# 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 180

^{o}.

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 180

^{o})
{Intensity I is proportional
to A

^{2}. 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 v

_{max}= 2πaf = 2π (2)(0.16) = 0.64π ms^{-1}Maximum kinetic energy = ½ mv

_{max}

^{2}= 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 F

_{A}and average force acting sphere B is F_{B}.
Forces acts for time t

_{A}on sphere A and time t_{B}on sphere B.
(i) State relationship between

1. F

_{A}and F_{B}
2. t

_{A}and t_{B}
(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. F

_{A}= - F_{B}
2. t

_{A}= t_{B}
(ii) Change in momentum, Δp = F

_{A}t_{A}= - F_{B}t_{B}**(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.**

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