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Nuclear Chemistry,Nuclear fission, Nuclear fusion

noreply@blogger.com (Study Physics) — Sat, 16 Jul 2011 15:48:00 +0000

Nuclear Chemistry

An Introduction

Traditional chemical reactions occur as a result of the interaction betweenvalence electrons around an atom's nucleus (see our Chemical Reactionsmodule for more information). In 1896, Henri Becquerel expanded the field of chemistry to include nuclear changes when he discovered that uranium emitted radiation. Soon after Becquerel's discovery, Marie Sklodowska Curie began studying radioactivity and completed much of the pioneering work on nuclear changes. Curie found that radiation was proportional to the amount of radioactive element present, and she proposed that radiation was a property of atoms (as opposed to a chemical property of a compound).Marie Curie was the first woman to win a Nobel Prize and the first person to win two (the first, shared with her husband Pierre and Becquerel for discovering radioactivity; the second for discovering the radioactive elements radium and polonium).

Radiation and nuclear reactions

In 1902, Frederick Soddy proposed the theory that "radioactivity is the result of a natural change of an isotope of one element into an isotope of a different element." Nuclear reactions involve changes in particles in an atom's nucleus and thus cause a change in the atom itself. All elements heavier than bismuth (Bi) (and some lighter) exhibit natural radioactivity and thus can "decay" into lighter elements. Unlike normal chemical reactions that form molecules, nuclear reactions result in the transmutation of one element into a different isotope or a different element altogether (remember that the number of protons in an atom defines the element, so a change in protons results in a change in the atom). There are three common types of radiation and nuclear changes:

Alpha Radiation (α) is the emission of an alpha particle from an atom'snucleus. An α particle contains two protons and two neutrons (and is similar to a He nucleus: ). When an atom emits an a particle, the atom's atomic mass will decrease by four units (because two protons and two neutrons are lost) and the atomic number (z) will decrease by two units. The element is said to "transmute" into another element that is two z units smaller. An example of an a transmutation takes place when uranium decays into the element thorium (Th) by emitting an alpha particle, as depicted in the following equation:

238
92

4
2

234
90

Beta Radiation (β) is the transmutation of a neutron into a proton and an electron (followed by the emission of the electron from the atom's nucleus: ). When an atom emits a β particle, the atom's mass will not change (since there is no change in the total number of nuclear particles), however the atomic number will increase by one (because the neutron transmutated into an additional proton). An example of this is the decay of the isotope of carbon named carbon-14 into the element nitrogen:

14
6

0
-1

14
7

Gamma Radiation (γ) involves the emission of electromagnetic energy(similar to light energy) from an atom's nucleus. No particles are emitted during gamma radiation, and thus gamma radiation does not itself cause the transmutation of atoms, however γ radiation is often emitted during, and simultaneous to, α or β radioactive decay. X-rays, emitted during the beta decay of cobalt-60, are a common example of gamma radiation.

Half-life

Radioactive decay proceeds according to a principal called the half-life. The half-life (T_½) is the amount of time necessary for one-half of the radioactive material to decay. For example, the radioactive element bismuth (210Bi) can undergo alpha decay to form the element thallium (206Tl) with a reaction half-life equal to five days. If we begin an experiment starting with 100 g of bismuth in a sealed lead container, after five days we will have 50 g of bismuth and 50 g of thallium in the jar. After another five days (ten from the starting point), one-half of the remaining bismuth will decay and we will be left with 25 g of bismuth and 75 g of thallium in the jar. As illustrated, the reaction proceeds in halfs, with half of whatever is left of the radioactive element decaying every half-life period.

Radioactive Decay of Bismuth-210 (T_½ = 5 days)

The fraction of parent material that remains after radioactive decay can be calculated using the equation:

Fraction remaining =

1
2ⁿ

(where n = # half-lives elapsed)

The amount of a radioactive material that remains after a given number of half-lives is therefore:

Amount remaining = Original amount * Fraction remaining

The decay reaction and T_½ of a substance are specific to the isotope of the element undergoing radioactive decay. For example, Bi210 can undergoa decay to Tl206 with a T_½ of five days. Bi215, by comparison, undergoes bdecay to Po215 with a T_½ of 7.6 minutes, and Bi208 undergoes yet another mode of radioactive decay (called electron capture) with a T_½ of 368,000 years!

Stimulated nuclear reactions

While many elements undergo radioactive decay naturally, nuclear reactions can also be stimulated artificially. Although these reactions also occur naturally, we are most familiar with them as stimulated reactions. There are two such types of nuclear reactions:

1. Nuclear fission: reactions in which an atom's nucleus splits into smaller parts, releasing a large amount of energy in the process. Most commonly this is done by "firing" a neutron at the nucleus of an atom. The energy of the neutron "bullet" causes the target element to split into two (or more) elements that are lighter than the parent atom.

The Fission Reaction of Uranium-235

The Fission of U235

Concept simulation - Illustrates a nuclear fission reaction.

(Flash required)

During the fission of U235, three neutrons are released in addition to the two daughter atoms. If these released neutrons collide with nearby U235 nuclei, they can stimulate the fission of these atoms and start a self-sustaining nuclear chain reaction. This chain reaction is the basis of nuclear power. As uranium atoms continue to split, a significant amount of energy is released from the reaction. The heat released during this reaction is harvested and used to generate electrical energy.

Two Types of Nuclear Chain Reactions

Concept simulation - Reenacts controlled and uncontrolled nuclear chain reactions.

2. Nuclear fusion: reactions in which two or more elements "fuse" together to form one larger element, releasing energy in the process. A good example is the fusion of two "heavy" isotopes of hydrogen (deuterium: H2 and tritium: H3) into the element helium.

