Thursday, March 12, 2015

Physics 9702 Doubts | Help Page 83

  • Physics 9702 Doubts | Help Page 83



Question 435: [Waves]
Fig shows variation with distance x along a wave of its displacement d at a particular time. 

Wave is a progressive wave having speed of 330ms-1.
(a)
(i) Use Fig to determine wavelength of the wave:
(ii) Hence calculate frequency of the wave:

(b) Second wave has same frequency and speed as wave shown in Fig but has double the intensity. Phase difference between the 2 waves is 180o.
On axes of Fig, sketch graph to show variation with distance x of displacement d of this second wave:

Reference: Past Exam Paper – June 2004 Paper 2 Q2



Solution 435:
(a)
(i) Wavelength, λ = 0.6m
(ii) Frequency, f (= v / λ) = 330 / 0.60 = 550Hz

(b)
The amplitude is shown as greater than a but less than 2a and constant. It should have a correct phase (graph is inverted - phase difference of 180o)
{Intensity I is proportional to A2. If intensity I is doubled, the new amplitude would become 2 A 1.4A}











Question 436: [Waves]
Graph represents a sinusoidal wave in the sea, travelling at speed of 8.0 m s–1, at one instant of time.
Maximum speed of oscillating particles in the wave is 2πaf, where a is the amplitude and f is the frequency.

Object P of mass 2.0 × 10–3 kg floats on the surface.
What is maximum kinetic energy of P due to wave? Assume that its motion is vertical.
A 0.026 mJ                  B 4.0 mJ                      C 39 mJ                       D 64 mJ

Reference: Past Exam Paper – November 2007 Paper 1 Q23



Solution 436:
Answer: B.
Speed of a wave: v = fλ

From the graph, the wavelength λ = 50m.
Frequency f = v / λ = 8 / 50 = 0.16Hz

From the graph, the amplitude a = 2m.
Maximum speed vmax = 2πaf = 2π (2)(0.16) = 0.64π ms-1

Maximum kinetic energy = ½ mvmax2 = 0.5 (2.0 × 10–3) (0.64π)2 = 0.0040 = 4.0mJ











Question 437: [Dynamics > Newton’s laws of motion]
(a)
(i) Define force.
(ii) State Newton’s third law of motion.

(b) 2 spheres approach one another along line joining their centres, as illustrated. 

When they collide, average force acting on sphere A is FA and average force acting sphere B is FB.
Forces acts for time tA on sphere A and time tB on sphere B.
(i) State relationship between
1. FA and FB
2. tA and tB

(ii) Use answers in (i) to show that change in momentum of sphere A is equal in magnitude and opposite in direction to change in momentum of sphere B.

(c) For spheres in (b), variation with time of momentum of sphere A before, during and after the collision with sphere B is shown. 

Momentum of sphere B before collision also shown.
Complete Fig to show variation with time of momentum of sphere B during and after collision with sphere A.

Reference: Past Exam Paper – June 2010 Paper 22 Q3



Solution 437:
(a)
(i) Force is defined as the rate of change of momentum.
(ii) Newton’s third law of motion states that the force on a body A is equal in magnitude to the force on a body B (from body A). The forces are in opposite directions and are of the same kind.

(b)
(i)
1. FA = - FB
2. tA = tB

(ii) Change in momentum, Δp = FAtA = - FBtB

(c)
For the graph, the momentum change occurs at the same time for both spheres. The final momentum of sphere B is to the right and of magnitude 5Ns.











Question 438: [Energy]
Solid rubber ball has diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface. It falls vertically and then bounces back up so that maximum height reached by the top of the ball is 45 cm, as shown.

If kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy just after it leaves the surface?
A 0.36 J                       B 0.39 J                       C 0.40 J                       D 0.42 J

Reference: Past Exam Paper – June 2013 Paper 11 Q17



Solution 438:
Answer: B.
This is a tricky question. It is necessary to deal with the bottom of the ball at 72 cm at the start and 37 cm at the end.

From the conservation of energy, as the ball falls, its potential energy is being converted to kinetic energy.

