JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT JC2 CT1 2013 H2 Physics Paper 2 solutions Qn Suggested solution 1(ai) A forced oscillation is one where the system is subjected to a periodic external force. The body then oscillates at the same frequency as the external periodic force. (aii) Resonance / Light damping From the graph, at maximum amplitude of 1.6 cm, f = 12 Hz (bi)
Maximum linear speed vmax = wx0 = (2πf)(x0) = 2 x π x 12 x 1.6 x 10-2 = 1.21 m s-1 bii Maximum linear acceleration amax = w2x0 = (2πf )2(x0) = (2 x π x 12)2 x 1.6 x 10-2 = 91.0 m s-2 Time interval between a maximum speed and maximum acceleration is 0.25T biii Time interval = ¼ T = ¼ x (1/12) = 0.0208 s
Remarks 1 mark for the 2 underlined points 1 1 for correct reading of amplitude
1 for substitution 1 for answer 1 for substitution 1 for answer 1 for deduction 1 end answer
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Qn (c)
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT JC2 CT1 2013 H2 Physics Paper 2 solutions Suggested solution
Remarks
1 mark for showing lower amplitude at all points and 1 mark for the decrease in resonant frequency.
2 (a) A longitudinal wave is one in which the direction of propagation of energy is parallel to the direction of oscillation of the particles in the wave. (b) Since polarisation is the restriction of the plane of oscillation of the particles in a wave, and particles in a longitudinal wave are oscillating parallel to the direction of propagation of energy of the wave, their oscillations cannot be restricted to a plane, so longitudinal waves cannot be polarised. ci 1.5λ
1 1 mark for each underlined point
1 mark, 1.5 not accepted
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JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT JC2 CT1 2013 H2 Physics Paper 2 solutions Suggested solution
Qn cii displacement
Remarks 1 mark for Correct sine curve and 1 mark for correct wavelength
Figure 2.1b 1 or zero
ciii excess pressure
Figure 2.2
3(a) Any two of the following: 1)
The sources must be coherent, i.e. they must maintain a constant phase difference with respect to each other.
[1] per condition
2) The 2 waves should have approximately the same amplitude. 3) The 2 waves must meet. 4)
The 2 waves must be of the same type so that principle of superposition applies.
5)
The 2 waves must be polarised in the same plane, or unpolarised.
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JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT JC2 CT1 2013 H2 Physics Paper 2 solutions Suggested solution
Qn (b)
Remarks
[1] for resultant amplitude within correct “box” [1] for general shape
(c)(i) The path difference of waves from the antennae to point P (equidistant from the two antennae) is zero. The phase difference at point P must be the same as the phase difference at the sources. Since point P experiences a minimum, implying destructive interference, the sources must be in anti-phase. (c)(ii) Since the two waves are broadcasted in anti-phase, in order for them to meet in constructive interference at point Q, path difference has to be an odd integer number of ½ wavelengths.
m
For the third maximum, 5 path difference = λ 2 5 path difference = ( 60 ) 2 path difference = 150
(c)(iii No change or min throughout ) (c)(iv Waves from S1 and S2 are of the same type and move opposite to each other and ) overlap.
[1]
[1]
[1] for
5 path diff = λ o 2 r substitution [1] for ans
[1] [1]
Interference occurs and a stationary wave is set up.
[1]
High and low intensities correspond to the anti-nodes and nodes.
[1]
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JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT JC2 CT1 2013 H2 Physics Paper 2 solutions Suggested solution
Qn 4(a) Electric field strength at a point in an electric field is defined as the force per unit positive charge acting on a small charge placed at that point.
Remarks [1]
(b) Work done by e-field in bringing a charge from a higher potential to a lower potential [1]* for is equal to W = qV. explanation for
W = qV Since the separation between the plates is d, thus the work done by the electric [1]* for force = F x d explanation for W=Fxd [1] for equating the work done and correct expression
F x d = qV
F V = q d From definition in (a), the uniform electric field strength between the plates is E =
(c)(i) ⎡ 10 ⎤ 200 ) ⎥ 1. Potential difference between point P and +300 V plate = ⎢ 40 ( ⎣ ⎦
⎡ 10
⎤
Potential at P = 300 - ⎢ ( 200 )⎥ ⎣ 40 ⎦
V d
*not awarded for provision of equations without explanations [1] for p.d.
