultant force = ma = 2322.38 a = 3.10 m s-2 (up the slope)
2011 J2 H2 9646
2
(a)(i)
(12) 2 Resistance of lamp = = 6.0 9 24
(ii)
Energy transferred to lamp = 24 x 1800 = 43 200 J = 4.32 x 104 J
(iii)
Current = 12/6.0 =2.0 A Number of electrons passing = (2.0 x 1800)/1.6 x10-19 = 2.25 x 1022
(b)(i)
The resistance of the lamp can be obtained from the ratio of V/I, with value of V and corresponding value of I read from the graph.
(ii)
(iii) The voltage supply has an internal resistor. The energy supplied by the voltage supply is shared between the lamp, the variable resistor and possibly an internal resistor in the voltage supply. Due to the internal resistance of the voltage supply, the p.d across the lamp cannot be 12V when variable resistor is at 0 9. When the variable resistor is at its maximum resistance, there is still current flowing and potential difference across it is less than 12 V and hence pd across lamp cannot be 0 V.
2011 J2 H2 9646 3
a
(i)
Diffraction is the phenomenon of bending or spreading out of waves when they travel through a small opening or when they pass round a small obstacle.
(ii)
The phase difference (φ) between two particles or two waves tells us how much a particle (or wave) is in front of or behind another particle (or wave). The value ranges from 0 to 2π radians.
(ii)
Two waves are considered to be coherent when they maintain a constant phase difference between them. Note: Coherence should be explained as a property of the waves as opposed to the light source. For light waves, it is impossible to get coherent light waves unless they start from the same source. There are frequent irregularities in the wave created by any light source. These irregularities cannot be matched even in the case of two waves coming from seemingly identical but separate laser beam sources.
b
(i)
Since the two waves from M1 and M2 are of the same type, are coherent and presumably have approximately the same amplitude, we expect superposition to occur thus forming an interference pattern on the screen AB. Let the position of the detector on the screen AB be D. Assuming that the two waves are emitted in phase. When the detector is positioned on the screen AB such that the path difference M1D − M2D = n where n = 0,1, 2, 3… , we expect constructive interference to occur thus giving a peak in the intensity detected. When the detector is positioned on the screen AB such that the path difference M1D − M2D = (n−1/2) where n = 1, 2, 3…, we expect destructive interference to occur thus no signal will be detected at this position. Note: State any assumption used for e.g. the waves are emitted in phase since this affects the conditions for constructive and destructive interference. Note that the experiment involves microwaves which can be seen, good to specify what we mean by “bright” fringes etc.
(ii)
1. The position of the “bright” fringes, where there is a peak in intensity, will remain unchanged. At the “bright” fringes, the intensity detected will increase. At the “dark” fringes where the intensity detected remains zero.
2011 J2 H2 9646
2. When the two microwaves from M1 and M2 are emitted out of phase, we expect : the positions on the screen where there was originally a “bright” fringe to now exhibit a “dark” fringe with zero intensity since a path difference M1D − M2D = n now leads to destructive interference between the two waves. Positions on the screen where there was originally a “dark” fringe will now exhibit a “bright” fringe since a M1D − M2D = (n−1/2) now leads to constructive interference between the two waves. The intensity detected at the “bright” fringes now is the same as that before. 4(a) (b)
Magnetic flux through a plane surface is defined as the product of the magnetic flux density normal to the surface and the area of the surface.
: = NBA cos 8 = 500 x 5 x 10-2 x 2.5 x 10-2 = 0.625 Wb
(ci)
As the coil is rotated, the component of the magnetic flux density normal to the surface changes. Therefore, the magnetic flux linkage changes.
(cii)
Average emf =
t (0.625 - 0)
= 0.25 × 10
-3
= 2500 V (ciii) According to Faraday’s law of electromagnetic induction, ε = −
dΦ . The maximum e.m.f. dt
can be determined from the gradient of the graph where the slope is the steepest, for instance, at t = 0.25 ms.
2011 J2 H2 9646
5. (a)(i)
Absolute zero is the temperature when all substances have minimum internal energy.
(a)(ii) Temperature of an ideal gas is directly proportional to the kinetic energy of the ideal gas. (b)(i)
Using # ;<3 # ; <3 $ %$ $ %$ ! ! = ;
(b)(ii)
;
% %
.( ; .( !!&> # <3 # ? .( <3 %$ $ $ %$ ! &>
% %
Mass of gas needed to escape ?
