CIE 9702 Physics Paper 41 MJ 2015 Mark Scheme Answers
Q1(a)
Gravitational force between two point masses is directly proportional to product of their masses and inversely proportional to square of their separation.
Note: The law can only be applied on mass of a single dot in space. Alternatively, you can comment that the size of the objects is much less than their separation.
Note: The law can only be applied on mass of a single dot in space. Alternatively, you can comment that the size of the objects is much less than their separation.
Q1(b)
Gravitational force provides the centripetal force
G MN m / r2= m r ω2
GMN = r3 (2π/ T)2
Note: State the physics concept from which you derive the equation.
G MN m / r2= m r ω2
GMN = r3 (2π/ T)2
Note: State the physics concept from which you derive the equation.
Q1(c)
GM = 4π2r3/ T2M α (r3/T2)
MN/ MU = (3.553/ 5.92) / (5.833/ 13.52)
= 1.18
Note: Ratio must be always stated in decimal number, not fraction.
Q2(a)
It is the sum of the random distribution of potential energy and kinetic energy of the molecules in a system.Note: must mention atoms/molecules/particles of a system'.
Q2(b)
(i) p V = n R T1.2 x 105(4x10-3) = n (8.31) (290)
n = 0.20
(ii) ( V / T )A= ( V / T )c
4.0 / 290 = 7.8 / Tc
Tc= 566 K
Q2(c)
There is a decrease in the volume at constant pressure, work must have been done on the system. There is also a decrease in temperature, therefore a decrease in the kinetic energy of the molecules. Since this is an ideal gas, it's internal energy has zero potential energy, hence change in kinetic energy will cause a similar change in internal energy.Note: Thermal Dynamic 1st Law is not needed here, as thermal energy release or absorb is unknown
Q3(a)
It is the amount of thermal energy per unit mass required for a substance to change state/phase at constant temperature.Note: No need to make reference to vaporization or fusion. Do not use '1 kg of mass', it is a mixed of unit and quantity.
Q3(b)
(i) 0.26 g s–1;0.38 g s–1(ii)
1.
Power by the heater + rate of heat gained from surrounding = rate of melting
P + h = mL
70 + h = 0.26 L..................(1)
110 + h = 0.38 L................(2)
(2) - (1) 110 - 70 = (0.38 - 0.26) L
L = 330 J kg–1
Note:Heat from surrounding will be absorbed as the apparatus is not insulated. Ignoring this will result in an inaccurate answer.
2.
Substitute L into (1) or (2), h = 15 to 17 W
Q4(a)
(i) The frequency at which the oscillating body is forced/made to oscillate.Note: For example, a spring-mass system that is attached to an automatic vibrator (the driving oscillator), preset to a certain vibrating frequency.
(ii) The frequency at which the body oscillates freely, there is no resistive force acting on it.
Note: For example, a spring-mass system that is displaced, and released to oscillate in a vacuum.
(iii) It is a phenomena whereby the vibration is at maximum amplitude. This will happen when the driving frequency (forced frequency) is same as the natural frequency of the body.
Q4(b)
The vibration of the piezoelectric crystal that can be used to generate large amplitude ultrasound.Q4(c)
Vibration of bridge like the case of the Tahoma Bridge. The tension in the suspension cable shall withstand the strong blowing wind across the bridge.
Q5(a)
Electric field strength is equal to negative of potential gradient.The gradient of the tangent line at 4 cm = 4.5 x 104 N C-1
Note: This method has a risk of getting an answer that is beyond the allowed range, Make sure the tangent line is carefully chosen and the values are carefully identified.
Alternative solution
E = Q / 4πεox2 ; V = Q /4πεox ; so E = V / x
E = 1.8 kV / 0.04 = 4.5 x 104 N C-1
Note: This method is only for case that has only a single charge, and it will not tell you about the direction.
Q5(b)
(i) 3.6 x 103 V
Note: surface is at 2.0 cm, therefore it is the value of the horizontal part of the graph.
(ii) C = Q / V
= 8.0 x 10-9 / 3.6 x 103
= 2.2 x 10-12 F
Note: surface is at 2.0 cm, therefore it is the value of the horizontal part of the graph.