Nuclear Fusion of Two Hydrogen Isotopes

Nuclear Fusion

Concept simulation - Reenacts the fusion of deuterium and tritium inside of a tokamak reactor.

Fusion reactions release tremendous amounts of energy and are commonly referred to as thermonuclear reactions. Although many people think of the sun as a large fireball, the sun (and all stars) are actually enormous fusion reactors. Stars are primarily gigantic balls of hydrogen gas under tremendous pressure due to gravitational forces. Hydrogen molecules are fused into helium and heavier elements inside of stars, releasing energy that we receive as light and heat.

Atomic Theory II

Ions, Isotopes and Electron Shells

by Anthony Carpi, Ph.D.

In Atomic Theory I: The Early Days (see our Atomic Theory I module), we learned about the basic structure of the atom. Normally, atoms contain equal numbers of protons and electrons. Because the positive and negative charges cancel each other out, atoms are normally electrically neutral. But, while the number of protons is always constant in any atom of a givenelement, the number of electrons can vary.

Ions

When the number of electrons changes in an atom, the electrical chargechanges. If an atom gains electrons, it picks up an imbalance of negatively charged particles and therefore becomes negative. If an atom loses electrons, the balance between positive and negative charges is shifted in the opposite direction and the atom becomes positive. In either case, the magnitude (+1, +2, -1, -2, etc.) of the electrical charge will correspond to the number of electrons gained or lost. Atoms that carry electrical charges are called ions (regardless of whether they are positive or negative). For example, the animation below shows a positive hydrogen ion (which has lost an electron) and a negative hydrogen ion (which has gained an extra electron). The electrical charge on the ion is always written as a superscript after the atom's symbol, as seen in the animation.

Hydrogen Ion Simulation

Isotopes

The number of neutrons in an atom can also vary. Two atoms of the sameelement that contain different numbers of neutrons are called isotopes. For example, normally hydrogen contains no neutrons. An isotope of hydrogen does exist that contains one neutron (commonly called deuterium). Theatomic number (z) is the same in both isotopes; however the atomic massincreases by one in deuterium as the atom is made heavier by the extra neutron.

Hydrogen Isotope Simulation

Electron Shells

Ernest Rutherford's view of the atom consisted of a dense nucleussurrounded by freely spinning electrons (see our Atomic Theory I module). In 1913, the Danish physicist Niels Bohr proposed yet another modification to the theory of atomic structure based on a curious phenomenon called line spectra.

When matter is heated, it gives off light. For example, turning on an ordinary light bulb causes an electric current to flow through a metal filament that heats the filament and produces light. The electrical energyabsorbed by the filament excites the atoms' electrons, causing them to "wiggle". This absorbed energy is eventually released from the atoms in the form of light.

When normal white light, such as that from the sun, is passed through a prism, the light separates into a continuous spectrum of colors:

Figure 1: Continuous (white light) spectra

Bohr knew that when pure elements were excited by heat or electricity, they gave off distinct colors rather than white light. This phenomenon is most commonly seen in modern-day neon lights, tubes filled with gaseous elements (most commonly neon). When an electric current is passed through the gas, a distinct color (most commonly red) is given off by the element. When light from an excited element is passed through a prism, only specific lines (or wavelengths) of light can be seen. These lines of light are called line spectra. For example, when hydrogen is heated and the light is passed through a prism, the following line spectra can be seen:

Figure 2: Hydrogen line spectra

Each element has its own distinct line spectra. For example:

Figure 3: Helium line spectra

Figure 4: Neon line spectra

To Bohr, the line spectra phenomenon showed that atoms could not emitenergy continuously, but only in very precise quantities (he described the energy emitted as quantized). Because the emitted light was due to the movement of electrons, Bohr suggested that electrons could not move continuously in the atom (as Rutherford had suggested) but only in precise steps. Bohr hypothesized that electrons occupy specific energy levels. When an atom is excited, such as during heating, electrons can jump to higher levels. When the electrons fall back to lower energy levels, precise quanta of energy are released as specific wavelengths (lines) of light.

Under Bohr's theory, an electron's energy levels (also called electron shells) can be imagined as concentric circles around the nucleus. Normally, electrons exist in the ground state, meaning they occupy the lowest energy level possible (the electron shell closest to the nucleus). When an electron is excited by adding energy to an atom (for example, when it is heated), the electron will absorb energy, "jump" to a higher energy level, and spin in the higher energy level. After a short time, this electron will spontaneously "fall" back to a lower energy level, giving off a quantum of light energy. Key to Bohr's theory was the fact that the electron could only "jump" and "fall" to precise energy levels, thus emitting a limited spectrum of light. The animation linked below simulates this process in a hydrogen atom.

Bohr's Atom: Quantum Behavior in Hydrogen

Concept simulation - Reenacts electron's "jump" and "fall" to precise energy levels in a hydrogen atom.

occupy specific energylevels, he also predicted that those levels had limits to the number of electrons each could hold. Under Bohr's theory, the maximum capacity of the first (or innermost) electron shell is two electrons. For any element with more than two electrons, the extra electrons will reside in additional electron shells. For example, in the ground state configuration of lithium (which has three electrons) two electrons occupy the first shell and one electron occupies the second shell. This is illustrated in the animation linked below.

The Lithium atom

For further details, the table linked below shows the electron configurations of the first eleven elements.