!! &>
2011 J2 H2 9646
6(a)
From the graph, when x = 17.5 cm, Temperature = 51.5° C
(b)(i)
The actual relationship for heat transport along a conductor is given by: , where Q/t: rate of heat flow along the rod, k: thermal conductivity, characteristic of the material A: cross sectional area of the rod ∆θ: temp difference across length ∆x Since ∆θ = (100 – 45)°C and ∆x = (17.5 – 0) cm, then !"" # $% & !'(% # " )* This relationship is no longer explicitly in the H2 syllabus, yet the Examiner’s feedback was “the majority of answers were correct.” This suggests, for 1 mark, they may have been looking for a simpler answer such as: “It is constant”
(ii)
+
,--./0 ,1(0.-
Temperature gradient, +
= 3.14 °C cm-1.
00 ,1(0
(iii)
There is heat loss by the rod to the surrounding for the rod in fig 6.2. For the insulated rod in fig 6.4, the insulation prevents/reduces heat loss to the surrounding, resulting in it having a great temperature. In addition, the rod in fig 6.2 is exposed to the surrounding at room temperature (~30°C), whereas the lower temperature end of the rod in fig 6.4 is maintained at 45 ° C.
(c)
If 2 x/cm 0 15 30
,
θx = constant. θ/°C 100 49 31
θx/°C cm 0 735 930
Since the product θx is not constant, θ cannot be inversely proportional to x. (d)(i)
x/cm 5.0
θ/°C 77
θE/°C 57
[θE = θ - θ1 For x = 0, 80 = 100 - θ1
ln(θE/°C) 4.04
θ1 = 20 °C]
2011 J2 H2 9646 (d)(ii) 2. Since θE = θoe-µx, Linearizing by taking natural logarithm of both sides, we get ln θE = -µx + ln θo Thus if the proposed equation is indeed true, then a graph of ln θE against x should yield a straight line, which in does indeed. Gradient: -µ Vertical intercept: ln θo 3. Using the points (1.00, 4.32) and (25.00, 2.70), Gradient = … = -0.675
µ = 0.0675 cm-1
From the graph, Vertical int = … = 4.39, thus ln θo = 4.39 θo = 80.6 ° C (e)
Expected marking points 1.Non-linear curve 2. steeper and below the original line. [other notes: Exact form of the curve is unknown. Horizontal asymptote is θ = 20 °C (not 30 ° C), as obtained from (d)(i)]
2011 J2 H2 9646 Q7 Suggested Solutions (Disclaimer: These solutions do not represent the only acceptable answer. Other methods, apparatus, procedures, precautions may also be acceptable. It is best you check with your tutor regarding the acceptability of your solutions if they differ from what is presented below.) Problem interpretation:
Note: • • •
Problem Interpretation and the full list of apparatus not required for assessment/exams Give clear descriptions on the procedures for determining the IV, DV and constants. Read the instructions carefully: you should describe the procedure for determining the power output of the solar panel.
Independent variable: The power supplied to the high intensity lamp
intensity of the radiation on the solar panel
Dependent variable: Increase in temperature of water the solar panel.
thermal energy absorbed by the water
power output of
Control variables: Fixed distance between the lamp and the solar panel, rate of flow of water
Aim of the experiment: To investigate how the power output of a solar panel depends on the intensity of the incident infra-red radiation.
2011 J2 H2 9646 Experimental setup: Note:: All key elements should be clearly labelled
Variable Power Supply Thermometer, T1 High intensity lamp
Tap with flow regulator
Constant height, h
Rubber tubing
Intensity meter Thermometer, T2 Water outlet Solar panel
Beaker to collect the water
Procedure ( R denotes reliability precautions))
1. Connect the e high intensity lamp to the variable power supply. Use a retort stand to hold the high intensity lamp at a fixed height, h,, above the glass front of the solar panel as shown in the diagram. Using a metre rule, measure h and ensure that it remains constant throughout the experiment. 2. Turn on the tap until a streamline, continuous flow of water stream is obtained. The tap should remain on throughout the experiment to ensure constant flow rate. 3. Connect the tap to the inlet of the solar panel with a rubber tubing. Water from the tap is passed through the inlet of the solar panel at constant rate. The temperature of the water as it enters and leaves is measured by thermometers T1 and T2 . 4. Starting from the minimum mum power setting, slowly increase the power supply until there is an observable difference between T1 and T2. 5. Place the intensity meter on the glass surface directly below the lamp. Record the intensity I. Remove the intensity meter from the glass surface surface. 6. When the apparatus settles down to a steady state (i.e. when the thermometers show no further change in temperature R ), record the temperature T1 and T2. 7. Place lace an empty beaker at the outlet and start the stopwatch. After time t = 20 seconds, stop the he stopwatch and remove the beaker from the outlet.
