### Welcome to Talha's Physics Academy

To Help Teachers and Students.

Talha's Physics Academy is an exploration environment for concepts in physics which employs free Physics Books and other linking strategies to facilitate smooth navigationThe entire environment is interconnected with thousands of links, reminiscent of a neural network.

New content for Talha's Physics Academy will be posted as it is developed,It is my intent to keep this material continuously available except for brief maintenance times.

All the Branches of Physics are covered.

Share with others.

### Projectile Motion : Practice Problems + Solutions

Solve the following questions using what you know about projectile motion.

1.      A roadrunner runs directly off a cliff with an initial velocity of 3.5 m/s.
a)      What are the components of this velocity?
Vx = 3.5 m/s                            Vy = 0 m/s

b)      What will be the horizontal velocity 2 seconds after the bird leaves the cliff?
3.5 m/s – horizontal velocity is unchanging

c)      If the cliff is 300 m high, at what time will the roadrunner reach the ground?
h = dy = ½ * 10 * t2 = 300
300 * 2 / 10 = t2 = 60
t = 7.75 s

d)     How far from the cliff will this bird land?
dx = 3.5 * 7.75 = 27.125 m

e)      If there is a small pond which begins 25m away from the cliff and extends 2.5 meters from there; will the roadrunner land in the pond?
Yes, the pond is from 25 m to 27.5 m, so the roadrunner will land in the pond.

f)       What is the final vertical velocity at which the roadrunner is traveling? [The vertical velocity at the time when the bird reaches the ground]
Vy = 10 * 7.75 + 0 = 77.5 m/s

g)      What is the final horizontal velocity at which the roadrunner is traveling? [The horizontal velocity at the time when the bird reaches the ground]
Vx = 0 + 3.5 = 3.5 m/s

h)      What is the total final velocity of this motion? [magnitude and direction]
V2 = 77.52 + 3.52 = 6018.5
V = 77.579 m/s
q = tan-1 (77.5 / 3.5) = 87.41o below the horizontal

2.      An object (any object) is dropped from a height of 300m
a)      How long does it take this object to fall to the ground?
h = dy = ½ * 10 * t2 = 300
300 * 2 / 10 = t2 = 60
t = 7.75 s

b)      Compare this answer with you answer from question 1, part c). What are the reasons for any similarities or differences?
They are the same. This is because their vertical motions are identical. All objects fall with the same gravitational acceleration, so two objects at the same height with the same initial vertical velocity will reach the ground at the same time.

3.      The intent of a bean bag toss game is to get your bean bag to land on the ‘bull’s-eye’ of a target. The target is set up parallel to the ground and is the same height above the ground as your hand is when you let go of the bean bag. The game’s rules further require you to be 5 m from the center of the target when you release the bag.
a)      Evaluate the following questions for both an angle of 32o and an angle of 58o if the bean bag is thrown with an initial velocity of 6 m/s:
32o                                                                   58o
i.        What are the components of velocity?
Vx32: Cos 32o = Vx32/6                                                Vx58: Cos 58o = Vx58/6
6 * Cos 32o = Vx32 = 5.09 m/s                                    6 * Cos 58o = Vx58 = 3.18 m/s

Vy32: Sin 32o = Vy32/6                                                 Vy58: Sin 58o = Vy58/6
6 * Sin 32o = Vy32 = 3.18 m/s                                     6 * Sin 58o = Vy58 = 5.09 m/s

ii.      What is the maximum height of the bean bag’s motion?
tTOP32 = Vy/10 = 3.18/10 = 0.318s                   tTOP58 = Vy58/10 = 5.09/10 = 0.509s
hMAX32 = ½ * 10 * 0.3182 = 0.506 m               hMAX58 = ½ * 10 * 0.5092 = 1.295 m

iii.    How long will the bean bag be in the air?
tTOTAL32 = 2 * tTOP32 = 0.636 s                        tTOTAL58 = 2 * tTOP58 = 1.018 s

iv.    How far away from you will the bag land?
dx32 = 5.09 * 0.636 =   3.24 m                         dx58 = 3.18 * 1.018 = 3.24m

v.      If the center of the bull’s-eye ranges from 4.9 m to 5.1 m away from you, does your bean bag win?
No                                                                   No

4.      A stunt driver drives a red mustang convertible up a ramp and off a cliff. The car leaves the ramp at a velocity of 60 m/s at an angle of 45o to the horizontal; the cliff and ramp combined cause the car to begin its projectile motion at a height of 315m above the ground. If you were coordinating this stunt, how far away would you put a landing surface so that your stunt driver was not injured?

