Tuesday, November 4, 2014

Physics 9702 Doubts | Help Page 7

  • Physics 9702 Doubts | Help Page 7



Question 40: [Mechanics > Dynamics > Forces and motion]
One horse pulls, with a force of X N, a cart of mass 800kg along a horizontal road at constant speed. Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8ms-2. Find the time it would take two horses to increase the speed of the cart from 2ms-1 to 5ms-1, given that each horse pulls with a force of X N, and that the resistance to motion has the same constant value at all times.



Solution 40:
From the question, resistance to motion (frictional force) is constant at all times.
Consider
One horse pulls, with a force of X N, a cart of mass 800kg along a horizontal road at constant speed.
With a force of X N, the cart moves with constant velocity. That is, acceleration is zero. The resultant force is thus zero. Let the force of friction be FR.
X – FR = ma = 0
So, FR = X N   (Frictional force = X N)

Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8ms-2
Resultant force = 3X – FR = 3X – X = 2X N
This causes the cart of mass of 800kg to have an acceleration of 0.8ms-2.
2X = ma = 800(0.8)
Force X = 800(0.8) / 2 = 320 N

When 2 horses are pulling,
Resultant force = 2X – FR = 2X – X = X = 320 N
320 = 800a
Acceleration = 320 / 800 = 0.4ms-2

Initial velocity, u = 2ms-1 and Final velocity, v = 5ms-1
v = u + at
Time, t = (v – u) / a = (5 – 2) / 0.4 = 7.5s







Question 41: [Waves > Progressive & Stationary]
(a) Transverse progressive wave travels along stretched string from left to right. Shape of part of string at particular instant is shown.
Frequency of wave is 15 Hz. For this wave, use Fig to determine
(i) Amplitude
(ii) Phase difference between points P and Q on string
(iii) Speed of wave

(b) Period of vibration of wave is T. Wave moves forward from position shown in Fig for a time 0.25 T. On Fig, sketch new position of wave:

(c) Another stretched string is used to form stationary wave. Part of this wave, at particular instant, is shown.
Points on string are at their maximum displacement.
(i) State phase difference between particles labelled X and Y:
(ii) Explain following terms used to describe stationary waves on string: Antinode, Node
(iii) State number of antinodes shown for this wave
(iv) Period of vibration of wave is τ. On Fig, sketch stationary wave 0.25 τ after instant shown

Reference: Past Exam Paper – June 2011 Paper 23 Q6 & Specimen Paper 2016 Paper 2 Q7



Solution 41:
(a)
(i)
{Amplitude is the maximum displacement}
Amplitude = 7.6mm

(ii) Phase difference = 180o / π rad

(iii) Speed of wave, v = f λ = 15 x 0.8 = 12ms-1

(b)
The sketch should be correct with the peak moved to the right. The curve should be moved by the correct phase angle / time period of 0.25T
{For time = period T, the wave travels a distance equal to the wavelength, λ. So, for a time = 0.25T = T/4, the distance moved by each point on the wave is λ / 4 = 80 / 4 = 20cm.}





(c)
(i)
{For a stationary wave, points in the same segment are in phase, while points in adjacent segments are in anti-phase (180o). This is because, apart from nodes, the points of a stationary wave can only move upwards or downwards. Here, point X and Y will both move downwards since the position displayed is that where the points are at their maximum displacement}
Phase difference = zero (rad)

(ii)
Antinode: maximum amplitude
Node: zero amplitude / displacement

(iii)
Number of antinodes = 3

(iv)
The sketch contains a horizontal line through the central section of the wave



In the time = period of vibration τ, the new position of the points of the stationary wave will be the same as displayed. [Consider the antinode between X and Y. In 1 period, the point would have to move downwards to the horizontal line, then downwards to the maximum amplitude. Then, the point moves upwards to horizontal line again and finally to upwards to its original position. So in a time = τ, a point travels a distance equal to 4 x amplitude.] So, for time = τ /2, the new position would be the reflection about the horizontal line. For time = 0.25τ = τ /4, the new position be at the horizontal line}
 






Question 42: [Waves > Phase difference]
How to calculate the phase difference between 2 points on a wave?

Solution 42:
Consider 2 points on the wave which are separated by a distance equal to the wavelength, λ of the wave. These 2 points would be in phase (they move the same way). So, a separation of λ corresponds to phase difference of 2π rad (they are in phase).
Therefore, the phase difference between 2 points on a wave can be related to the separation apart.