2011 J2 H2 9646 8. Using a weighing scale, measure the mass of the water, m collected in the beaker. 9. Compute the mass flow rate m/t to verify that the flow rate is indeed constant. 10. Compute the power output of the solar pan panel using the equation P = mc(T2-T1) / t, where c is the specific heat capacity of water. 11. Increase the power supply and repeat steps 5 to 10 to collect 7 further sets of readings for the power output, P and the intensity, I. 12. Plot a graph of P against inten intensity I to obtain the relationship between the power output of the solar panel and the intensity of the radiation on the solar panel panel. P/W
I / W m-2
Other Reliability Measures:
R
1. Enclose the high intensity lamp and the solar panel in a Styrofoam box with its internal int walls painted black to minimize the effect due to ambient light. 2. Measure the intensity of the infra-red red radiation at several points on the solar panel and take average.
3. Using a marker, mark out the position to place the intensity meter with small sma crosses on the glass surface to ensure consistency in the placement of the intensity meter. Safety Concerns: 1. Avoid prolonged exposure to the infra infra-red radiation as it may cause injury to the skin. 2. Exercise proper electrical safety when handling the variable power supply which is plugged into the 240 V mains electricity supply, such as not handling with wet hands and checking to ensure wires are properly insulated.
2011 J2 H2 9646
2011 J2 H2 Physics 9646/03 H2 Physics Paper 3 Suggested Solutions (Term 4) 1a) The latent heat of vaporization is the quantity of heat required to change a given mass of substance from liquid phase to vapour phase without any change in temperature. 1bi) It is to allow for heat losses to the environment to be eliminated through calculation. Since the temperature of water and time taken is the same for both experiments, the heat loss should be constant. 1bii)
E = mL + h Pt = mL + h
where h represents the heat loss to the environment (140)(5x60) = (14.1) L + h ------ (1) (95)(5x60) = (8.2) L +h ------(2) Eqns (1) – (2): (140-95)(5x60) = (14.1 - 8.2) L L = 2290 J g-1
2(a)(i) According to N2L (for constant mass), Fnet = ma. Since the ball undergoes circular motion, the net force constitutes a centripetal force (Fnet = Fc = mv2/r) Taking downwards as positive, and since both T and mg act downwards, Fnet = (T) + (mg) = Fc, Fc > T
(ii)
Fc = 3mg
2011 J2 H2 9646 (iii)
Applying N2Lon the ball at the lowest point, taking upwards as positive. Fnet = ma T + (-mg) = Fc = 3mg (|Fc|= 3mg throughout since uniform circular motion) T = 4mg
(b)(i) Fc = mrω2 3mg = m(0.72)ω2 ω= 6.39 (ii) Fc = mv2/r 3mg = mv2/0.72 v = 4.60 m s-1 (c)(i)
Let t represent the top (highest point) and b represent the bottom (lowest point) If no work is done, GPEt + KEt = GPEb+ KEb KEb > KEt Since GPEt > GPEb , it would Since the speed and thus KE is constant, Work is done. [from t to b, work done is –ve to remove energy from the system. From b to t, work done is +ve to supply energy to the system.
(c)(ii)
Since |v| and thus KE is constant, W = ∆GPE = mg∆h W = (0.240)(9.81)(-2x0.72) = -3.4 J !
"
#
$
3ai) A field of force is a region of space where a body experiences a force. aii) Changes in the strength of a force-field are represented by closeness of the lines of force. The closer the lines, the stronger the strength of the field. bi) Mass of the object bii) A positive charge. biii) The north pole of a magnet
2011 J2 H2 9646 Note: the direction of the magnetic force on a moving charge is not in the direction of the magnetic field.
c)
Note: diagram should be clearly labelled. Directions for both the B field and E field should be shown clearly. FB= Bqv
S B into paper
+q
E
Charged particles with velocity v= E/B
FE= qE
The diagram above shows positively charged particles entering a region where both a uniform electric field and a uniform magnetic field act perpendicular to one another. The charges experience an electrostatic force, FE = qE acting vertically downwards and an electromagnetic force, FB = Bqv acting vertically upwards. Only particles with speed v = E/B will travel in a straight line undeflected since the forces exerted by the two fields are equal in magnitude and opposite in direction. By varying the magnitudes of E and B, we can select charges with one specific velocity to emerge from the slit S.