First let’s think about strategy. The question is basically asking how far away from the cliff the car will land. In order to find horizontal distance we need horizontal velocity and time. We can find both horizontal and vertical velocity from the initial conditions, but we’ll have to calculate the time it will take for the car to reach the ground. So first we’ll find the components of velocity:

Vx: Cos 45o = Vx/60                                       Vy: Sin 45o = Vy/60
60 * Cos 45o = Vx = 42.43 m/s                       60 * Sin 45o = Vy = 42.43 m/s

Note: Remember that horizontal velocity is constant, but the vertical velocity we calculated above is only the initial vertical velocity.
From here we need to use the initial vertical velocity to find the time it takes the car to reach the top of its path and fall to the ground. Let’s think about this in two parts; the time it takes to reach hMAX first:

tTOP = 42.43/10 = 4.243 s

Now what about the time it takes to fall from the maximum height? Well first we need to know the maximum height:

hTOP = ½ * 10 * 4.2432 = 90.02 m

hMAX = hTOP + ho =  90.02 + 315 = 405.02 m

Now we calculate the time it takes to fall from a height of 405.02 m:

405.02 = ½ * 10 * tDOWN2

tDOWN2 = 81.004

tDOWN = 9.000 s

Putting these two times together, we have the total time it takes the car to travel up to its maximum height and then fall back down. This is the total time in the air and this is the time we will want to use to solve for horizontal distance.

tTOTAL = tTOP + tDOWN = 4.243 + 9.000 = 13.243 s

dx = 42.43 * 13.243 = 561.9 m

You will want to make sure that the landing surface is centered 561.9 m from the base of the cliff.

### Chapter 6: Work and Energy

Practice Problems
 1
Just as a car tops a 38 meter high hill with a speed of 80 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill?

 2
A pendulum has a mass of 3.6 kg, a length of 1.7 meters, and swings through a (half)arc of 29.4 degrees. What is its amplitude to the nearest centimeter?

 3
To the nearest tenth of a Joule, what is its maximum kinetic energy of the pendulum in problem 2?

 4
To the nearest tenth of a Joule, what is the total energy of the pendulum in problem 2?

 5
A 2 kg metal plate slides down a 13-meter high slope. At the bottom its speed is 9 m/s. To the nearest Joule, what was the magnitude of the work done by friction?

 6
If the slope in the above problem is 23 degrees, what is the coefficent of friction (to 2 decimal places)?

 7
An unstretched spring with spring constant 36 N/cm is suspended from the ceiling. A 3.6 kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?

 8
If the mass in the previous problem is attached to the spring and slowly let down, to the nearest tenth of a centimeter, how far does it stretch the spring?

 9
A mass of 2.2 kg is dropped from a height of 4.7 meters above a vertical spring anchored at its lower end to the floor. If the spring constant is 33 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?

 10
If the top of the spring in the preceding problem is 1.34 meters above the ground when the mass is released, what is the ball's kinetic energy, to the nearest Joule, just before the mass strikes the spring?

Problem 1  The correct answer is 63.
Problem 2  The correct answer is 87.
Problem 3 The correct answer is 7.7.
Problem 4  The correct answer is 7.7.
Problem 5  The correct answer is 174.
Problem 6  The correct answer is 0.29.
Problem 7  The correct answer is 2.
Problem 8  The correct answer is 1.
Problem 9  The correct answer is 25.4.
Problem 10  The correct answer is 101.

### Cosmic Rays

High energy electrons, protons, and complex nuclei can be produced in a number of astronomical environments. Such particles travel throughout the universe and are called cosmic rays. Some of these particles reach our Earth. As these objects hit our atmosphere, other particles called pions and muons are produced. These particles then slow down or crash into other atoms in the atmosphere. Since the atmosphere slows down these particles, the higher we travel, the more cosmic radiation we see. When you visit the mountains or take an airplane ride, you will encounter more cosmic radiation than if you stayed at sea level.
Most cosmic radiation is very energetic. It can easily pass through an inch of lead. Since cosmic radiation can cause genetic changes, some scientists believe that this radiation has been important in driving the evolution of life on our planet. While cosmic radiation can cause some damage to individuals, it also has played an important role in creating humans. Our atmosphere is naturally shielding us from harmful effects. However, if we were to leave the earth and travel to some planet, we could be subjected to very high levels of radiation. Future space travelers will have to find some way to minimize exposure to cosmic rays.