Nuclear Physics

noreply@blogger.com (Study Physics) — Fri, 15 Jul 2011 16:41:00 +0000

Chapter 1: The atom
( Study Physics Online )

An atom consists of a positive charged atomic nucleus where you can find protons and neutrons and it consists of a negative charged atomic shell with electrons. In every atom the number of the electrons is equal to the number of the protons so it is neutral. The number of the protons decides which chemical element the atom is. The first element in the "Periodic table of the elements" is hydrogen. The elements in the "Periodic table of the elements" are sorted by the number of the protons. The atomic nucleus of a hydrogen atom consists of only one proton. But there are a few isotops of every element. Isotops are atoms with the same number of protons, but another number of neutrons. The different isotops of one element do not differ in their chemical properties. There are for example three isotops of hydrogen. The first isotop is the one I wrote about. The second isotop of hydrogen is deutrium with one proton and one neutron in his atomic nucleus and the third isotop is tritium which has got one proton and two neutrons in his atomic nucleus. In the atomic nucleus of a tritiumatom there is no balance between the protons and the neutrons so it is instable and decays. The particle which is emited from this decay is radioaktiv and it is charged. You can make ions of atoms. We can say that an ion is an atom which has got less or more electrons than protons. An ion is not neutral an so it is radioactif.

Figure: Study Nuclear Physics

Chapter 2: Radioactivity

Radioactivity means that atoms decays. The reason for this decays is that they are instable. A atomic nucleus is instable when he is to heavy or when a balance is missing between the protons and the neutrons. Every atom which has got a higher number of nucleons (protons and neutrons togehter) than 210 is instable. There are three types of decays: alpha decay, beta decay and gamma decay. Because it is impossible today to say which atomic nucleus will be the next who decays there statistics. We can say how many atomic nucleus will decay in a certain time. This is the princip for half lifes. After one half life a half of the atomic nucleus of a certain material decayed. Plutonium-239 for example has got a half life 24,000 years, radium-228 has got a half life of 6.7 years, thorium-232 has got a half life of 14,000,000,000 years and polonium-212 has got a half life 0.0000003 seconds. There are many physical properties, but I will talk about the acivity now. The activity is the number of decays devided by a certain time. the unit of the activity is becquerel. 1 becquerel is one decay per second. So 20 becquerels are 20 decays per second. To prove these decays there is a geiger counter. It consists of a closed tube which is often filled with argon. At the end of the tube there is a wire, which is not allowed to touch the other end of the tube or the walls. The wire is charged positive and the walls are charged negative. A radioactive particle which flows into the tube ionizes one or a few gas atoms. The out-pushed electrons go to the wire. The consequence is a voltage surge. This voltage surge is shown on an output device as a decay. On the photo there shown a geiger counter.

Figure: Study Nuclear Physics

The alpha decay

When we talk about the alpha decay then it means that a twice positive charged heliumion (helium atomic nucleus) is emited from the atomic nucleus. Then we find two protons ans two neutrons less in this atomic nucleus, so it is lighter. The alpha radiation is the most dangerous of the three types of radiation, but a sheet of paper is enough to protect oneself. The skin protects us also from alpha radiation.

Figure: Study Nuclear Physics

The beta minus decay

There are two types of the beta decay. The one is the beta minus decay and the other is the beta plus decay. When we talk about the beta minus decay a neutron decays into a proton, an electron and an antineutrino. The electron and the antineutrino are emited. The radioactive particle is the electron. The number of nucleons do not change, but we have got one proton more than before the decay. 2 or 3 cm of wood are enough to protect oneself.

Figure: Study Nuclear Physics

The beta plus decay

When we talk about the beta plus decay a proton decays into a neutron, a positron (the antiparticle of the electron) and a neutrino. The positron and the neutrino are emited. The radioactive particle is the positron.

Figure: Study Nuclear Physics

The gamma decay

When we talk about the gamma decay high-energy electromagnetic waves are emited from the atomic nucleus. This waves are photons, which have got a higher frequency and less wave long than light. A gamma decay can happen after an alpha decay or a beta decay, because the atomic nucleus is very energitif. You need a big wall of lead to protect yourself from gamma radiation.

Figure: Study Nuclear Physics

Chapter 3: The applications of radioactivity

Everyone knows that strong radiation is not good fot the health, but we use radioactive materials for nuclear power plants ans nuclear weapons (Chapter 4) for example. But there are good sides for radioactivity, too. There for example nuclear medicine. An X-ray instrument sends X-Rays throught our body onto a photo plate. Where the photo plate becomes black the X-rays goes throught our body, there where the photo plate stays transparent the X-rays do not pass our body. Another positive aspect is the radiotherapy. It is used to destroy cancer. In old clocks which have illuminated you can find radium and thorium which were used to bring the zinc sulfite to illuminate. The glowing trunk for camping lamps contained thorium. The energy source for the batteries for cardiac pacemaker is plutonium-238. There is not any nuclear fission in those batteries, because the energy source is the natural nuclear decay. Radionuclide batteries are also used for space probes like Voyager I, Voyager II and Cassini who are very long in space and so they need radionuclide batteries who are an energy source for a long time. In the next chapter I will talk bout nuclear power plants and nuclear weapons.

Chapter 4: Nuclear reactions and their applications

There are many nuclear reactions, but I will only discribe the nuclear fission and the nuclear fusion. For a nuclear fission in a nuclear power plant or for an explosion of a nuclear bomb you need plutonium-239 or uranium-235 as a split material. To make a nuclear fission it is necessary to bombard the split material with thermal neutrons. After the fission there there are two new atoms and and two or three free neutrons. This free neutrons make a fission of other atoms and so it is a nuclear chain reaction.

