FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Tuesday, October 7, 2014

9702 November 2007 Paper 2 Worked Solutions | A-Level Physics

  • 9702 November 2007 Paper 2 Worked Solutions | A-Level Physics


Paper 2


Question 1
(a)
Distinguish between systematic and random errors:
For systematic error: Choose any 1: e.g.
There is a constant error (in all the readings)
It cannot be eliminated by averaging
The error is in the measuring instrument

For random error: Choose any 1: e.g.
The readings are scattered (equally) about the true value
The error is due to the observer
The error can be eliminated by averaging

(b)
Cylinder of length L has circular cross-section of radius R. Volume V of cylinder given by            V = πR2L
Volume and length of cylinder are measured as V = 15.0 ± 0.5 cm3; L = 20.0 ± 0.1 cm.
Radius of cylinder, with uncertainty:

{R2 = πL / V. So 2(ΔR / R) = (ΔL / L) + (ΔV / V)}
15 = πR2 (20)
R = 0.4886cm (accept any number of sf)
% uncertainty in V = 3.3% (or =0.5/15)
% uncertainty in L = 0.5% (or =0.1/20)
% uncertainty in R = 1.9% (half of the sum of % uncertainties)
R = 0.489 ± 0.009cm



Question 2
Girl G is riding bicycle at constant velocity of 3.5 ms–1. At time t = 0, she passes boy B sitting on bicycle that is stationary. At time t = 0, boy sets off to catch up with girl. He accelerates uniformly from time t = 0 until he reaches speed of 5.6 ms–1 in a time of 5.0 s. He then continues at constant speed of 5.6 ms–1. At time t = T, boy catches up with girl. T measured in seconds.
(a)
In terms of T, distance moved by girl before boy catches up with her:
Distance = (Speed x Time =) 3.5T

(b)
For boy,
(i)
Distance moved during his acceleration:
Distance = Average Speed x time
(a = (5.6 – 0) / 5)
Distance, s = ut + ½ at2 = 0 + ½ [(5.6 – 0) / 5] x 52 = 14m

(ii)
Distance moved during time that he is moving at constant speed. Give answer in terms of T:
(Time he is moving with constant speed = T - 5)
Distance = 5.6 (T – 5)             (or 3.5T – 14)

(c)
Use answers in (a) and (b) to determine time T taken for boy to catch up with girl:
3.5T = 14 + 5.6(T – 5)
Time, T = 6.7s

(d)
Boy and bicycle have combined mass of 67 kg.
(i)
Force required to cause acceleration of boy:
Acceleration, a = (5/6 / 5 =) 1.12ms-2
Force = ma = (67 x 1.12 =) 75N

(ii)
At speed of 4.5 ms–1, total resistive force acting on boy and bicycle is 23 N. Output power of boy’s legs at this speed:
Power = (force x speed =) [75+23] x 4.5 = 440W



Question 3
(a)
(i)
Potential energy is defined as the stored energy available to do work

(ii)
Difference between gravitational potential energy and elastic potential energy:
Gravitational potential energy is due to height / the position of a mass OR distance from a mass OR moving a mass from one point to another.
Elastic potential energy is due to the deformation / stretching / compressing

(b)
Small sphere of mass 51 g suspended by light inextensible string from fixed point P. Centre of sphere is 61 cm vertically below point P, as shown. Sphere is moved to one side, keeping string taut, so that string makes an angle of 18° with vertical.
(i)
Gain in gravitational potential energy of sphere:
The height raised, h = [61 – 61cos(18) =] 3.0cm
Gain in energy = [mgh = 0.051(9.8)(0.030) =] 1.5x10-2J 

(ii)
Moment of weight of sphere about point P:
Moment = Force x perpendicular distance
Moment = (0.051 x 9.8) (0.61sin(18)) = 0.094Nm





Question 4
{Detailed explanations for this question is available as Solution 1025 at Physics 9702 Doubts | Help Page 214 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-214.html}





Question 5

{Detailed explanations for this question is available as Solution 240 at Physics 9702 Doubts | Help Page 38 - http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-38.html}




Question 6
{Detailed explanations for this question is available as Solution 1020 at Physics 9702 Doubts | Help Page 213 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-213.html}





Question 7
(a)
Evidence for nuclear atom provided by α-particle scattering experiment. Results of this experiment:
Most of the α-particles are deviated through small angles
A few α-particles are deviated through angles greater than 90o.

(b)
Estimates for diameter for
(i)
Atom: allow 10-9m to 10-11m

(ii)
Nucleus: allow 10-13m to 10-15m



6 comments:

  1. Why cant I use P=VI in no. 6 c) i ?

    ReplyDelete
    Replies
    1. The explanation has been updated. Check again.

      Delete
  2. is it because minimum area of the MATERIAL has to be at least 2 minimum, so bubble can be anywhere between 0 and 1.2 as it fits inside the wire??
    Also, what does the difference in extensions on a thicker rod matter for its breaking?

    ReplyDelete
    Replies
    1. The explanation has now been updated. Check again.

      Delete
  3. Replies
    1. See solution 1097 at
      http://physics-ref.blogspot.com/2016/02/physics-9702-doubts-help-page-234.html

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | 9702 November 2007 Paper 2 Worked Solutions | A-Level Physics