[1]
(c)(i) Electric potential energy = Vq = (250)( +1.5 × 10−17 C ) 2. = 3.75 x 10-15 J
[1] for ans
(c)(ii) Loss in gravitational potential energy 1. = mg∆h = (1.4 x 10-15)(9.81)(55 x 10-2)
[1]
(c)(ii) Final KE of oil drop 2. = gain in KE since initial KE is zero = loss of electric potential energy and gravitational potential energy = (3.55 x 10-15 - 1.50 x 10-15) + (7.55 x 10-15) = 9.80 x 10-15 J
5(a)
i) Gravitational field ii) Electric field iii) Magnetic field
[1] for change in e.p.e.
[1] ans
[1] for any 2 correct ans Or [2] for all 3 correct ans Page 5 of 10
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT JC2 CT1 2013 H2 Physics Paper 2 solutions Qn Suggested solution Since the proton deflects downwards, the magnetic force acting on the proton is (b)(i) downwards. With velocity towards right, the direction of the magnetic field is out of the plane of paper (not upwards) according to Fleming’s Left Hand rule. (b)(ii) The magnetic force provides the centripetal force.
Fc = ma c Bqv =
mv 2 r
0.12 × 1.6 × 10 −19 =
Remarks [1]
[1] for statement [1] sub
1.67 × 10 −27 × 4.5 × 10 6 r
[1] ans
r = 0.391 m
[1] ans
(c)(i)
E
With magnetic force acting downwards, the electric force acting on the proton must be upwards. Thus the electric field should be acting upwards. Arrow pointing vertically upwards
(c)(ii)
When the proton passes undeviated, Fn = 0 FE = FB Eq = Bqv E = 0.12 × 4.5 × 106 = 5.40 × 105 V m-1
[1] sub [1] ans
(d) The magnitude of gravitational force is negligible as compared to the magnitude of [1] the electric or magnetic forces. (‘mass of proton is small’ is an insufficient answer) 6(a) Weight of Hywind structure, W = ρgV= 1025 x 9.81 x 5300 = 53.3 MN(3 sf)
[1] sub [1] ans
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JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT JC2 CT1 2013 H2 Physics Paper 2 solutions Qn Suggested solution (b)(i) When v= 11 ms-1, p = 1500000 W lg (11/ ms-1) = 1.041 , lg (1500000/W) = 6.176 = 6.18 (3 sf)
Remarks [1] correct point plotted [1] correct bestfit line plotted
(b) Using points from the graph: (0.93,5.84), (1.13,6.44) (ii) Gradient = (6.44 – 5.84)/(1.13 – 0.93) = 0.6/0.2 = 3.0
[1] sub [1] ans
Fig. 6.3 (b) P = kv 3 (iii) lg p = 3 lg v + lg k where gradient of the line is 3. Since the experiment data shows a straight line graph with gradient equals 3.0, the relationship P = kv 3 is valid. (c) Some of the possible problems: • Wind speed which is excessively high leading to over current produced which will damage the generator. •
Corrosion of the structure due to the salty environment.
•
Possible lightning strike due to the tall structure and the blade tip.
•
Power loss due to long distance transmission.
•
Strong tide leading to instability of the structure.
•
Difficulty in maintenance as it is out in the open sea.
[2] correct comment with correct explanation 1 mark for state and 1 mark for explain
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Question 7 Defining the problem (3 marks) P1
A is the independent variable and V is the dependent variable or vary A and measure V.
[1]
P2
Keep the number of turns on coil Y or coil X constant.
[1]
P3
Keep the current in coil X constant.
[1]
Methods of data collection (4 marks) M1
Alternating power supply/signal generator connected to coil X.
[1]
M2
Coil Y connected to voltmeter/c.r.o. in a workable circuit.
[1]
M3
Measure diameter/radius/lengths with a ruler/vernier callipers.