If point A and B on a wave of wavelength λ are separated by a distance Δx, the phase difference, ϕ between them is given by
ϕ = (Δx / λ) 2π rad


Alternatively, if the start of the wave [of wavelength, λ] (which is at equilibrium position) is taken as the reference, then the phase of point A which is at a distance x1 from the point of origin is given by
ϕ1 = (x1 / λ) 2π rad

Similarly, the phase of a point B which at a distance x2 from the origin is given by
ϕ2 = (x2 / λ) 2π rad


Then the phase difference between the 2 points is
Δϕ = ϕ2 – ϕ1 = {(x2 / λ) 2π} – {x1 / λ) 2π} = (Δx / λ) 2π rad

To directly obtain the phase difference in degrees, use 360o instead 2π rad.







Question 43: [Current of Electricity > Variable resistor]
(a) Variable resistor is used to control current in circuit, as shown. 

Variable resistor is connected in series with 12 V power supply of negligible internal resistance, ammeter and 6.0 Ω resistor. Resistance R of variable resistor can be varied between 0 and 12 Ω.
(i) Maximum possible current in circuit is 2.0 A. Calculate minimum possible current:
(ii) On Fig, sketch variation with R of current I1 in circuit:

(b) Variable resistor in (a) now connected as potential divider, as shown. Calculate maximum possible and minimum possible current I2 in ammeter:

(c)
(i) Sketch I-V characteristic of filament lamp:
(ii) Resistor of resistance 6.0 Ω is replaced with filament lamp in circuits of Fig. 5.1 and Fig. 5.3. Give an advantage of using circuit of Fig. 5.3, compared to circuit of Fig 5.1, when using circuits to vary brightness of the filament lamp:

Reference: Past Exam Paper – June 2011 Paper 21 Q5



Solution 43:
(a)
(i)
{Current is maximum when the resistance of the variable resistor is minimum. Current is minimum when resistance of variable resistor is maximum}
Minimum current, I = 12 / (6 + 12) = 0.67A

(ii)
The graph should have a correct start (0.0, 2.0) and finish points (12.0, 0.67).  The curve should have a correct shape with a decreasing gradient




{Note that the curve passes through the starting point (0.0, 2.0) and ends at (12.0, 0.67) and that the curve has a decreasing gradient. Your graph should have the same shape as above. Some of the other points on which the curve passes have been calculated below. A better graph than the above can be drawn by calculating more points (but I don’t think such accuracy is required for these 2 marks. Note also that the curve should not go beyond R = 12).

To calculate the other points, R is the resistance chosen for the variable resistor (these values are between 0 and 12). The value of current I is obtained by I = 12 / (6+R). So, for example, at R = 6, I = 1.0A. At R = 4, I = 1.2A and at R = 8, I = 12 / 14 = 0.86A. Additionally, at R = 10, I = 12 / 16 = 0.75A and at R = 2, I = 12 / 8 = 1.5A}

(b)
This part has previously already been explained as question 4 at http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-1.html

(c)
(i)
The graph is a smooth curve starting at (0, 0) with a decreasing gradient. The end section should not be horizontal



(ii)
With the circuit of Fig 5.3 (parallel combination), the full range of current / p.d. is possible
OR The currents / p.d. can go down to zero
OR The brightness (of the lamp) ranges from off to full brightness

 





Question 44:   [Measurements > Uncertainty]
Measurements made for sample of metal wire are shown.
Quantity          Measurement               Uncertainty
Length             1750mm                      ±3mm
Diameter         0.38mm                       ±0.01mm
Resistance       7.5Ω                            ±0.2Ω
(a) State appropriate instruments used to make each of these measurements.
(i) Length
(ii) Diameter
(iii) Resistance:

(b)
(i) Show that resistivity of the metal is calculated to be 4.86 × 10–7 Ω m:
(ii) Calculate uncertainty in resistivity:
(iii) Use answers in (b) to express resistivity with its uncertainty to appropriate number of significant figures:

Reference: Past Exam Paper – June 2011 Paper 21 Q1



Solution 44:
(a)
(i) Length: metre rule / tape
(ii) Diameter: micrometer (screw gauge) / digital caliper
(iii) Resistance: an ammeter and a voltmeter / ohmmeter / multimeter on ‘ohm’ setting

(b)
(i)
Resistivity ρ of metal = RA / L = [7.5 x π(0.38x10-3)2 / 4] / 1.75 = 4.86x10-7Ωm