The animation of a nuclear fission:

Figure: Study Nuclear Physics

In a nuclear bomb there is a globe made of plutonium-239 or uranium-235. In this globe there is a neutron source which only effective when the TNT (trinitrotoluene) exploses. Because of the compression of the explosion the critical mass of the split material is overstepped. There are nuclear bomb which are build otherwise, but the princip is always the same. This both materials are very expensive, because on earth we find very little plutonium so it means that we must produce plutonium. To produce plutonium it is necessary to bombard the natural and very cheap uranium-238 with neutrons to make uranium-239. Uranium-239 decays to neptunium-239 and neptunium-239 decays after a certain time to plutonium-239. You can find uranium-235 in nature, but only in uranium-238. To split this uranium-235 from uranium-238 is very expensive, because their chemical properties are the same so it is not possible to split them in a chemical way. A nuclear bomb like this can have an explosion force of 20 kilotons (20000 tons). This means that an explosion of such a bomb is as effective as the explosion of 20 kilotons of TNT.

Hydrogen bombs can reach an explosion force of 20 megatons (20 million tons). This bombs are also knows as three-phase fuzes. The fission like in a nuclear bomb is only the first phase. In the second phase there is a fusion between deutrium and tritium. The temperatur in the second phase behave 200 to 300 million degrees celsius (much hoter than the core of the sun). The third phase is the fission of uranium-238 which is of the outer side of the bomb. Under this conditions the fission of uranium-238 is possible. The princip of power plants is the same like in nuclear bombs, but without using TNT. The reason why nuclear power plants do not exploses is that there are control rods to control the number of the neutrons in the reactor. This is a controlled nuclear chain reaction in the opposite of an uncontrolled nuclear chain reaction in nuclear bombs. The nuclear power plants in the future will be fusion reactors which do not crack heavy atomic nucleus, but fuses light atomic nucleus. Fusion are today possible but energy which you need for a fusion is higher than the energy you get and this is not the sense of nuclear fusions. With fusions the last elements of the "Periodic table of the elements" have been created, because their are not on earth. In 1999 a few physicists thought that they have discovered the element 118 but two years later in 2001 they said that it was a mistake, so element 114 is the last know element. In stars there are also fusions. In our sun it is the proton proton cycle which you can find on the website of astronomy and astrophysics. Now I will give an answer why we get energy from this nuclear reactions. We must begin which Einstein's famous formula: E=mc2 (E stands for energy, m stands for mass and c stands for the speed of light in the vacuum). This formula makes it possible transform masse in energy. Atomic nucleus have got different binding energy. The binding energy is the energy which holds the nucleons together. Because of this fact there is in every atomic nucleus a mass defect. A free proton and a free neutrons weighs more than deutrium (heavy hydrgen, consists of one proton and one neutron). Iron has got the highest binding energy and stands in the middle of the "Periodic table of the elements". When somebody goes closer to this middle with fissions or fusions a part will be transformed into energy.

The animation of a nuclear fusion:

Figure: Study Nuclear Physics

Interpherence, Young's Double-slit Formula

noreply@blogger.com (Study Physics) — Fri, 05 Nov 2010 06:33:00 +0000

Q. Interpherence, Young's Double-slit Formula

Question.
A police cruiser has an unusual radar speed trap set up. It has two transmitting antennae at the edge of a main road that runs north and south. One antenna is 2.0 m[west] of the other, and they are both fed from a common transmitter with a frequency of 3.0 x10^9 Hz. These antennae can be considered as point sources of continuous radio waves. The trap is set for cars travelling south. A student drives his car along an east-west road that crosses the main road at a level intersection 100 m[north] of the radar trap. The student's car has a "radar detector", so he hears a series of beeps as he drives west through the intersection. If the time interval between successive quiet spots is 1/5 s, as he crosses the main road, what is his speed?

Answer.
This is the problem of Young's double-slit interference experiment.

d = 2.0 m
D = 100 m
λ = c/f = 3x10^8 / 3 x 10^9 = 10^(-1) m
distance between successive destructive interference,
x = λD/d = 10^(-1) * 100/2 = 5 m
=> speed of the car
= 5/(1/5) m/s
= 25 m/s
= 25 * 3.6 km/h
= 90 km/h.

Refraction of light

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 18:34:00 +0000

Topic :Light

Refraction of light

Question.
What is the final y-position of the ray?

Answer.
For the green slab, the angle of refraction, r, is given by
sinr / sin35° = 1/1.3
=> r = 26.1813°
=> y-position at interface
= 5 tan(26.1813°) cm
= 2.4583 cm

For the blue slab, the angle of refraction, r', is given by
sinr' / sinr = 1.3/1.8
=> r' = (1.3)/1.8) * sin(26.1813°)
= 18.5815°
=> final y-position
= 2.4583 cm + 5 tan(18.5815°) cm
= (2.4583 + 1.6809) cm
= 4.1392 cm.

Total internal reflection, refraction of light, prism

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 18:27:00 +0000

Topic :Light

Total internal reflection, refraction of light, prism.

Question.

Refer to the figure as under. Monochromatic light is incident normally on the face OY of a prism made of glass whose index of refraction is such that the critical angle is 42 degrees.

Redraw this figure showing the path of light through the prism, any reflections as well as refractions, until some of the light emerges from the prism.

Answer.

As the incident ray is normal to the hypotenuse, it passes undeviated and hits the vertical surface of the prism at 30 degrees with the vertical or 60 degrees with the norm.al at 8 cm below O.
As the critical angle is 42 degrees, the ray wiull undergo total internal reflection as its incident angle is 60 degrees which is more than the critical angle and hit the bottom surface at some point with the incident angle of 30 degrees and come out with refraction.

If r = angle of refraction
sin i / sin r = 1/sinC
=> sin r = sin i / sinC = sin30 / sin42
=> sinr = 0.7472
=> r = 48.35 degrees.