(ii) C = Q / V
= 8.0 x 10-9 / 3.6 x 103
= 2.2 x 10-12 F
Q6(a)
(i) gravitational field
(ii) gravitational field, electrical field
(iii) gravitational field, magnetic field and electrical field
(ii) gravitational field, electrical field
(iii) gravitational field, magnetic field and electrical field
Q6(b)
(i) out of the page
(ii) Magnetic force supplies the centripetal force
B q v = m v2 / r
B (1.6 x 10-19) = 3.32 x 10-26 x 7.60 x 104 / (12.2 x 10-2 / 2)
B = 0.26 T
= (6.63 x 10-34) / (6.50 x 10-12)
= 1.02 x 10-22
(ii) energy = h c / λ
= (6.63 x 10-34 x 3.00 x 108) / (6.50 x 10-12)
= 3.06 x 10-14
θ = 30.7o
(ii) During the deflection, the electron will also get deflected and start moving. It must have gain some energy from the photon. So the deflected photon will have less energy, therefore a longer wavelength. Hence the change is always positive.
(ii) ΔN / Δt
(iii) ΔN / N
(iv) ΔN / NΔt
Note: Answer (iv) is derive from A = λN.
From (ii) A = ΔN / Δt,
so ΔN / Δt = λN
λ = ΔN / NΔt
Note: The curve shall look like a flip graph of Fig. 9.1. When the sample reduces by half during one half-life, S should increase by half during one half-life. So, at the second half-life, it is expected that S is 1.5 times of the first half-life.
pd across R1.2 = [ 1.2 / (1.2 + 4.2) ] x 4.5
= 1.0 V
Since the other end of the resistor is earthed, i.e. 0 V, so the potential at the inverting input is 1.0V.
(ii)
When Vin is greater than 1.0 V, output of the op-amp is positive and saturated, and the diode is forward biased, so pd across R is + 5 V.
When Vin is less than 1.0 V, output of the op-amp is negative and saturated, but the diode is reverse biased, so pd across R is 0 V
(ii) It can be used as a regenerator amplifier for digital signal
Note: Question mentioned 'X-ray tube', therefore your explanation must be related to the properties of the tube, i.e. the tube voltage.
(ii) the value of μ of both blood and muscle is very similar, which mean the absorption of X-ray is similar. This will produce an image with poor contrast, because contrast is the difference in degree of blackening.
(ii) television, audio amplifier (electrical devices that channel high quality information
(iii) mobile phones, satellite TV channel
(ii) 38 = 10 lg (P/N)
= 10 lg [ P / (9.2 x10-6) ]
P = 58 mW
(ii) Magnetic force supplies the centripetal force
B q v = m v2 / r
B (1.6 x 10-19) = 3.32 x 10-26 x 7.60 x 104 / (12.2 x 10-2 / 2)
B = 0.26 T
Q6(c)
According to the equation in 6b, m is directly proportional to r, less mass means smaller r.
So, path of Y should be a semicircle with a smaller diameter.
So, path of Y should be a semicircle with a smaller diameter.
Q7(a)
Alternating voltages can be changed efficiently. It allows stepping up to high voltages during transmission, and allows stepping down to different voltages to suit the different needs of factories, shop lots or housing areasQ7(b)
The use of high voltages will reduce the power lost, as current will be small for the same power. Power lost in transmission cable due to heating = (current)2 x resistance, lower current will cause less power lost.Q8(a)
(i) p = h / λ= (6.63 x 10-34) / (6.50 x 10-12)
= 1.02 x 10-22
(ii) energy = h c / λ
= (6.63 x 10-34 x 3.00 x 108) / (6.50 x 10-12)
= 3.06 x 10-14
Q8(b)
(i) [ 6.84 x 10-12 - 6.50 x 10-12 ] = 6.63 x 10-34 / 9.11 x 10-31 x 3.00 x 108) / (1 - cos θ)θ = 30.7o
(ii) During the deflection, the electron will also get deflected and start moving. It must have gain some energy from the photon. So the deflected photon will have less energy, therefore a longer wavelength. Hence the change is always positive.