[1]
M4
Method to determine area.
[1]
Method of analysis (1 mark) A
Plot a graph of V against A. Relationship valid if straight line through origin.
[1]
Alternatively : Plot a graph of lg V against lg A. Relationship valid if straight line with gradient = 1.
[1]
Safety considerations (1 mark) S
Precaution linked to (large) current in coil/heating, e.g. switch off when not in use to avoid overheating coil; do not touch coil because it is hot. [1]
Additional detail (3 marks) D
Relevant points might include [maximum : 3 marks] 1
Use large current in coil X/large number of turns/high frequency a.c. to produce measurable e.m.f.
2
Detail on measuring e.m.f., e.g. height × y-gain on CRO.
3
Keep frequency of power supply constant.
4
Use of rheostat to keep current constant in coil X.
5
Monitor with a.c. ammeter.
6
Avoid other alternating magnetic fields.
7
Repeat measurement for r or d or lengths and average.
Do not allow vague computer methods.
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Suggested answer for Planning Question Objective: To investigate if the relationship between the induced e.m.f (V) in coil Y and the cross-sectional area (A) of coil X, V is directly proportional to A, is valid. Variables: (3 marks) Independent variable: the cross-sectional area of coil X, A Dependent variable: e.m.f induced in coil Y, V Controlled variables: 1) the number of turns on coil Y or coil X to be kept constant. 2) current in coil X to be kept constant. Diagram
diameter, d
Coil X should be wound tightly around the cardboard tube
Procedures: (4 marks) (Type of apparatus used to measure/ vary the variables) 1. Measure the length and width of the cardboard using a metre rule. 2. Measure the diameter, d, of the cross-sectional area of coil X using a vernier caliper and to calculate πd2 the cross-sectional area, A = . 4 3. Measure the e.m.f induced in coil Y, using an A.C. voltmeter / C.R.O. 4. Use an alternating power supply/ signal generator connected to coil X 5. Use a variable resistor to adjust the current in coil X and an A.C. ammeter to measure and ensure a constant current in the circuit. (How to vary the independent variable) 6. Prepare cardboards of same width but rolled up to form cylindrical tubes of different diameters. 7. Set up the apparatus as shown in the diagram. 8. Coil X must be wound tightly around the cardboard tube. Page 9 of 10
9. Use a vernier caliper to measure the diameter of Coil X at 3 different positions along the tube. Take average reading as d. (Measuring the dependent variable) 10. Close the circuit for Coil X and ensure a constant current is supplied by the a.c. power supply. 11. Measure the induced e.m.f., V, across coil Y. Note : If using C.R.O., state that the maximum e.m.f. is measured on the y‐axis of the screen using e.m.f. V = (height) x (y‐gain on C.R.O.). 12. Repeat steps 7 to 8 for different values of A, and record the corresponding e.m.f. V for 6 sets of readings.
Analysis (1 mark) 1. Given that V is directly proportional to A, V = kA 2. Plot a graph of V against A. 3. The relationship is true if a straight line graph is obtained which passes through the origin. Alternatively : 1.
Assume that V and A are related by the equation: V = kAn , where k and n are constants.
2. Taking lg on both sides, lg(V) = n lg(A) + lg (k). 3. Plot a graph of lg (V) against lg (A) 4. If V is directly proportional to A, a straight line graph will be obtained where the gradient is equal to 1 and the y-intercept is equal to lg(k).
Safety Precautions (1 mark) 1. Do not touch the coil as it is hot / Use gloves when removing the coil. 2. Switch off the circuit when not in use to avoid over heating of coil. Steps taken to produce reliable results (3 marks, Any 3) 1
Use large current in coil X / large number of turns / high frequency a.c. to produce measurable e.m.f.
2
Detail on measuring e.m.f., e.g. height × y-gain on CRO.
3
Keep frequency of power supply constant.
4
Use of rheostat to keep current constant in coil X.
5
Monitor with a.c. ammeter.
6
Avoid other alternating magnetic fields.
7
Repeat measurement for r or d or lengths and average.
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