Tuesday, September 16, 2014

9702 June 2004 Paper 2 Worked Solutions | A-Level Physics

  • 9702 June 2004 Paper 2 Worked Solutions | A-Level Physics



Paper 2



Question 1
(a)
Difference between scalar quantity and vector quantity:
A scalar quantity has a magnitude only while a vector quantity has both a magnitude and a direction

(b)
2 forces of magnitude 6.0N and 8.0N act at point P. both forces act away from point P and angle between them is 40o. Fig shows 2 lines at angle of 40o to one another. On Fig, draw vector diagram to determine magnitude of resultant of 2 forces:
The diagram has a correct shape (parallelogram or triangle) with the arrows in the correct directions (parallel to the given ones)
(Note that the dotted lines do not represent the size of the 2 vectors mentioned – you need to choose the appropriate scale)
Resultant = 13.2 ± 0.2 N



Question 2
{Detailed explanations for this question is available as Solution 435 at Physics 9702 Doubts | Help Page 83 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-83.html}



Question 3
Student has been asked to determine linear acceleration of toy car as it moves down a slope. He sets up apparatus as shown.
Time t to move from rest through a distance d found for different values of d. Graph of d (y-axis) plotted against t2 (x-axis) as shown.
(a)
Theory suggests that graph is a straight line through origin. Name feature on Fig that indicates presence of
(i)
Random error:
The scatter of points (about the line)

(ii)
Systematic error:
The intercept (on the t2 axis)

(b)
(i)
Gradient of line of graph in Fig:
Gradient = Δy / Δx = (100 – 0) / (10.0 – 0.6) = 10.6 (cms-2)

(ii)
Use answer to (i) to calculate acceleration of toy down slope:
s = ut + ½ at2
(½ a = gradient) So, acceleration = 2 x gradient
Acceleration = 0.212ms‑2
 


Question 4
Ball has mass m. it is dropped onto horizontal plate as shown. Just as the ball makes contact with plate, it has velocity v, momentum p and kinetic energy Ek.
(a)
(i)
Expression for momentum p in terms of m and v:
p = mv

(ii)
Show that kinetic energy given by Ek = p2 / 2m:
Ek = ½ mv2
Algebra leading to (e.g. p = mv and v = p/m)
Ek = p2/2m

(b)
Just before impact with plate, ball of mass 35g has speed 4.5ms-1. It bounces from plate so that its speed immediately after losing contact with plate is 3.5ms-1. Ball is on contact with plate for 0.14s.
For time that ball is in contact with plate,
(i)
Average force, in addition to weight of ball, that plate exerts on ball:
EITHER
Δp = 0.035(4.5 + 3.5) = 0.28Ns
Force = Δp / Δt (= 0.28 / 0.14) = 2.0N
OR
a = (4.5 – (-3.5)) / 0.14 = 57.1ms-2
F = ma (= 0.035 x 575.1) = 2.0N

(ii)
Loss in kinetic energy of ball:
Loss in kinetic energy = ½ (0.035) (4.52 – 3.52) = 0.14J

(c)
Explain whether linear momentum conserved during bounce:
The plate (and Earth) gain momentum which is equal and opposite to the change for the ball. So, momentum is conserved.



Question 5
2 forces, each of magnitude F, form a couple acting on the edge of disc of radius r, as shown.
(a)
Disc is made to complete n evolutions about axis through its centre, normal to plane of disc. Expression:
(i)
Distance moved by point on circumference of disc:
Distance = n(2πr) = 2πnr

(ii)
Work done by 1 of the 2 forces:
Work done = F x (2πnr)

(b)
Using answer to (a), show that work W done by couple producing torque T when it turns through n revolutions given by          W = 2πnT:
Total work done = 2 x F x 2πnr
But the Torque T = 2Fr
Hence, work done = 2πnT

(c)
Car engine produces torque of 470Nm at 2400 revolutions per minute. Output power of engine:
Power = Work done / time (= 470 x 2π x 2400 / 60) = 1.2x105W



Question 6
Fig shows wavefronts incident on, and emerging from, double slit arrangement. Wavefronts represent successive crests of wave. Line OX shows 1 direction along which constructive interference may be observed.
(a)
The principle of superposition states that when two (or more) waves meet, the resultant displacement is the sum of the individual (displacements).