Q9(a)
During radioactive decay, the unstable nuclei release alpha particles, beta particles or gamma ray, randomly and spontaneously, to become more stable nuclei.Q9(b)
(i) N – ΔN(ii) ΔN / Δt
(iii) ΔN / N
(iv) ΔN / NΔt
Note: Answer (iv) is derive from A = λN.
From (ii) A = ΔN / Δt,
so ΔN / Δt = λN
λ = ΔN / NΔt
Q9(c)
Note: The curve shall look like a flip graph of Fig. 9.1. When the sample reduces by half during one half-life, S should increase by half during one half-life. So, at the second half-life, it is expected that S is 1.5 times of the first half-life.
Q10(a)
(i) Potential divider rule:pd across R1.2 = [ 1.2 / (1.2 + 4.2) ] x 4.5
= 1.0 V
Since the other end of the resistor is earthed, i.e. 0 V, so the potential at the inverting input is 1.0V.
(ii)
When Vin is greater than 1.0 V, output of the op-amp is positive and saturated, and the diode is forward biased, so pd across R is + 5 V.
When Vin is less than 1.0 V, output of the op-amp is negative and saturated, but the diode is reverse biased, so pd across R is 0 V
Q10(b)
(i)
(ii) It can be used as a regenerator amplifier for digital signal
Q11(a)
Hardness of the X-ray beam can be controlled by changing the tube voltage. This will change the kinetic energy of the electrons that strikes on the anode, causing emission of X-ray photons with various frequencies.Note: Question mentioned 'X-ray tube', therefore your explanation must be related to the properties of the tube, i.e. the tube voltage.
Q11(b)
(i) I= Io e -μx(ii) the value of μ of both blood and muscle is very similar, which mean the absorption of X-ray is similar. This will produce an image with poor contrast, because contrast is the difference in degree of blackening.
Q12(a)
(i) loudspeaker, doorbell, telephone (things that you usually have in your house that transmit electrical signal)(ii) television, audio amplifier (electrical devices that channel high quality information
(iii) mobile phones, satellite TV channel
Q12(b)
less prone to interference, more secure, greater bandwidth...Q12(c)
(i) P = input power, N = noise power, O = output power
P/N = O/N x P/O Note: This is a mathematical technique to find the answer
lg(P/N) = lg( O/N x P/O )
lg(P/N) = lg(O/N) + lg(P/O) Note: This is the rule of log function
10lg(P/N) = 10lg(O/N) + 10lg(P/O) [ times 10 to all terms so that the unit is in dB ]
= 25 dB + (0.21 x 62) dB
= 38 dB
P/N = O/N x P/O Note: This is a mathematical technique to find the answer
lg(P/N) = lg( O/N x P/O )
lg(P/N) = lg(O/N) + lg(P/O) Note: This is the rule of log function
10lg(P/N) = 10lg(O/N) + 10lg(P/O) [ times 10 to all terms so that the unit is in dB ]
= 25 dB + (0.21 x 62) dB
= 38 dB
= 10 lg [ P / (9.2 x10-6) ]
P = 58 mW
Q13(a)
(i) the large constant magnetic field is used to align the nuclei to the same direction as the field. This cause the nuclei to process, at Larmour frequency in the radio frequency region.
(ii) the non-uniform magnetic field set the nuclei to precess at different Larmor frequencies, because Larmour frequency is directly proportional to magnetic field strength. By knowing the frequency of photon emitted from the nuclei, those nuclei can be located in the body
hf = 2.82 x 10-26 B
6.63 x 10-34 x 42 x 106 = 2.82 x 10-26 B
B = 0.99 T
(ii) the non-uniform magnetic field set the nuclei to precess at different Larmor frequencies, because Larmour frequency is directly proportional to magnetic field strength. By knowing the frequency of photon emitted from the nuclei, those nuclei can be located in the body
Q13(b)
E = 2.82 x 10-26 Bhf = 2.82 x 10-26 B
6.63 x 10-34 x 42 x 106 = 2.82 x 10-26 B
B = 0.99 T
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