### Antimatter

In 1930, Paul Dirac developed the first description of the electron that was consistent with both quantum mechanics and special relativity. One of the remarkable predictions of this theory was that an anti-particle of the electron should exist. This antielectron would be expected to have the same mass as the electron, but opposite electric charge and magnetic moment. In 1932, Carl Anderson, was examining tracks produced by cosmic rays in a cloud chamber. One particle made a track like an electron, but the curvature of its path in the magnetic field showed that it was positively charged. He named this positive electron a positron. We know that the particle Anderson detected was the anti-electron predicted by Dirac. In the 1950's, physicists at the Lawrence Radiation Laboratory used the Bevatron accelerator to produce the anti-proton, that is a particle with the same mass and spin as the proton, but with negative charge and opposite magnetic moment to that of the proton. In order to create the anti-proton, protons were accelerated to very high energy and then smashed into a target containing other protons. Occasionally, the energy brought into the collision would produce a proton-antiproton pair in addition to the original two protons. This result gave credibility to the idea that for every particle there is a corresponding antiparticle.
A particle and its antimatter particle annihilate when they meet: they disappear and their kinetic plus rest-mass energy is converted into other particles (E = mc2). For example, when an electron and a positron annihilate at rest, two gamma rays, each with energy 511 keV, are produced. These gamma rays go off in opposite directions because both energy and momentum must be conserved. The annihilation of positrons and electrons is the basis of Positron Emission Tomography (PET) discussed in the section on Applications (Chapter 14). When a proton and an antiproton annihilate at rest, other particles are usually produced, but the total kinetic plus rest mass energies of these products adds up to twice the rest mass energy of the proton (2 x 938 MeV).
Antimatter is also produced in some radioactive decays. When 14C decays, a neutron decays to a proton plus an electron and an electron antineutrino, . When 19Ne decays, a proton decays to a neutron plus a positron, e+, and an electron neutrino, .
14C --> 14N + e- + 19Ne --> 19F + e+ +
The neutrino and electron are leptons while the antineutrino and positron are anti-leptons. Leptons are point-like particles that interact with the electromagnetic, weak and gravitational interaction, but not the strong interaction. An antilepton is an antiparticle. In each reaction, one lepton and one antilepton is produced. These processes show a fundamental law of physics - that for each new lepton that is produced there is a corresponding new antilepton.
Although from a distance matter and antimatter would look essentially identical, there appears to be very little antimatter in our universe. This conclusion is partly based on the low observed abundance of antimatter in the cosmic rays, which are particles that constantly rain down on us from outer space. All of the antimatter present in the cosmic rays can be accounted for by radioactive decays or by nuclear reactions involving ordinary matter like those described above. We also do not see the signatures of electron-positron annihilation, or proton-proton annihilation coming from the edges of galaxies, or from places where two galaxies are near each other. As a result, we believe that essentially all of the objects we see in the universe are made of matter not antimatter.
Elementary particle physicists create massive particles by accelerating lower mass particles close to the speed of light, and then smashing them together. The mass/energy of the colliding particles becomes the mass of the created particles. One method includes taking positrons and electrons, accelerating both of them, and smashing them into each other. Out of this energy, very massive particles such as quarks, tau-particles, and the Z0 can be created. Studies of such electron-positron annihilations are carried out at the Stanford Linear Accelerator and at the LEP facility at CERN. A similar technique is used at the Fermi National Accelerator Laboratory except that it involves colliding protons with anti-protons. Collisions of this kind were recently used to produce the sixth type of quark, known as the top. This particle has a rest mass energy of approximately 160,000 MeV, which is nearly the same as the mass of nucleus of a gold atom!
Atoms of anti-hydrogen, which consist of a positron orbiting an antiproton, are believed to have been created in 1995 at the CERN laboratory in Europe. Physicists are now searching for very small differences between the properties of matter atoms and antimatter atoms. This will help confirm or confound our understanding of the symmetry between matter and anti-matter.

### The Tao of Physics - F. Capra

Contents:
I THE WAY OF PHYSICS
1 Modern Physics-A Path with a Heart? 17
2 Knowing and Seeing 26
3 Beyond Language 45
4 The New Physics 52
II THE WAY OF EASTERN MYSTICISM
5 Hinduism 85
6 Buddhism 93
7 Chinese Thought 101
8 Taoism 113
9 Z e n 121
III THE PARALLELS
10 The Unity of All Things 130
11 Beyond the World of Opposites 145
12 Space-Time 161
13 The Dynamic Universe 189
1 4 Emptiness and Form 207
15 The Cosmic Dance 225
16 Quark Symmetries-A New Koan! 247
17 Patterns of Change 261
18 Interpenetration 285

### Chapter #10 :Geometrical Optics : Short Q/A / C.R.Q's

Q.Why do thick lenses possess chromatic and spherical aberrations? Suggest remedies for the rectification of these defects.
Ans.Chromatic aberration occurs because lenses have a different refractive index for different wavelengths of light (the dispersion of the lens) and spherical aberration is an optical effect observed in an optical device that occurs due to the increased refraction of light rays when they strike a lens or a reflection of light rays when they strike a mirror near its edge, in comparison with those that strike nearer the centre.
Chromatic aberration was reduced by increasing the focal length of the lens where possible.

