CIE 9702 Physics Paper 22 MJ 2016 Mark Scheme Answers

Q1(a)

Acceleration is defined as the rate of change of velocity.

Q1(b)

(i) a = v - u / t
        = 36 - 0 / 19
        = 1.89

(ii) s = 1/2 (u+v) t
         = 0.5 x 36 x 19
         = 342

(iii)
1. Change in KE = final - initial KE
                            = 1/2 m v² - 0
                            = 0.5 x 95 x 36²
                            = 6.16 x 10⁴ J

2. Change in PE = mgh
                           = 95 x 9.81 x 342 sin 40^o
                           = 2.05 x 10⁵ J

(iv) Lost of energy = PE - KE = 1.43 x 10⁵ J
       frictional force x distance moved = Lost of energy
                         f      x       342              = 1.43 x 10⁵
                                            f                = 418 N


(v) frictional force - downward force due to gravity = net force from B to C
              f               -           mg sin 20^o                       = m a
              f               -        95 x 9.81 x sin 20^o             = 95 x 3.0
              f  = 604 N
Note:

Q2(a)

Pressure, P = force / cross sectional area
                   = weight of liquid / A

where density of liquid = mass / volume  ,  and  ,  weight = mass x g  ,  and  ,  volume = A x h
                                   ρ = [ weight / g ] / [ A x h ]
So,                     weight = ρ x g x A x h

Therefore, P = ρ x g x A x h / A
                  P = ρ x g x h
Note:

Q2(b)

(i) It is because at h = 0, pressure is not zero at the base of the cylinder, there is atmospheric pressure due to the open upper section of the container.

(ii) Take the point of P = 2.0 x 10⁵, gives h = 0.75 m
      However, the pressure cause by the liquid column is P = (2.0 - 1.0) x 10⁵
      So,          P = ρ x g x h
         1.0 x 10⁵ = ρ x 9.81 x 0.75
                      ρ = 1.36 x 10⁴

Q3(a)

Young modulus is defined as stress divided by strain

Q3(b)

(i)
Comparison question:
1. Equation: YM = F L / e A

2. Tabulate the quantities into a table:

Steel	Copper
YMs	YMc
es	ec
A	A
L	L
F	F

3. Move all constants to one side:

                   YM = F L / e A
                   YM x e = FL / A = constant
                   YM_s x e_s = YM_c x e_c
       
4. Now you can rearrange to the ratio required
                    YM_s x e_s = YM_c x e_c
                     e_c / e_s = YM_s / YM_c 
                               = 1.9 x 10¹¹ / 1.2 x 10¹¹

                               = 1.58
(ii) Draw two straight lines from (0,0) with S having a steeper gradient than C

Q4(a)

(i) Vibration of the particles in a longitudinal wave is parallel to the direction of the propagation of energy.
(ii) Vibration of the particles in a transverse wave is perpendicular to the direction of the propagation of energy.
Note:

Q4(b)

Equation of I = Energy / time / area
Unit of I = kg ms^-2 m / s m²
               = kg s^-3
Unit of v = m s^-1
Equation of density = mass / volume
Unit of density = kg m^-3
Unit of f = s^-1
Unit of A = m

Combined units on the left hand side = kg s^-3
Combined units on the right hand side = m s^-1 kg m^-3 s^-2 m²
                                                              = kg s^-3


Note:

Q4(c)

(i)  It is a phenomenon where there is a change in the observed/apparent frequency when the source is moving (relative to the observer).
(ii) The wavelength of the light increases, i.e. red shift happens.
      or frequency of the light decreases.

Q4(d)

observed f = fs v / ( v - vs )
           550 = 510 (340) / (340 - vs)
             vs = 24.7

Q5(a)

When the waves arrive at each slit, diffraction happens, that the waves spreads.
These waves overlap after passing through the slits. At a point where their path difference is one λ, the first order is formed. This is the interference.

Q5(b)

d sinθ = nλ
d = (2 × 486 × 10^-9) / sin (59.4°/2)
   = 1.962 × 10^-6
number of lines, N = 1 / d
                               = 510 mm^-1

Q6(a)

Draw 6 horizontal lines, equally space, between the two plates

Q6(b)

Electrical potential energy is being converted into kinetic energy
q x V (J) = 15 (keV)
2 e x V (J) = 15000 x e (J)     [q=2e because alpha particle has two protons]
V = 7500 V

E = V / d
   = 7500 / (16 x 10^-3)
    = 4.69 x 10⁵ V m^-1

Note: be careful with the unit conversion of eV and J, will be good to bracket them to avoid confusing yourself.   

Q7(a)

Charged is quantised means charge exists only in discrete amounts
Note: Charge of any particle has only a certain amount which is the elementary charge (1.6 x 10^-19) or the multiple of this number. Example, charge of helium-4 nucleus, 3.2 x 10^-19, due to the two protons in the nucleus. 

Q7(b)

(i) V = IR
     9.0 = I (0.25 + 0.15 + 2.7 + 0.15)
         I = 2.77 A

(ii) V = IR
          = 2.77 (0.15 + 2.7 + 0.15)
          = 8.31

or

       V = E - Ir
           = 9.0 - 2.77(0.25)
           = 8.31
Note: Pd across internal resistance shall be ignored when finding pd across the battery.

Q7(c)

(i) I = Anvq
     2.77 = (2.5 x 10^-6) (8.5 x 10²⁹) v (1.6 x 10^-19)
          v = 8.15 x 10^-6

(ii)
Resistance is inversely proportional to area (or diameter squared).
So resistance of Z is 4 times of X, because area of Z is 4 times smaller than X.

Mathematically, total external resistance will increase by less than 4 times, and so current decreases less than 4 times.

Based on the equation above, drift speed is directly proportional to current, and inversely proportional to area, so decrease in 'area' over weighs decrease in 'current', therefore drift speed increases.
Note: This is a very challenging question, as it involves variation of multiple quantities. Explanation cannot solely based on the equation used in (i), because not only A has been changed, current also has been changed. 

Q8(a)

electron: lepton
neutron: hadron
neutrino: lepton
proton: hadron

Q8(b)

(i) 


(ii)  uud --> udd
(iii) weak nuclear force