CIE 9702 Physics Paper 43 MJ 2016 Mark Scheme Answers
Q1(a)
Gravitational potential is defined as zero at infinity (beyond which, the effect of gravity has vanished).Since gravity is always a pulling force, object in a gravitational field will always be attracted inwards.
Hence, work done by the gravity got out causing object to lose potential energy when it is pulled from infinity.
Therefore, gravitational potential is always negative.
Q1(b)(i)
Change in kinetic energy = change in potential energy= mass x change in potential
= 180 x (14 -10) x 108
= 7.2 x1010 J
increase, because potential energy is being converted into kinetic energy.
Q1(b)(ii)
To reach the other end, it will just need enough kinetic energy to get pass the point with the highest potential, i.e. 0.52 x 1012 m. In other words, it needs kinetic energy equivalent to the change in potential energy to reach this point. Beyond this point, gravity of star B will do the work. Change in kinetic energy = change in potential energy= mass x change in potential
1/2 m v2= m x(10 - 4.4) x108
v = 3.3 x 104
Q2(a)
Any of the two below:- time of collisions negligible compared to time between collisions
- no intermolecular forces (except during collisions)
- random motion (of molecules)
- large numbers of molecules
- total volume of molecules negligible compared to volume of containing vessel
Q2(b)(i)
Helium-4 means one mole of helium-4 weight 4 grams6.02 x 1023atoms = 4 grams
1atom = x grams
Ratio method will give x = (1 x 4) / 6.02 x 1023
= 6.6 x 10-34 grams
Q2(b)(ii)
Mean KE of an ideal gas = 1/2 mc2= 3/2 kT1/2 x 6.6 x 10-34= 3/2 x 1.38 x 10-23 x (27+273)
r.m.s speed = 1.4 x 103
Q3(a)
Simple harmonic motion is an oscillatory motion with its acceleration directly proportional to the displacement from a fixed point, and is always opposite direction to the displacement.Q3(b)
1. the graph is not a straight line2. the graph has a different maximum displacement
Q3(c)
(i) ω = 2 π / T= 2 π / 0.8
= 7.9 rad s-1
(iii) Change in KE = ½ mω2(xi2 – xf2)
= ½ × 120 × 10-3 × 7.92 × {(1.5 ×10-2)2 – (0.9 ×10-2)2}
= 5.3 ×10-4
Note:
The first three periods last at t = 2.4 s.
Within this time,determine the initial maximum kinetic energy and the final maximum kinetic energy.
The equation to find maximum kinetic energy is KEmax = 1/2 mv2max ,whereby v2 = ω2 (xo2 - x2),
Remember that maximum KE happens when x = 0, therefore v2 max = ω2xo2
= ½ × 120 × 10-3 × 7.92 × {(1.5 ×10-2)2 – (0.9 ×10-2)2}
= 5.3 ×10-4
Note:
The first three periods last at t = 2.4 s.
Within this time,determine the initial maximum kinetic energy and the final maximum kinetic energy.
The equation to find maximum kinetic energy is KEmax = 1/2 mv2max ,whereby v2 = ω2 (xo2 - x2),
Remember that maximum KE happens when x = 0, therefore v2 max = ω2xo2
Substituting this into the earlier equation gives KEmax = 1/2 m ω2xo2
Identify the first max displacement and the last max displacement.
Substitute them as in the solution shown above.
Q4(a)(i)
Specific acoustic impedance of a medium is the product of density of the medium and speed of sound wave in the medium.Q4(a)(ii)
a: ratio of reflected intensity and incident intensity, it is also known as the intensity reflection coefficient Z1 and Z2: specific acoustic impedances of media on each side of boundary where the wave travels through.Q4(b)
Add on some labels to represent the different stages that the intensity changes, like those shown in red in the diagram below.
Form a mathematical solution:
I0 = I1 / I0 x IT / I1
Break down the situations:
I1 / I0 , intensity changes due to absorption in the muscle
I0 = I1 / I0 x IT / I1
Break down the situations:
I1 / I0 , intensity changes due to absorption in the muscle
So,
I1 / I0 = e-23 (0.034) = 0.457
1T / I1 , Intensity changes due to reflection at boundary
So 1T / I1 = 1 - 12 / I1
= 1 - (6.3 – 1.7)2 / (6.3 + 1.7)2
= 1 - 0.33
= 0.67
Substitution: IT / I0 = 0.457 x 0.67
= 0.31
Q5(a)
(i) 1011(ii) 6 = 0110 ; 8 = 1000 ; 14(round down) = 1110 ; 5 = 0101 ; 3 = 0011 ; 1 = 0001
Q5(b)
Q5(c)
1. increase sampling frequency so that step width is reduced2. increase number of bits so that step height is reduced
Q6(a)
Electrical potential is constant inside a charged metal sphere, and is inversely proportional to distance after R.