(ii)
{ρ = RA / L. So, (Δρ /ρ) x 100% = [(ΔR / R) + (ΔA / A) + (ΔL / L)] x 100%}
% Uncertainty in R (= ΔR/R x 100%) = [0.2 / 7.5] x 100% = 2.7%
And % Uncertainty in L (= ΔL/L x 100%) = [3 / 1750] x 100% = 0.17%

{A = πr2 = πd2 / 4. So, (ΔA / A) x 100% = 2(Δd / d) x 100%}
% Uncertainty in A (= 2(Δd / d) x 100%) = 2[0.01 / 0.38] x 100% = 5.3%

Total uncertainty (= 2.7 + 0.17 + 5.3) = 8.17%
{(Δρ /ρ) x 100% = 8.17%. So, Δρ = (8.17 / 100) x ρ}
Uncertainty (= (8.17/100) x 4.86x10-7) = 0.397x10-7Ωm

(iii)
Resistivity ρ of metal = (4.9 × 10–7 ± 0.4 × 10–7) Ω m



26 comments:

  1. May/june 2013 variant 22
    Q2 (b)
    please show and explain how the graphs are to be drawn

    ReplyDelete
    Replies
    1. The graphs have been added. More explanations have been included

      Delete
  2. oct/nov 2013 variant 23
    Q2 (b)
    please show the steps in calculation

    ReplyDelete
    Replies
    1. I have added the steps there. Check again

      Delete
  3. may/june 2011 variant 21
    Q1 b (ii)
    pls show the steps

    ReplyDelete
  4. Replies
    1. which year? if it's already posted, plz specify your doubt at the respective page

      Delete
    2. view question 49 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-8.html

      Delete
  5. in june2014 variant 21
    Q1 (c)
    why are both the arrows at the left?

    shouldn't the acceleration be in the direction of motion?

    ReplyDelete
    Replies
    1. I have explained this part there. See if you now understand

      Delete
  6. june 2014 variant 21
    Q3 (b)
    how have yu decided tht the starting point is (0,100) and ending point is (5,980)?

    ReplyDelete
    Replies
    1. Explanations for the graph are now included there

      Delete
  7. nov2013 paper23
    Q4 (b) (ii)
    I saw the paper you've solved but I'm still in doubt
    air resistance opposes motion so the acceleration decreases and hence the time taken should increase

    ReplyDelete
    Replies
    1. details have been added at the page. check it again

      Delete
  8. For solution 41 here, I got why for time = τ, a point travels a distance equal to 4 x amplitude. But I dont understand why for time = τ /2, the new position would be the reflection about the horizontal line and why for time = 0.25τ = τ /4, the new position be at the horizontal line. Please explain :(

    ReplyDelete
    Replies
    1. in a time τ/2, a points travels a distance 2 x amplitude. so, if a point is at its highest position, a displacement = 'amplitude' causes it to move to the equilibrium position. And the remaining displacement = 'amplitude' causes it to now be at the lowest position

      Delete
  9. kindly please solve p2 of o/n 2014 v22 :((

    ReplyDelete
    Replies
    1. Question 2 is explained as solution 527 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-102.html

      Specify the specific doubts you have and I'll try to answer. I'm not explaining complete papers right now.

      Delete
  10. Thank you so much for your help, I have a doubt in question 41 shouldn't the graph be like this(link below), corresponding with each point?

    The link-
    http://postimg.org/image/q2hrljpln/

    Like the lower portion of the graph shouldn't be shown as there is no distance before 0 given for original wave?

    Thanks,

    ReplyDelete
    Replies
    1. It is the WAVE on the string that moves forwards. The string will still there. Remember that with a wave, energy is transferred without any movement of the material.

      The particles on the string are still there, and they move up and down. So, the part before 20cm will still be part of the wave as I drawn above.

      Delete
  11. For question 41, c (I), does this mean that if one point X (which is above equilibrium), if X was compared with another point that was below the equilibrium position, then they would be said to be out-of-phase? And so, are there only 2 possible phase differences on stationary waves? (i.e. 0 or 180)

    ReplyDelete
  12. For solution 41 I don't understand why 10ms is considered t/4 shouldn't it be t/2 because t is 20ms so t/2 should be 10ms?

    ReplyDelete
    Replies
    1. Sorry, but I cannot locate where they are having problems. Could you be more specific? Thanks

      where is it mentioned that t/4 = 10ms???
      what is t?

      Delete

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