2011 J2 H2 9646 4.
When switch A and switch B are open, the equivalent circuit is: R2
R1 X
R3
Y
R1 + R2 + R3 = 12 9
When switch A is open and B is closed, R3 is shorted and the equivalent circuit is: R2
R1 X
Y R1 + R2 = 10 9
When switch A is closed, both R2 and R3 are shorted R1 X
R1 = 6 9
Y
2011 J2 H2 9646 4ai)
69
aii)
R1 + R2 = 10 9 R2 = 10 – 6 = 4 9 R1 + R2 + R3 = 12 9
aiii)
b)
R3 = 12 – 10 = 2 9 R2
R1 X
Z R3
1 1 + R2 R3 =
1 1 + 4 2
= 7
1 3
=7.339
−1
+ R1 −1
+6
2011 J2 H2 9646 5a) The work function of a surface is the minimum energy of the individual photons in the incident radiation necessary to result in the emission of photoelectrons from the surface of the material.
E=
b)
hc @
(6.63 × 10 )(3.0 ×10 ) (540 × 10 ) = -34
8
-9
=3.68 x 10-19 J
(3.68 × 10 ) ( ) = 1.60 × 10 -19 -9
= 2.3 eV Since the energy of the photons is 2.3 eV, it is lower than the work function of 2.5 eV and electrons are not emitted from the surface. c) The conclusion in (b) is not affected by the intensity of the light since the intensity of light only changes the rate of arrival of photons at the surface but not the energy of the individual photons.
6(a)
(in S.I. units) Force is defined as the rate of change of linear momentum, 34 linear momentum, 64 *74.
+54 , +
where
[Newton’s 2nd law provides for basis for the definition of force. Newton’s 2nd law does not quite define it because it only describes a proportional relationship.] (b)(i)
Energy stored = area bounded by the line and the length (or extension) axis.
2011 J2 H2 9646
(b)(ii) EPE = Work done in stretching spring (by distance of x beyond its original length) EPE = 8- 3 9:
8- : 9:
[Alt, EPE = W = x. = ½ Fmax = ½ kx
, ;
:;
since F = kx, = ½ Fmax = ½ kx.
Thus EPE = (½kx) x = ½ kx2. (c)(i)
When 4.0 N hung, L = 9.2 cm When stretched an additional 0.80 cm, total length = 9.2 + 0.80 = 10.0 cm
(c)(ii)
1. Change in GPE = |mg∆h| = (4.0)(0.80x10-2) = 0.032 J [mg∆h = (- 0.032 J) actually since spring stretched downwards] 2. EPE = ½ k x2. < 0 = 125 Nm-1 k= ,-.= >,-?@ When 4.0 N hung, L = 9.2 cm x1 = 9.2 – 6.0 = 3.2 cm When extended an additional 0.80 cm, x2 = 10.0 – 6.0 = 4.0 cm Thus change in EPE = ½ k :;; # :,; = ½ (125)(0.0402 – 0.0322) = 0.036 J
(d)(i)
TE of oscillation = 4.0 x 10-3 J [reasoning: 1. Lowest point EPE/J ½ k (0.040)2 = 0.100 GPE/J 0 KE/J 0 Tot 0.100 mech energy
2. Equilibrium ½ k (0.032)2 = 0.064 0.032 KEmax 0.100
k = 125 Nm-1 Assume GPE = 0 at lowest point
Thus KEmax = Energy of oscillation = 0.100 – 0.064 – 0.032 = 4.0 x 10-3 J (d)(ii) 1. TE of oscillation = KEmax = ½ m(vmax)2 = 4.0 x 10-3 J M = W/g = 4.0/9.81= |vmax| = 0.140 m s-1
2011 J2 H2 9646 2. v = ±ω :A; # : ; (provided in formula list) |vmax| = ωxo (vmax occurs when x = 0) Thus ω = 2πf = |vmax|/xo f = 2.78 Hz (e)
The period of oscillation for a loaded spring (spring-mass) system is given by B (formula not required to be memorized in H2 syllabus),
CD
E
where m normally represents the mass of the
load for a light/massless spring. When the spring has non-negligible mass, m in the equation represents the effective mass of the system, and as m ↑ → T↑ (or equivalently ω, f ↓).