Heat:

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 18:18:00 +0000

Topic : Heat / Thermodynamics

Heat:

Question.
An insulated beaker with negligible mass contains a mass of 0.345 kg of water at a temperature of 60.4 deg. Celsius.
How much ice at a temperature of -11.2deg. Celsius must be dropped in the water so that the final temperature of the system will be 20.2deg. Celsius?
Take the specific heat for water to be 4190 J/(kg*K), the specific heat for ice to be 2100 J/(kg* K), and the heat of fusion for water to be 334 kJ/kg.

Answer.

Let the mass of ice required = x kg.
Now, heat lost by 0.345 kg water when its temperature falls from 60.4 to 20.2 degrees
= 0.345 *1000*( 60.4 - 20.2 ) = 13869 cal.

Heat gained by x kg of ice when its temperature rises from - 11.2 to 0 degree
= x*1000*0.5*11.2 = 5600x cal.

Heat gained by x kg of ice at 0 degree to melt to water at 0 degree
= x*1000*80 = 80000x cal

Heat gained by x kg of water when its temperature rises from 0 to 20.2 degrees
= x*1000*1*20.2 = 20200x cal.

Total heat gained by ice
= ( 5600 + 80000 + 20200 ) x cal.
= 105800 x cal.

Heat gained by ice = heat lost by water
=> 105800x = 13869
=> x = 0.131 kg.

Heating using electric heater

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 18:14:00 +0000

Topic : Heat / Thermodynamics

Heating using electric heater.

Question.
Mr Smith has a 3kW. water heater containing 120kg. of water initially as 20°C, and a clock radio which consumes 5W. No other appliances are using any electricity. There is 41p. of credit remaining on his electricity meter.
The alarm is set to sound after 8 hours, but if the meter runs out sooner than that then Mr Smith will oversleep.
What is the hottest temperature that his thermostat could be set to without running the electricity meter out of credit?
Specific Heat Capacity of water = 4200 J/(kg.K), Price of electricity = 8p/kWh.
Assume the water heater is thermally insulated, and neglect its own heat capacity.

Answer.

Let T = the maximum temperature to which the thermostat is set, then the temperature of water should not reach this value in 8 hours.
In 41 p, amount of electricity available
= 41/8
= 5.125 kWh
Of this, amount of electricity consumed by clock
= 5 x 8
= 40 Wh
= 0.040 kWh
Balance available for consumption by water heater
= 5.125 - 0.040 kWh
= 5.085 kWh
= (5.085) * (3600000) J
This should equal the energy for heating 120 kg of water
=> 120*(T - 20)*(4200) = (5.085) * ( 3600000)
=> T - 20 = (5.085) * (3600) / [(12) * (42)]
=> T - 20 = 36.32
=> T = 56.32° C.

Thermodynamics, Carnot Engine

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 18:10:00 +0000

Topic : Heat / Thermodynamics

Q. Thermodynamics, Carnot Engine.

Question
A reversible heat engine receives heat from two thermal reserviors at 870K and 580K,and rejects 50KW of heat to a sink at 290K.If the engine output is 85KW,make calculations for the energy efficiency and heat supplied by each reservior.

Answer
Let Q1 = heat supplied by the first reservoir
and Q2 = heat received by the 2nd reservoir
=> Q1 + Q2 = work done + heat rejected
=> Q1 + Q2 = 85 + 50 = 135 kW ... (1)

Efficiency of the engine for heat received from the first source
= (870 - 290) / (870) = 2/3
Efficiency of the engine for heat received from the 2nd source
= (580 - 290) / (580) = 1/2

=> (2/3)Q1 + (1/2)Q2 = 85
i.e., 4Q1 + 3Q2 = 510 ... (2)

Solving eqns. (1) and (2),
Q1 = 105 kW and
Q2 = 30 kW

Efficiency
= W / (Q1 + Q2) x 100
= 85 / (105 + 30) x 100
= 62.96 %.

Application of Kirchhoff's laws

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 18:02:00 +0000

Q. Application of Kirchhoff's laws

Question.

In the circuit below, R1 and R2 are in parallel and R3 is in series with this combination. There is no immediate place to calculate a voltage drop by Ohm's Law because none of the resistors has both a known current and resistance. Neither can the total resistance be obtained because R2 is unknown. So how can to find the current in R3 by applying Kirchhoff's current Law to node A?

Answer.
Let current in R1 = i and in R3 = i'
Applying Kirchoff's first law to node A, i + .001 = i'
Applying Kirchhoff's second law to the outermost loop,
76 = 56000i + 56000 ( i + 0.001)
=> i = 20 /112000 = 0.00018 A = 0.18 mA
=> i' = 1.18 mA

Work done by a variable force

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:56:00 +0000

Q. Work done by a variable force

Question.

When a particle is located a distance x meters from the origin, a force of cos(πx/3) newton acts on it. How much work is done in moving the particle from x = 1 to x = 2 ? Interpret your answer by considering the work done from x = 1 to x = 1.5 and from x = 1.5 to x = 2.

Answer.

Work done
= integral (x1 to x2) cos (πx/3) dx
1)
= (3/π) sin (πx/3) [ x = 1 to x = 2 ]
= (3/π) [sin (2π/3) - sin (π/3)]
= (3/π) (1/2 - 1/2) = 0
2)
= (3/π) sin (πx/3) [ x = 1 to x = 1.5 ]
= (3/π) [sin(π/2) - sin(π/3)]
= (3/π) (1 - 1/2)
= 3/(2π)
3)
= (3/π) sin (πx/3) [ x = 1.5 to x = 2.0 ]
= (3/π) [sin(2π/3) - sin(π/2)]
= (3/π) (1/2 - 1)
= - 3/(2π)

Interpretation:
This is the case of a variable force. The direction of force changes at x = 1.5. So work done is positive from x = 1 to x = 1.5. Same amount of negative work is done as the force is negative for x = 1.5 to x = 2. Hence, total work done for x = 1 to x = 2 is zero.

