CIE 9702 Physics Paper 41 ON 2015 Mark Scheme Answers

Q1(a)(i)

Gravitational force provides the centripetal force to keep satellite in orbit

GM ms / r² = msv² / r            where r is equivalent to x

                                                msv² = GM ms / x
                                          1/2 msv² = GM ms / 2x

Note: Do not write: 'gravitational force equals to centripetal force'. It will mislead that there are two balanced forces.

Q1(a)(ii)

from list of formula: grav potential = - GM / r
Ep = ms x grav potential
Ep = - GMms / x

Q1(a)(iii)

ET = EK + Ep

     = GMms / 2x - GMms  / x

     = - GMms / 2x

^Q1(b)

^{(i) decreases
Note: Resistive force causes energy lost, therefore total energy decreases}

(ii) decreases

Note: Refering to the equation in a(iii), decrease in x will decrease the total energy (more negative value), i.e. the satellite has a smaller orbit.

^{(iii) decreases}

Note: Its getting closer to the Earth, potential energy will decrease. 

^{(iv) increases}

Note: It has to move faster to supply a greater centripetal force.

Summary: 
To stay in a higher orbit, a satellite will move slower. It has higher potential energy nad total energy. To stay in a lower orbit, it needs to move faster. It will have lower potential energy and total energy.

Q2(a) 

An ideal gas obeys the equation pV = nRT, p = pressure, V = volume, n = number of mole, R = molar gas constant, and T = thermodynamic temperature

Note: All symbols must be explained. Not enough to mention T = temperature, has to specify that it is in kelvin.

Q2(b)(i) 

80 - 32 = 48 K

Q2(b)(ii) 

From 1/2 m <c²> = 3/2 k T
<c²> is directly proportional to T (in kelvin)
<c²> / (1.9 x 10⁶) = (80+273) / (32+273)
<c²> = 2.21 x 10⁶
root of <c²> = 1480

Q3(a)

Increase in the internal energy equals to the heat supplied to the system plus the work done on the system (or: minus the work done by the system).

Q3(b)(i) 

When ice melts, change in volume will be very small volume actually decreases)

, therefore work done on the system is negligible. Thermal energy will be absorbed to break the lattice structure. Thus, in total, the internal energy increases.

Q3(b)(ii) 

There is an increase in volume, so work done by the system. But the increase happens in a very short time (suddenly), indicating that there is no thermal energy enter or leave the gas. Thus, internal energy decreases.

Q4(a)

In a free oscillation, there will be no loss of energy due to zero resistive force.
In a forced oscillation, the body is made to move by a periodical force, which is continuously supplying energy to the oscillation.

Q4(b)(i)

2.1 Hz. Resonance happens at this frequency as the amplitude of the vibration is the highest.

Q4(b)(ii)

Yes, there are frictional forces acting on the trolley because the peak is not very sharp. 

Note: If there is no resistive force, the peak can be infinite.

Q4(c)

From formula list, v =  ω√x_o² - x²

maximum speed occurs at x = 0, so                                 

                              v = 2 π (2.1) √(4.7x10^-2)²         

                                 = 0.62

Q5(a)(i)

Coulomb's law states that force between point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation.

Note: The idea of point charge is important, that is, charge is assumed to be concentrated at a single point at the center of the charge/charged sphere.

Q5(a)(ii)

1. direction of forcce is radially away from the sphere

2. at x = r

3. F inversely proportional to square of r

    F_r (r)² = F_4r(4r)²

    F_r / F_4r =  (4r)² / (r)² 

                 = 16

Q5(b)

    From coulomb's law equation and definition of field strength,
    Field strength, E, is directly proportional to charge of the field provider, Q
    So, E1 / Q1 = E2 / Q2,
          (1.5 x 10⁶) / (6.0 x 10^-7) = (2.0 x 10⁶) / Q2
          Q2 = 8 x 10^-7
          Additional charge = Q2 - Q1
                                        = 2.0 x 10^-7

Q6(a)

    (i) It will experience a force of magnitude mg along the direction as the field
   (ii) It will experience zero force

Q6(b)

(i) Field into the page. This is to provide an upward force to balance the downward electrical force. According to Fleming's left hand rule, a positive charge moving in the direction in the diagram will experience an upward force when a uniform magnetic field applied into the page

(ii) Force due to magnetic field = B q v, and force due to electric field = q E
      when this two forces are equal, B q v = q E, it can be simplified as: v = E / B

Q6(c)

When the speed is 2v, magnetic force will be doubled, which will be greater than the electrical force, therefore it should bend upwards due to an upward resultant force.

Q7(a)

Threshold frequency is the minimum frequency of the incident electromagnetic radiation, to a metal surface, to cause emission of an electron.