Just before it strikes the surface, the kinetic energy is 0.75J (all potential energy has been converted to KE).
mgh = KE
m (9.81) (0.72) = 0.75
Mass m of the ball = 0.11kg

Now, for the rebound, the all of the kinetic energy just after the ball leaves the surface is converted to potential energy at the maximum height reached.
KE = 0.11 (9.81) (0.37) = 0.3854J.

Alternatively, this could also be solved without actually calculating the mass. Before striking the surface, a height of 72cm results in a kinetic energy of 0.75J. After striking the surface, the kinetic energy should also be proportional to the new maximum height reached. This gives (37/72) × 0.75 J = 0.3854 J.









Question 439: [Current of Electricity]
Circuit is set up with LDR and a fixed resistor as shown.

Voltmeter reads 4 V.
Light intensity is increased.
What is a possible voltmeter reading?
A 3 V                         B 4 V                         C 6 V                          D 8 V

Reference: Past Exam Paper – June 2007 Paper 1 Q34



Solution 439:
Answer: A.
The resistance of an LDR decreases as the light intensity incident on it increases.

From Ohm’s law: V = IR
The p.d. across a component is dependent on its resistance. Since the resistance of the LDR decreases, the p.d. across it will also decrease.




20 comments:

  1. In the formula, v= IR, is the voltage the potential DIFFERENCE?

    ReplyDelete
    Replies
    1. yes. Ohm's law always give the potential difference.

      Delete
  2. can you solve june 2007 paper 2 first question? the calibrating

    ReplyDelete
    Replies
    1. it's explained as solution 556 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-109.html

      Delete
  3. Q 436, speed of vibrating particles is different from that of that of the wave, right?

    ReplyDelete
    Replies
    1. Yes, they are different.
      At some positions, the speed of a particle is zero, and at other position it may be maximum, ...

      We do not usually use this speed.

      We use the speed of propagation of the wave.

      Delete
  4. A lot of people recommended this blog and I found it wonderful.
    Since the solutions of all papers are not available, I need help on these questions. It would be great if you can take out some time to help me. Thank you.

    O/N 07 P1 Q35, MJ 08 P1 Q30, ON 08 P1 Q32.

    ReplyDelete
    Replies
    1. For O/N 07 P1 Q35, check solution 770 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-155.html

      For MJ 08 P1 Q30, check solution 579 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-114.html

      For ON 08 P1 Q32, check solution 14 at
      http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-2.html

      Delete
  5. Can you please explain these questions?
    MJ 2005_P1_Q24 and Q29
    MJ_2006_P1_Q24
    Thank you.

    ReplyDelete
    Replies
    1. For MJ_2006_P1_Q24, see solution 214 at
      http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-34.html

      Delete
  6. Can you please explain these questions?
    M/J/09 p01 Q2 and O/N/10 p12 Q3.
    Thanks.

    ReplyDelete
    Replies
    1. For M/J/09 p01 Q2, see solution 791 at
      http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-159.html

      Delete
  7. Can you please explain these questions?
    O/N/10 p12 Q3 and M/J/09 p01 Q2.
    Thanks.

    ReplyDelete
    Replies
    1. For O/N/10 p12 Q3, go to
      http://physics-ref.blogspot.com/2014/11/9702-november-2010-paper-12-worked.html

      Delete
  8. Muhammad Mooneeb HussainDecember 30, 2015 at 3:30 PM

    Please answer a query:
    Question 438:
    "It is necessary to look at the bottom of the ball."

    Why do we need to take in consideration the bottom of the ball?

    ReplyDelete
    Replies
    1. We are given the KE just before it strikes the surface. At this position, if we consider in terms of points on the ball, the bottom of the ball has zero PE – all its potential energy has been converted to KE. However, the top of the ball is still at some distance above the ground, so it has some PE. But we are given the KE just before it strikes the ground, so to use it in calculations, we need to consider the bottom of the ball.

      Delete
  9. in solution 435 i don't understand how new amplitude is 1.4A please show full calculation.

    ReplyDelete
    Replies
    1. You need to understand proportionality.

      Intensity I is proportional to A2. So, A is proportional to √(I). Let’s say √I = A

      If intensity I is doubled (I becomes 2I), the new amplitude would become √(2I) = √2 √I = √2 A ≈ 1.4 A

      Delete

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