Electromagnetism

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:54:00 +0000

Q. Electromagnetism

Question.
A particle having mass m and charge q is released from the origin in a region in which magnetic field and electric field are given by B vector = - Bo*j vector and E vector = Eo* k vector. Find the speed of the particle as a function of its Z - coordinate.

Answer.

Lorentz force: F = (Fx, Fy, Fz) = q(E + V x B)
Here E = (0, 0, Eo), V = (Vx, Vy, Vz) and B = (0, -Bo, 0) ==>
V x B = (Vz*Bo, 0, -Vx*Bo)
m*(dV/dt) = q(Vz*Bo, 0, Eo -Vx*Bo)
(m/qBo)*(dV/dt) = (Vz, 0, Eo/Bo -Vx) ==>

(m/qBo)*(dVx/dt) = Vz ..... (1)
(m/qBo)*(dVz/dt) = Eo/Bo -Vx ..... (2)

Differentiate (1) wrt t ==> (m/qBo)*(d²Vx/dt²) = dVz/dt ..... (3)
Substitute dVz/dt from (3) in (2) ==>
(m/qBo)² * (d²Vx/dt²) = Eo/Bo -Vx
(m/qBo)² * (d²Vx/dt²) + Vx = Eo/Bo ..... (4)
The solution of the differential equation (4) is
Vx = (Eo/Bo)*[1-cos(qBo*t/m)] ..... (5)
From (1) ==>
Vz = (Eo/Bo)*sin(qBo*t/m) ..... (6)

From (5) and (6) ==>
X = (Eo/Bo)*[t - (m/qBo)sin(qBo*t/m)] ..... (7)
Z = (mEo/qBo²)*[1 - cos(qBo*t/m)] ..... (8)

From (7) and (8) ==>
The particle is moving along the cycloid in the xz coordinate plane.

V² = Vx² + Vz² = (Eo/Bo)² * [2 - 2cos(qBo*t/m)]
V² = (Eo/Bo)² * 2[1 - cos(qBo*t/m)] ..... (9)
From (8) and (9) ==> V² = 2qEo*Z/m
V = √(2qEo*Z/m)

Laws of Motion

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:50:00 +0000

Q. Laws of Motion

Question.
Two blocks are in contact and pushing on one another. The blocks are on a frictionless surface. The mass of the blocks are 3.9kg and 9.9kg. The block with the mass of 3.9kg is being pushed into the 9.9kg block with a force of 69N. The 9.9kg block Is being pushed into the 3.3kg block with a force of 11N. What is the magnitude of the force exerted on the 3.9kg block by the 9.9kg block.
Answer.
Let R = force of action-reaction exerted by each block on the other.
Net force on block of mass 3.9 kg
= 69 - R N
and its acceleration will be
= (69 - R) / 3.9 m/s^2

Net force on block of mass 9.9 kg
= R - 11 N
and its acceleration will be
= (R - 11) / 9.9 m/s^2.

Both blocks will move with the same acceleration.
=> (69 - R) / 3.9 = (R - 11) / 9.9
=> R = 52.6 N.

Heat

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:48:00 +0000

Q. Heat

Question.
An insulated beaker with negligible mass contains a mass of 0.345 kg of water at a temperature of 60.4 deg. Celsius.
How much ice at a temperature of -11.2deg. Celsius must be dropped in the water so that the final temperature of the system will be 20.2deg. Celsius?
Take the specific heat for water to be 4190 J/(kg*K), the specific heat for ice to be 2100 J/(kg* K), and the heat of fusion for water to be 334 kJ/kg.

Answer.
Let the mass of ice required = x kg.
Now, heat lost by 0.345 kg water when its temperature falls from 60.4 to 20.2 degrees
= 0.345 *1000*( 60.4 - 20.2 ) = 13869 cal.

Heat gained by x kg of ice when its temperature rises from - 11.2 to 0 degree
= x*1000*0.5*11.2 = 5600x cal.

Heat gained by x kg of ice at 0 degree to melt to water at 0 degree
= x*1000*80 = 80000x cal

Heat gained by x kg of water when its temperature rises from 0 to 20.2 degrees
= x*1000*1*20.2 = 20200x cal.

Total heat gained by ice
= ( 5600 + 80000 + 20200 ) x cal.
= 105800 x cal.

Heat gained by ice = heat lost by water
=> 105800x = 13869
=> x = 0.131 kg.

Electrostatics

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:46:00 +0000

Q. Electrostatics

Question.
A charge 4.59 x 10-6 C is held fixed at origin. A second charge 3.40 x 10-6 C is released from rest at the position (1.42 m, 0.746 m). If the mass of the second charge is 6.35 g, what is its speed when it moves infinitely far from the origin?

Answer.
Distance between the charges,
r = sqrt ( 1.42^2 + 0.746^2 )
= 1.60 m

Potential energy of the second charge w.r.t. the first charge,
kqq' / r gets fully converted into kinetic energy,
(1/2) mv^2 at an infinite distance.
=> kqq'/r = (1/2) mv^2
=> v = sqrt (2kqq' / ( r*m )
= √[2 (9*10^9) (4.59*10^-6) (3.40*10^-6) / (1.6) (0.00635)
= 3.72 m/s

Collision, Projectile Motion

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:43:00 +0000

Q. Collision, Projectile Motion.

Question.

Mass 1 hits mass 2 and causes the two masses to

fall on the floor as shown in the figure. The surfaces

are NOT frictionless.