Q7(b)

Only the surface electron will have the maximum kinetic energy. The emitted electrons are from different depth inside the metal. They use different amount of energy from the photon to overcome the surrounding bonds before coming to the surface and escaping with the remaining kinetic energy. Therefore they will have lower kinetic energy compared to the surface electrons.

Q7(c)

(i) The threshold frequency can be determined when the Ekmax is zero, the graph touches x-axis at 1.85 x 10⁶, the 1 / λ₀. From c = f λ, f₀ = 3.00 x 10⁸ x 1.85 x 10⁶ 

                                                                =5.55 × 10¹⁴ Hz

(ii) Φ = h f₀
          = 6.63 × 10^-34 × 5.55 × 10^-14
          = 3.68 × 10^-19 J

Q7(d)

From Φ = h c / _λ0 , Φ = 2.2 × 10^-19 J will give 1 / λ₀ = 1.1 x 10⁶

so the graph will be a straight line touching x-axis at 1.1 x 10⁶ with the same gradient.

Note: To draw a correct graph, you will need to first form a straight line equation to match Einstein equation of photoelectric. i.e Ekmax = - hc (1/λ0) + E          (E = energy of photon)

Then you will notice that the gradient is '-hc' , and it will constant for all cases of E or metal with different λ0. 

Q8(a)

Nucleus of an atom is a small core where the mass of the atom concentrated. It consists of protons and neutrons. 

A proton or a neutron in the nucleus is also known as nucleon.

Q8(b)(i)

1. decay constant = 1n 2 / half life
                             = ln 2 / (3.8 x 24 x 60 x 60)
                             = 2.1 × 10^-6 s^-1 2. A = λN
                        97 = 2.1 × 10^-6 × N 
                         N = 4.6 × 10⁷

Q8(b)(ii) Volume of air is directly proportional to number of air molecule, according to p V = n R T.                                                   V1 / N1 = V2 / N2                          
                        2.5 × 10^-2 / 6.02 × 10²³ = 1.0 / N2  
Number of air molecule in 1.0 m³, N2 = 2.408 × 10²⁵  
                                        N radon / N2 =  4.6 × 10⁷ /2.408 × 10²⁵                                             
                                                              = 1.9 × 10^-18

Q9(a)
(i) V = 1/2 of 6.0 V since both resistors have the same resistance
         = 3.0 V
(ii) VB = (2.0 / 2.8+2.0) x 6.0 = 2.5 V
Note: Take p.d. across 2.0 resistor so that you get the potential at point B straight away, because p.d on the other side of 2,0 resistor is 0 V.
(ii) VB = (2.0 / 1.8+2.0) x 6.0 = 3.2 V

Q9(b)
Vout changes from -9.0 V to 9.0 V, a sudden switch happen when VB=VA
At 10 ^oC, VB = 2.5V, < VA, Vout will be -9.0V,
At 20 ^oC, VB = 3.2V, > VA, Vout will be +9.0V,
Note: As the question has mentioned that it is an ideal op amp, you should estimate a very large gain, which causes the output to saturate easily.

Q10(a)
Sharpness means clarity of the edges in the image
Contrast means difference in degree of blackening

Q10(b)
(i) '80 keV X-rays' refers to the X-rays produced when a cathode ray beam (or a beam of electrons) hits the metal target after being accelerated by a high potential difference of 80 kV.

(ii) I_T / I = e^{-3.0 x 1.4}
               = 0.015

Q10(c)
For good contrast, μx must be very different.
From Fig 10.2, there is a large the difference between the two values of μ, i.e. about 10 times.
Even though there is a comparatively large thickness of muscle, it will hardly overweight the effect of μ of bone. 
So, image will have good contrast.
Note: the explanation on the thickness of muscle is tricky, it may contradict your comment on the effect of the coefficient.

Q11(a)
Frequency modulation means the frequency of the carrier wave varies in synchrony with the displacement of the signal.

Q11(b)
(i) 5.0 V
Note: don't get confused by the 1.5 V of the signal.
(ii) 20 kHz x 1.5 = 30 kHz, so min f is 750 kHz - 30 kHz = 720 kHz.
(iii) similarly, 750 kHz + 30 kHz = 780 kHz
(iv) which is related to the time taken for the signal to change from a positive amplitude to next positive amplitude. This means the period of the sinusoidal signal, also means it's frequency, 7.5 kHz 

Q12(a)
(i) gradual loss of power/intensity/amplitude
(ii) e.g. noise can be eliminated because pulses can be regenerated
      e.g. much greater data carrying capacity because many messages can be carried at the same time (or greater bandwidth)
      e.g. more secure because it can be encrypted
      e.g. error checking because extra information bit can be added

Q12(b)
attenuation = 10 lg (145 / 29) = 7.0 
attenuation per unit length = 7.0 / 36
                                           = 0.19 dB km^-1