(b)
On Fig, draw lines to show
(i)
second direction along which constructive interference may be observed (label line CC):
Any correct line through the points of intersection of crests

(ii)
direction along which destructive interference may be observed (label line DD):
Any correct line through the intersections of a crest and a through

(c)
Light of wavelength 650nm incident normally on double slit arrangement. Interference fringes formed are viewed screen placed parallel to and 1.2m from plane of double slit, as shown. Fringe separation is 0.70mm.
(i)
Separation a of slits:
λ = ax / D                    or λ = asinθ     and θ = x/D
650x10-9 = (a x 0.70x10-3) / 1.2
Separation, a = 1.1x10-3m

(ii)
Width of both slits increased without changing their separation a.  State effect, if any, that change has on:
1.
Separation of fringes:
No change

2.
Brightness of light fringes:
Brighter

3.
Brightness of dark fringes:
No change



Question 7
Household electric lamp rated as 240V, 60W. Filament of lamp made from tungsten and is a wire of constant radius 6.0x10-6m. Resistivity of tungsten at normal operating temperature of lamp is 7.9x10-7Ωm.
(a)
For lamp at its normal operating temperature:
(i)
Current in lamp:
P = VI
So, current I = P / V = 60 / 240 = 0.25A

(ii)
Show that resistance of filament is 960Ω:
Resistance, R (= V/I) = 240 / 0.25 = 960Ω

(b)
Length of filament:
R = ρL / A       (wrong formula, 0/3)
960 = (7.9x10-7) L / (π{6.0x10-6}2)
Length of filament, L = 0.137m

(c)
Comment on answer to (b):
Example:
The filament must be coiled / it is long for a lamp



Question 8
Thermistor has resistance 3900Ω at 0oC and resistance 1250Ω at 30oC. Thermistor connected into circuit of Fig in order to monitor temperature changes. Battery of e.m.f. 1.50V has negligible internal resistance and voltmeter has infinite resistance.
(a)
Voltmeter is to read 1.00V at 0oC. Show that resistance of resistor R is 7800Ω:
V / E = R / Rtot                        or 0.5 = I x 3900
1/0/1.5 = R / (R+3900)            or 1.0 = 0.5R / 3900
R = 7800Ω                              or R = 7800Ω

(b)
Temperature of thermistor increased to 30oC. Reading on voltmeter:
EITHER
V = 1.5 x [7800 / (7800+1250)] = 1.29V
OR
I = 1.5 / (7800+1250)              (=1.66x10-4A)
V = IR (= (1.66x10-4)x7800) = 1.29

(c)
Voltmeter in Fig replaced with one having resistance of 7800Ω. Reading on this voltmeter for thermistor at temperature of 0oC:
Combined resistance of resistor R and voltmeter (= [1/7800 + 1/7800]-1) = 3900Ω
So, reading at 0oC (= [3900 / (3900+3900)]x1.5 ) = 0.75V




27 comments:

  1. Q. Nov2003/19, Jun2004/21, Nov2010/20/P11 Thank You!

    ReplyDelete
    Replies
    1. They are explained at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-131.html

      Delete
  2. Q. May2002/36 why not B, June2003/35 why not C, June2009/29, Nov2009/27/P11 Thank You.

    ReplyDelete
    Replies
    1. For May2002/36, see solution 677 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-136.html

      Delete
    2. How about other questions? Thank You.

      Delete
    3. For June2003/35 and June2009/29 see solutions 678 and 681 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-137.html

      Delete
    4. For Nov2009/27/P11, go to
      http://physics-ref.blogspot.com/2014/10/9702-november-2009-paper-11-worked.html

      Delete
  3. Replies
    1. State the specific questions for which you require help

      Delete
  4. June2009/10 thank you

    ReplyDelete
    Replies
    1. See question 321 at
      http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-55.html

      Delete
  5. Nov2011/P12/9 and 10 Thank You.

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/08/9702-november-2011-paper-12-worked.html

      Delete
  6. Replies
    1. See solution 779 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-157.html

      Delete
  7. June 05/p1/24, Nov 10/p11/25 June08/p1/25, June11/P21/25,26,27

    ReplyDelete
    Replies
    1. For June 05/p1/24, see solution 851 at
      http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-170.html

      For Nov 10/p11/25, see solution 140 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-25.html

      For June08/p1/25, see solution 146 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-26.html

      For June11/P21/25,26,27, go to
      http://physics-ref.blogspot.com/2014/09/9702-june-2011-paper-12-worked.html

      Delete
  8. Oct 2011/12/21, 28 Thank You.

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/08/9702-november-2011-paper-12-worked.html

      Delete
  9. 2010/Oct/13 Q11,28

    ReplyDelete
    Replies
    1. 2010/Oct/13 Q11 has been explained as solution 1039 at
      http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-217.html

      Delete
  10. can you please do w03qp4,Q4b) how is the mean potential difference 5.4?

    ReplyDelete
  11. and w03 qp 4,4c), i dont get the graph.

    ReplyDelete
    Replies
    1. Check solution 272 at
      http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-45.html

      Delete
  12. Thank you very much. Your help is deeply appreciated.
    P.S, Keep up the wonderful work.

    ReplyDelete

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