Using conservation of energy with consideration of the dissipated energy due to friction, calculate the velocity and momentum just before collision.

Calculate the momentum of mass1 and mass2 after collision.

Answer.

Let u and v be the velocity of mass 1 before and after collision respectively.

Let v' be the velocity of mass 2 after collison. Initially, mass 2 is at rest.

First we calculate v and v' from the projectile motion of the two masses.

Time taken by both the masses to reach the ground

= √(0.7/4.9) s.

= 0.378 s.

In this time mass 1 covers a distance of 0.2 m with velocity v and mass 2 covers a distance of 1.1 m with velocity v'

So, v = 0.2 / 0.378 = 0.53 m/s

and v' = 1.1 / 0.378 = 2.91 m/s

Now, by momentum balance,

25u + 0 = (25)*(0.53) + (15)*(2.91)

=> u = (13.25 + 43.65) / 25 = 2.276 m/s

Dissipated energy due to friction

= P.E. at top - K.E. at the bottom of curved surface

= (0.025)*(9.8)*(0.3) - (1/2)*(0.025)*(2.276^2) joule

= (0.025)*[ (9.8)*(0.3) - (0.5)*(2.276^2) ]

= (0.025)*[ 2.94 - 2.59 ]

= 0.00875 joule

Velocity of mass 1 before collision = 2.276 m/s

Momentum of mass 1 before collision

= (0.025)*(2.276) = 0.0569 (kg)(m)/s

Velocity and momentum of mass 2 before collision = zero.

Momentum of mass 1 after collision

= (0.025)*(0.53) =0.0133 (kg)(m)/s

Momentum of mass 2 after collision

= (0.015)*(2.91) = 0.04365 (kg)(m)/s

Collision, Spring, Energy Balance

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:39:00 +0000

Q. Collision, Spring, Energy Balance

Question.
Body A, of mass 2kg, has a light spring attached to it as shown in the adjoining figure.
Body B has a mass of 5kg and is initially suspended by a string of length 0.1m. Body A moves with a velocity of 3.5m/s over a frictionless plane directly towards body B as shown in Fig. 1.1. The two bodies undergo an elastic collision, during which the spring on body A is compressed as shown in Fig. 1.2 and then the spring extended. The two bodies separate after collision as shown in Fig. 1.3
i)During the maximum compression, the initial kinetic energy of A is more than the total kinetic energy of A and B. Explain why this is possible.
ii)Find the velocities VA and VB after the collision.
iii)Calculate the tension of the string when B swings through an angle of 30 degrees and reaches point C (Take g=10m/s^2).

Answer .

i)
During the maximum compression, kinetic energy of A is converted into the kinetic energy of B plus potential energy stored in the spring. Thus, kinetic energy of A is zero and of its original kinetic energy only a part is converted into the kinetic energy of B and the balance is stored as the potential energy in the spring. This is the reason why original kinetic energy of A is more than the sum of the kinetic energy of A at maximum compression of the spring (which is zero) and the kinetic energy of B (which is only a part of the kinetic en energy of A).

(ii)
By the law of conservation of linear momentum,

2 * 3.5 = 2 * (- VA) + 5 * VB
=> VA = 2.5VB - 3.5 ... ( 1 )
As A moves on frictionless plane and collision is elastic, kinetic energy is also conserved.
=> (1/2) * 2 * (3.5)^2 = (1/2) * 2 * VA^2 + (1/2) * 5 * VB^2
=> 2.5 VB^2 = (3.5)^2 - VA^2
Plugging the value of VA from ( 1 ),
=> 2.5VB^2 = 12.25 - (2.5VB - 3.5)^2
=> 2.5VB^2 = 17.5VB - 6.25VB^2
=> 8.75VB = 17.5
=> VB = 2 m/s.
and VA = 2.5VB - 3.5 = 2.5 * 2 - 3.5 = 1.5 m/s.

(iii)
When B goes to point C, it rises through height, h
= (0.1) (1 - cos30°) and gains PE
= mgh
= 5*10*(0.1)(1 - cos30°) J
= 0.67 J
KE at C = KE at B - 0.67
=> (1/2) * 5 * v^2 = (1/2) * 5 * (2)^2 - 0.67
=> v^2 = 2.68 J

If T = tension in the string, balancing forces radially,
T - mgcos30° = mv^2/r = 5 *2.68/(0.1)
=> Tension, T
= 5*10*cos30° + 5*26.8 N
= 177.3 N.

Laws of Motion, Friction

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:34:00 +0000

Q. Laws of Motion, Friction

Question.
A 3.70 kg block starts from rest at the top of a 30.0° incline and slides 2.10 m down the incline in 1.50 s.
1)Find the coefficient of kinetic friction between the block and the incline.
2)Find the frictional force acting on the block.
3)Find the speed of the block after it has slid a distance of 2.10 m.
Answer .

s = ut + 0.5at^2
=> 2.1 = 0 + 0.5a(1.5)^2
=> a = 4.2 * (4/9) = 5.6/3 m/s^2
=> net force down the plane
= mass x acceleration
= (3.7) * (5.6/3) N
= 6.91

Now, net force = downward component of weight - frictional force
=> Frictional force
= mgsin30° - 6.91
= (3.7)*(9.8)*(0.5) - 6.91
= 18.13 - 6.91
= 11.22 N.

Normal force x coefficient of friction = frictional force
=> mgcos30° * μ = 11.22
=> μ = (11.22) /[3.7*9.8*cos30°]
= 0.36

Answers:
1) 0.36
2) 11.22 N
3) v^2 = u^2 + 2as
=> v^2 = 2*(5.6/3)*(2.1) = 1.4 * 5.6
=> v = 2.8 m/s.