Q. Does the chromatic aberration takes plane in mirror?
Ans. In optics, chromatic aberration (chromatic distortion) is a type of distortion in which there is a failure of a lens to focus all colors to the same convergence point. It occurs because lenses have a different refractive index for different wavelengths of light (the dispersion of the lens).But A/c to law of reflection” The angle of incidence is always equal to law of refraction”. So there is no chance of chromatic aberration in mirror

### Chapter #9 :Nature of Light ( Physical Optics) : Short Q/A / C.R.Q's

Q.Why the central spot in the Newton’s rings is dark?
Ans. The Newton’s rings are formed due to the phenomenon of thin film interference. here, the condition for constructive interference(the ring appearing bright) is that the optical path difference between interfering waves should be an integral multiple of the wavelength. the optical path difference is given by 2t-(l/2) if t is the thickness of the air film at that point and l is the wavelength of light. At the central point, the lens touches the surface so thickness t=0. thus the optical path difference is simply l/2, which is the condition for destructive interference, not constructive interference. so the central spot has to always be dark.

Q. Why the dark and bright fringes of Newton’s experiment are circular?
Ans. The Newton’s rings are formed due to the phenomenon of thin film interference. here, the condition for constructive interference(the ring appearing bright) is that the optical path difference between interfering waves should be an integral multiple of the wavelength. As the Plano convex lens is used in Newton’s rings so the thickness of the film is increasing and then decreasing that’s why the fringes are circular.

Q.24 Why the conditions of constructive and destructive interference are reversed in thin films?
Ans. Reflected light will experience a 180 degree phase change when it reflects from a medium of higher index of refraction and no phase change when it reflects from a medium of smaller index. This phase change is important in the interference which occurs in thin films, the design of anti-reflection coatings, interference filters, and thin film mirrors. So in thing film the phase change of 180 degrees occur that’s why the crests converts into troughs and troughs are converted into crests. therefore the conditions are reversed.

Q. An oil film over a wet footpath shows colors? Explain how does it happen?
Ans.This is known as thin-film interference, because it is the interference of light waves reflecting off the top surface of a film with the waves reflecting from the bottom surface. To obtain a nice colored pattern, the thickness of the film has to be on the order of the wavelength of light. Consider the case of a thin film of oil floating on water. Thin-film interference can take place if these two light waves interfere constructively:
1. the light from the air reflecting off the top surface
2. the light traveling from the air, through the oil, reflecting off the bottom surface, traveling back through the oil and out into the air again.

Q. Why Polaroid sun glasses are better than the ordinary sun glasses?

Ans. There are four things that a good pair of Polaroid sunglasses should do for you:

Sunglasses provide protection from ultraviolet rays in sunlight. Ultraviolet (UV) light damages the cornea and the retina. Good sunglasses can eliminate UV rays completely.

Sunglasses provide protection from intense light. When the eye receives too much light, it naturally closes the iris. Once it has closed the iris as far as it can, the next step is squinting. If there is still too much light, as there can be when sunlight is reflecting off of snow, the result is damage to the retina. Good sunglasses can block light entering the eyes by as much as 97 percent to avoid damage.

Sunglasses provide protection from glare. Certain surfaces, such as water, can reflect a great deal of light, and the bright spots can be distracting or can hide objects. Good sunglasses can completely eliminate this kind of glare using polarization (we'll discuss polarization later).

Sunglasses eliminate specific frequencies of light. Certain frequencies of light can blur vision, and others can enhance contrast. Choosing the right color for your sunglasses lets them work better in specific situations.

Q. Can interference be without diffraction or vise versa?
Ans. You can  have diffraction without interference. Interference occurs when coherent light waves coming from two different sources interact. In single-slit diffraction, the two sides of the slit act as these two sources. If you make the slit much smaller than the wavelength of whatever you're diffracting, though, it effectively becomes a single point source, and no appreciable interference occurs and interference in the thin films is without diffraction.