Monday, May 26, 2014

9702 November 2013 Paper 23 Worked Solutions | A-Level Physics

  • 9702 November 2013 Paper 23 Worked Solutions | A-Level Physics



Question 1
Diameter of cylindrical disc = 28mm
Thickness of cylindrical disc = 12mm
Density of material of disc = 6.8 x 103 kgm-3

Density = mass, m /volume

Volume = π(14 x 10-3) 2 x 12 x 10-3 = 7.389 x 10-6 m3   
Mass, m = density x volume = 6.8 x 103 x 7.389 x 10-6 = 0.0502kg
Weight = mg = 0.0502 x 9.81 = 0.49N




Question 2
(a)
SI units for T : s, R : m and M : kg
K = T2M/R3     units : s2kgm-3

(b)
Uncertainty in T : 0.5%, R : 1% and M : 2%
K = T2M/R3    
{(ΔK / K) x 100% = [2(ΔT / T) + (ΔM / M) + 3(ΔR / R)] x 100%} 
% uncertainty in K = 2(0.5)% {for T} + 2% {for M} +3(1)% {for R} = 6%
K = [864002 x 6 x 1024] / (4.23 x 107) = 5.918 x 1011
{(ΔK / K) x 100% = 6%. So, ΔK = (6/100)  x K} 
(Uncertainty in K =) 6 % of K = 0.355 x 1011
K = (5.9 ± 0.4) x 1011 s2kgm-3




Question 3
(a)
(i)
Velocity is the rate of change of displacement OR displacement change per unit time taken.

(ii)
Acceleration is the rate of change of velocity OR velocity change per unit time taken

(b)
(i)
Initially , the car moves with constant velocity (gradient constant). Then the velocity decreases in the middle section (gradient decreases). Finally, in the last section, the car moves with a constant velocity, which is smaller than the initial constant velocity (gradient is constant but smaller).

(ii)
Velocity =  45/1.5 = 30ms-1

(iii)
At t = 6s, x = 122m and at t = 4s, x = 98m
Velocity at 4s = (122-98) / 2 = 12ms-1

at t = 1.5s, velocity = 30ms-1 and at t = 4s, velocity = 12ms-1
Acceleration = (12-30) / (4.0-1.5) = (12-30) / 2.5 = -7.2ms-2

(iv)
F = ma = 1500 x (-7.2) = -10800N ≈  -11000N




Question 4
{Detailed explanations for this question is available as Solution 390 at Physics 9702 Doubts | Help Page 72 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html}



Question 5
{Detailed explanations for this question is available as Solution 660 at Physics 9702 Doubts | Help Page 132 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-132.html}





Question 6
(a)
e.m.f is the total energy available per unit charge. The potential difference across the battery is not equal to the e.m.f of the battery because some of the available energy is used / lost / wasted / given out in the internal resistance of the battery. Hence the potential difference available is less than the e.m.f.

(b)
(i)
V =IR
Current in circuit, I = V/R = 6.9 / 5.0 = 1.38A ≈ 1.4A

(ii)
r = lost volts / current = (9 - 6.9) / 1.38 = 1.5(2)Ω

(c)
(i)
P = EI = 9 x 1.38 = 12 (12.4) W

(ii)
Efficiency of battery = output power / total power = VI /EI
                 = 6.9 / 9      or (9.52) / (12.4)  = 0.767 (76.7%)




Question 7
(a)
(i)
The 6 vertical lines from one plate to the other should be equally spaced with the arrow downwards 

(ii)
Electric field strength E = V /d = 1200 / 40 x 10-3 = 3.0 x 104Vm-1  (allow 1 sf)

(b)
(i)
Force acting on A, F = Ee = 3.0 x 104 x 1.6 x 10-19 = 4.8 x 10-15 N

(ii)
Torque of the couple acting on the rod = F x separation of charges
             =   4.8 x 10-15 x 15 x 10-3 = 7.2 x 10-17 Nm

(iii)
The rod will be vertically aligned, parallel to the electric field, with A will be at the top, close to the +ve plate and B will be at the bottom, close the –ve plate.

The forces are equal and opposite in the same line. So, there is no resultant force and no resultant torque. 

4 comments:

  1. can you please draw Q5(c)? i don't get where the strobe or videocamera has to be.

    ReplyDelete
  2. 6 question c part 2 explanation for efficiency i do not get it properly

    ReplyDelete
    Replies
    1. what exactly do you not understand

      Delete

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