[Even without the above equation, it can be reasoned that for the same restoring force F = kx, a greater total mass with more inertia will result in smaller acceleration by Newton’s 2nd law (a = F/m), which will result in smaller average velocity over the same displacement (e.g. from the amplitude position to the equilibrium position. A smaller average velocity over the same distance means greater time interval, and thus period over the entire oscillation.]
7. (a)
Progressive: A progressive wave is one that travels from one point to another, resulting in transfer of energy. Transverse:
The plane of oscillation is perpendicular to the direction of wave propagation. The direction of energy transfer is normal to the oscillation of the wave particles.
(b)
A polarised wave is a wave whose particles oscillates in a single plane normal to the direction of energy transfer while an unpolarised wave has particles that oscillate in more than one plane that is normal to the direction of energy transfer.
(c)(i)
Amplitude = A
(c)(ii)
Intensity = I
(c)(iii)
6
(d) Angle 8 180° 90° 60°
amplitude A 0
intensity I 0
6 (using A’ = Acos8)
(using
6 )
2011 J2 H2 9646 (e)(i)
See lecture notes.
(e)(ii)
As the incident sound wave travels to the other open end of the pipe, it is reflected backwards. The reflected wave will then interfere with the incident sound waves and undergo superposition to form a standing wave.
(f)(i)
67 cm For first loud sound for open pipes, L= @ @ = 2L (f)(ii)
v = f@ = 250 × (2 × 0.67) = 335 A 340 ms-1
(g)
The antinodes should be located outside of the pipe (end correction). Thus, wavelength of the soundwave is longer. Hence, using v = f@, the speed of sound is an underestimate.
2011 J2 H2 9646 2010 ‘A’ Level H2 P3 Q8 8 a (i) mα v α + mD v D = 0
4u (V ) + ( A − 4 ) u ( −v ) = 0 4V − ( A − 4 ) v = 0
(ii)
Although I have derived the equation above, you can easily write the equation directly. 1 mαV 2 KE of α particle = 2 KE of daughter nuclei 1 mDv 2 2 m V2 = α 2 mDv
=
A−4 4 ( A − 4) 4
2
using substition for
V2 from (a)(i) v2
A−4 4 1 = A −1 4 =
(iii) 212 83
Bi →
Tl + 24He + ∆E
208 81
Note that ∆E refers to the energy released which includes the KE of the thallium nucleus, B-particle and may include C-rays as well.
∆E = ( mBi − mTl − mα ) c 2
(
= ( 211.9459 − 207.9374 − 4.0015 ) × 1.66 × 10 −27 × 3.0 × 108 = 1.0458 × 10 −12 J = 6.536 MeV ≈ 6.54 MeV
)
2
2011 J2 H2 9646
From a(ii), KEα 1 = A −1 KED 4 1 (212) − 1 4 = 52 KEα = 52KED =
52 × 6.54 53 = 6.416 ≈ 6.42 MeV
∴ KEα =
(c)
(ii)
Note that our calculations assume that all the energy released manifests as the KE of the daughter nuclei and B-particle. C-rays may have been emitted in the decay process. Students may be tempted to say that in deriving the ratio in a(ii), we have used the nucleon numbers which deviate slightly from the actual mass of the nuclei (measured in a.m.u). Thus accounting for the difference in energy of the alpha particle emitted. This is not true as the discrepancy in energy of the alpha particle is far too large.
(ii)
Heat is also wrong because heat is a manifestation of the KE of the particles which we have accounted for and is too vague a term. The -rays (photons) emitted carry momentum. Unless the thallium nucleus, alpha-particle and straight line (which is very unlikely to happen),
(d) (i)
the thallium nucleus and alpha-particle cannot travel in opposite directions because we need momentum to be conserved two-dimensionally. ln 2 t 1= = 3648 s 2
(ii)
-rays are emitted along a
λ
≈ 61 min After two hours, the bismuth sample has undergone approximately 2 half-lives, 2 1 1 hence we expect = of the original Bi sample to be left . 2 4 Throughout these two hours, the thallium nuclei formed decays further. Given that its half-life is significantly shorter than that of bismuth i.e. thallium has a greater rate of disintegration, we expect the amount of thallium present after two hours to be much less than 3N/4.
2011 J2 H2 9646
