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 2Ï€af, 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 = 2Ï€af
= 2Ï€ (2)(0.16) = 0.64Ï€ ms-1
Maximum kinetic energy = ½ mvmax2 = 0.5 (2.0 × 10–3) (0.64Ï€)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.
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:
Go to
A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface.
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.
In the formula, v= IR, is the voltage the potential DIFFERENCE?
ReplyDeleteyes. Ohm's law always give the potential difference.
Deletecan you solve june 2007 paper 2 first question? the calibrating
ReplyDeleteit's explained as solution 556 at
Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-109.html
Q 436, speed of vibrating particles is different from that of that of the wave, right?
ReplyDeleteYes, they are different.
DeleteAt 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.
A lot of people recommended this blog and I found it wonderful.
ReplyDeleteSince 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.
For O/N 07 P1 Q35, check solution 770 at
Deletehttp://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
Can you please explain these questions?
ReplyDeleteMJ 2005_P1_Q24 and Q29
MJ_2006_P1_Q24
Thank you.
For MJ_2006_P1_Q24, see solution 214 at
Deletehttp://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-34.html
Can you please explain these questions?
ReplyDeleteM/J/09 p01 Q2 and O/N/10 p12 Q3.
Thanks.
For M/J/09 p01 Q2, see solution 791 at
Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-159.html
Can you please explain these questions?
ReplyDeleteO/N/10 p12 Q3 and M/J/09 p01 Q2.
Thanks.
For O/N/10 p12 Q3, go to
Deletehttp://physics-ref.blogspot.com/2014/11/9702-november-2010-paper-12-worked.html
june 2006 q6 please
ReplyDeletewhich paper?
DeletePlease answer a query:
ReplyDeleteQuestion 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?
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.
Deletein solution 435 i don't understand how new amplitude is 1.4A please show full calculation.
ReplyDeleteYou need to understand proportionality.
DeleteIntensity 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
hi can u explain Q437 why is it 5Ns?
ReplyDeletewe consider by how much the momentum of sphere A changes. Sphere B should also change by the same amount.
DeleteSO, by adding this amount to the original momentum of sphere B, it becomes 5 Ns
For the solid rubber ball question, shouldnt we take height as 0.76m? Bcoz gravity if considered to be at the centre of a mass right? So why did u take 72?
ReplyDeletethis is explained in one of the comments above, as a response to Muhammad Mooneeb Hussain.
DeleteSee if you understand.
Thnks
Hi, for Solution 438, if the KE is converted to GPE after rebound, then why is KE 0.385J (from what I see it is the GPE gained)? Shouldn't KE be 0.75J-0.385J?
ReplyDeleteThanks.
the explanation above has been updated. See if it helps
Delete