The smooth curve passes through correct point at 0.5Vs for 2R and 0.33Vs at 3R
Electric field strength is zero inside a charged metal sphere, and is inversely proportional to square of distance after R.
The smooth curve passes through correct point at 0.25Es for 2R and 0.11Es at 3R
Q7(a)
When the potential is zero, charge stored should be zero.When a graph is plotted, it is a straight line that does not pass through the origin.
Q7(b)
Capacitance = Charge / VoltageForm a straight line equation of the graph: Q = gradient x V + y-intercept
Therefore capacitance is the gradient of the graph = 2800 microfarad
Q7(c)
E = 1/2 CV2,Difference in energy = E final - E initial
= 1/2 C (V2 final - V2 initial)
= 1/2 (2800) (9.02 - 6.0)
= 6.3 x 10-2
Note: Do not assume that V in the equation E=1/2 CV2 is the difference in the potential.
Q8(a)
Typical op-amp has very high gain, resulting in a saturated output in most cases. For the output not to be saturated, the inverting and non-inverting input must be equal. Diagram shows that the non-inverting input is earthed, therefore having 0 V. Hence, inverting input must be 0 V, and that is why point P is called virtual earth.Q8(b)
Input resistance to op-amp is very high, current throughR2 = current through R1
VIN - VP / R2 = VP - VOUT / R1
VIN - 0 / R2 = 0 - VOUT / R1
VIN / R2 = - VOUT / R1
VOUT / VIN = - R1/ R2
Gain = - R1/ R2
Q8(c)
Note: Arrange the diode 1 such that it is forward biased when output is positive, and diode 2 such that the current is looped in a clockwise direction. Arrange the opposite when question asks for negative output.
Q9(a)
t = 0.10 mm (where the magnetic field passes through)Q9(b)
VH = (0.13 × 3.8) / (6.0 × 1028 × 0.10 × 10–3 × 1.60 × 10–19)= 5.1 × 10–7 V
Q10(a)
Alternating current generates a changing magnetic flux, that is cut by the copper core. Different magnitude of EMF are induced at different parts of the copper core, resulting in an Eddy current, I. Due to the resistance in copper, R, heat will be lost according to P = I2R. Therefore temperature rises.Q10(b)
Magnetic fields brought in by magnet B is cut by the aluminium tube inducing an emf ,and therefore, eddy current in the aluminium. This electrical energy is tranferred from the kinetic energy of magnet B, hence reducing its speed and causing to take longer time to fall.Q11(a)
Period = 15 ms, Frequency = 1/15m = 67 HzQ11(b)
Mean current = 0 A
Note: The shape of the sinusoidal wave is symetrical, therefore the average must be the equilibrium point of the wave, i.e. zero.
Q11(b)
R.M.S = Max current / surd 2 = 0.53AQ11(c)
Energy dissipated = IRMS2R x t = 0.532 x 450 x 30m = 3.8 JQ12(a)
In a solid, neighboring atoms are closer together. This changes and spreads the energy levels into a band.Q12(b)
In a constant temperature, increase in light intensity will bring more photons to the electron on semiconductor. This will increase the number of electrons in valence band being excited, upon receiving energy from the photons, to the conduction band. They leave a same number of holes in the valence band. Both electrons and holes are the charge carriers, that increase the current, Hence reducing the resistance.Q13(a)
1. There are background radiation2. Radiated particles from the sample escape through the space between the sample and the detector
3. The daughter nuclei emit particles too
Q13(b)
A = Ao exp ( –λt )1.21 × 102 = 3.62 × 104 exp( –λ× 42.0 )
λ = ln2 / T½
T½ = 5.1 minutes
Q13(c)
They have discrete energy levels.Note: Emission line spectra show narrow strips of colour that indicates photons of discrete energies are emitted from an atom. This also indicate discrete energy levels in the atom as the photons are emitted when excited electrons drop from higher energy levels.
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