CIE 9702 Physics Paper 21 MJ 2016 Mark Scheme Answers

Q1(a)

(i) (50 to 200) × 10^–3 kg
(ii) (50 to 300) cm³
Note: To estimate volume of a tennis ball, hold your fist, measure the diameter with a ruler, then calculate volume using uniform sphere equation

Q1(b)

(i) Density = Mass / Volume
     Volume, V = π (0.38 × 10^–3 / 2)² × 25.0 × 10^–2
                       = 2.835 × 10^–8 m³
      Density, ρ = (0.225 × 10^–3) / 2.835 × 10^–8
                       = 7940 kg m³
Note: Do not round off value of volume and density too early, it is good to keep more than 3 s.f.

             Δρ / ρ = 2Δd / d  +  ΔL / L  +  ΔM / M
                        = 2(0.01/0.38) + (0.1/25.0) + (0.001/0.225)
                        = 0.061
                   Δρ = 0.061 × 7940
                        = 480

            density = 7900 ± 500 or (7.9 ± 0.5) × 10³
REMEMBER: Uncertainty needs to be round up to 1 s.f., the actual value is then made to follow its d.p!!

Q2(a)

(i) horizontal component of U = 12 cos 50° = 7.7
(ii) vertical component of U = 12 sin 50° = 9.2
Note:

Q2(b)

Component to be used: vertical components
Assumption: upward is positive
SUVAT method: S = to prove , U = 9.2 , V = 0.0 , A = g = - 9.81, t=?
Equation to use: V² = U² + 2gS, because t is not given.
Substitute: 0.0² = 9.2² + 2(-9.81) S
                       S = 4.31 = 4.3
Note:

Q2(c)

Component to be used: vertical displacement and horizontal displacement 

horizontal displacement, S_H = horizontal speed, u_H x time taken to reach Q.
Use vertical components to find time taken,
Equation:      -g = V - U / t
                 -9.81 = - 9.2 / t
                        t = 0.94 s

                     S_H = 7.7 x 0.94 = 7.2

Displacement PQ  = [ S_H²+S²]^1/2
                              = [ 7.2²+ 4.3²]^1/2

                              = 8.4 m

Q3(a)

(i) N = 0.150 x 9.81 = 1.47 N

(ii) Normal force equals weight acting on ball, but acts in opposite direction, according to Newton's 3rd Law.

Q3(b)

(i) resultant force is directly proportional to rate of change of momentum

(ii) F = change of momentum / time
         = m (v - u) / t
         = 0.15 (2.5-(-6.2) / 0.12
         = - 10.9 N
answer is negative, so acting upwards
Note: - 6.2 because it is opposite direction with 2.5, so the assumption made is that downward is positive. 

(iii) Considering in this system of collision, there is the ball and the floor (the Earth) as the second object. The ball experiences a change in momentum in an upward direction, and the Earth experiences an equal but opposite direction of the change in momentum. Due to the relatively bigger mass, Earth does not seem to move. So, linear momentum is conserved.

Q4(a) 

Elastic potential energy refers to the energy stored in a body due to its extension/ compression/ deformation/ change in shape/size

Q4(b)(i)

two values of F/x are calculated which are the same

e.g. 10.4 / 40 = 0.26 and 6.5 / 25 = 0.26

or

ratio of two forces and the ratio of the corresponding two extensions are

calculated which are the same

e.g. 5.2 / 10.4 = 0.5 and 20 / 40 = 0.5

or

gradient of graph line calculated and coordinates of one point on the line used with straight line equation y = mx + c to show c = 0

(so) force is proportional to extension (and so Hooke’s law obeyed)

Note: Question says: 'Use data from Fig. 4.1...' means it is expecting some calculations using the data of the graph. Just a statement saying that force is proportional to extension is not sufficient.

Q4(b)(ii)

(1) spring constant = F / x = 26 N m^-1

(2) work done = area under the graph = ½ × (5.2 + 10.4) × 20 × 10^-2
                                                             = 1.6 J

Q4(c)

Remove the force and check whether the spring return to its original length.

Q5(a)

Period = 4.0 ms, frequency = 1/4.0m = 250 Hz

Q5(b)

intensity ∝ (amplitude)²  , so new amplitude = 2.8
same period
curve shifted 1.0 ms to left or to right of wave X

Q5(c)

(i) From the equation of double slit, x = λ D / a

     λ / a equals to the gradient.

     λ = gradient × a

     gradient =  (4.5 – 2.4) × 10^-3 / (3.25 – 1.75) 

                   = 1.4 × 10^-3

                λ = 1.4 × 10^-3× 0.45 × 10^-3

                   = 630 nm
Note: Although you can calculate the wavelength using the equation of double slit, and data of a single point, you are reminded to always follow the instruction of the question, i.e. 'Use the gradient of the line...'

(ii) λ / a equals to the gradient.
When a is increased, gradient will be smaller.
But all value of x will be smaller for all values of D, therefore the line will be entirely below the original graph.

Note: This question is very uncommon, it requires thorough thinking of the differences between the two graphs

Q6(a)

One coulomb is the amount of charge that carries by one ampere in one second

Q6(b)

(i)
total charge, Q = (n × volume) × charge of one electron = n × Al × e
current, I = Q / t

Q = I × t = n × A × l × e
             I = [ n × A × l × e ] / t

Since speed = distance / time, v = l /t
I = [ n × A × e ] × v
v =  I / [ n × A × e ]

Q6(c)(i)

Comparison question 
Steps of solution:

1. Identify the equation: v =  I / [ n × A × e ]
2. Tabulate the quantities into a table:

Y	Z
I	I
n	n
d	2d
e	e

3. Move all constants to one side:

                   v =  I / [ n × A × e ]
                vA = I/ne = constant
           v_Y A_Y = v_Z A_Z

4. Now you can rearrange to the ratio required
             v_Y / v_Z  = A_Z / A_Y  ; A = π d²
                          = d_Z² /d_Y²                        
                          = 4d²/ d²
                          = 4

Q6(c)(ii)

Comparison question
Steps of solution:

1. Identify the equation: R = ρl / A   ; ρ = constant
2. Tabulate the quantities into a table:

Y	Z
l	2l
d	2d

3. Move all constants to one side:
                          RA / l = constant
                          [ RA/l ]_Y = [ RA/l ]_Z

4. Now you can rearrange to the ratio required
                            R_Y / R_Z =  [ l / A ]_Y [ A / l ]_Z
                                          = [ l / d² ] [ 4d² / 2l ]
                                          = 2

Q6(c)(iii)

Potential divider rule:
V_Y = R_Y / [ R_Y +R_Z ] × 12
      = [ 2/3 ]× 12
      = 8.0 V

Q6(c)(iv)

P = I²R
P_Y / P_Z = R_Y / R_Z
             = 2

Q7(a)

hadron: neutron/proton lepton: electron/(electron) neutrino

Q7(b)

(i) proton: up up down or uud
(ii) neutron: up down down or udd

Q7(c)

(i) neutron → proton + electron + (electron) antineutrino
(ii) up down down (quarks) change to up up down (quarks)
or
down (quark) changes to up (quark)