Mechanics, Moments of forces

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:30:00 +0000

Q. Mechanics, Moments of forces

Question.

The workers and the platform together have a mass of 204kg and their combined centre of gravity is located directly above C. Given that the direction of the force exerted on the telescopic arm ABC at B by the single hydraulic cylinder BD is in the direction from D to B, find the force on the arm at A for the position when θ = 20°. You may neglect the weight of the telescopic arm in your calculation.

Answer

Let Fx and Fy be the forces at A towards +ve x- and y- directions.

Taking moments about B,
204*9.81*2.6cos20° + Fx*2.4sin20° + Fy*2.4cos20° = 0
=> 4889 + 0.821Fx + 2.255Fy = 0
=> Fx = - 5955 - 2.75Fy ... ( 1 )
Taking moments about D,
204*9.81*(5cos20° - 0.5) + 0.5Fy = 0.9Fx
=> 8402 + 0.5Fy = 0.9Fx=> Fx = 9336 + 0.556Fy ... ( 2 )

Subtracting ( 1 ) from ( 2 ),
3.306Fy = - 15291
=> Fy = - 4625 N
Plugging Fy in ( 1 ),
Fx = - 5955 – 2.75 * (-4625) = 6764 N

F = √[4625)^2 + (6764)^2] = 8194 N

Angle with the horizontal
= arctan (4625)/(6764) = 34.4°.

Mechanics Roller Coaster

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:26:00 +0000

Q. Mechanics Roller Coaster

Question

At an amusement park, a child gently releases a toy roller coaster, Toy B, at the top of a curved track.

During the journey on the track, the toy roller coaster, Toy B, is subjected to the frictional force due to the toy's motion on the track. The frictional force between the toy and the curved track may be assumed to have an average value of 1.26 x 10^-2 N throughout the entire track under normal condition and 2.1 x 10^-3 N when lubricated. Mass of Toy B=50g, distance H =0.82m.

i) If Toy B moves along the track under normal condition, and just missed reaching point Z, find the energy loss to friction.

ii) Hence find the total distance travelled by Toy B on the curved track from the point of release to just before point Z.

iii) If the track is lubricated, find the speed of Toy B at point Z.

iv) If the track is lubricated, Toy B will make a head-on elastic collision with the identical stationary toy Toy Z at point Z. Find the horizontal distance d, that Toy Z will hit the horizontal surface, which is at 0.4m below point Z. (assume toy Z moves off with constant speed after collision, before free fall)

Answer
i)

Kinetic energy gained during drop by height H is lost due to frction
=> Loss of energy due to friction
= mgH
= 50*10^-3*9.81*(0.82) J
= 0.402 J

ii)
Work done by frictional force
= Force x displacement opposite to the direction of force = 0.402
=> Distance travelled
= (0.402) / (1.26x10^-2) m
= 32 m.

iii)
If the track were lubricated, loss due to friction
= 32 * 2.1 x 10^-3 J
= 0.0672
KE = PE - Loss due to friction
=> (1/2) mv^2 = 0.402 - 0.0672 = 0.3348
=> v = √(2*0.3348)/(0.050) = 3.66 m/s.

iv)
In head-on elastic collision between two identical bodies, velocities get exchanged.
=> Toy Z will move with velocity = 3.66 m/s in the horizontal direction.
Time to move down by 0. 4 m with zero initial vertical velocity is given by
h = (1/2)*9.81t^2
=> t = √(0.8/9.81) = 0.286 s.
During this time toy Z moves with constant horizontal velocity of 3.66 m/s covering horizontal distance
= (0.286) * (3.66) m
= 1.047 m.

Buoyancy, Motion of balloon under gravity and buoyancy.

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:20:00 +0000

Q. Buoyancy, Motion of balloon under gravity and buoyancy.

Question
Before a small party balloon is inflated, it is placed on an electronic weighing scale which records its mass as 1.3g. The balloon is now inflated with air so that it is spherical in shape with a diameter of 22cm. If the density of air in the room is 1.21kg/m^3.

i) Calculate the mass of air displaced by the balloon,

ii) The inflated balloon in i) is placed on the scale which now gives a reading of 1.55g. Calculate the mass of   air in the balloon and suggest a reason as to why this value is more than the value you calculated in i),

iii) A 2.1g nut is now tied to the bottom of the inflated balloon using a light cotton thread. The balloon is     dropped from a height of 3m above the ground.
a) Show that the magnitude of the acceleration of the balloon and nut at the start of their descent is 3.45m/s^2
b) Explain why the value of this acceleration will approach zero as the balloon decends,
c) Calculate the magnitude of the drag force acting on the balloon and nut when they are moving at their terminal velocity of 2m/s,
d) The value of the drag force Fd is given by : Fd = (CdrouAv^2)/2 where Cd is the drag coefficient , rou is the density of air and A is the cross-sectional area of the balloon. Calculate the value of the drag coefficient Cd for the balloon,

iv)The same balloon used in i) and ii) is now inflated with helium (density of helium in the balloon = 0.166 kg/m^3) instead of air so that it has the same volume as the air balloon,
a) Calculate the downward force that would be required to hold this balloon alone (no nut attached) at a constant height of 3m above the ground,
b)The downward force is removed and the helium balloon is allowed to rise.
    Explain how the conservation of energy is applicable to the motion.

Answer
i)
Volume of air displaced
= (4/3)πr^3
= (4/3)π * (0.11)^3 m^3
= 0.0055753 m^3
Mass of air diaplaced
= 1210 * 0.0055753 g
= 6.74 g

ii)
Let m = mass of air in the balloon
=> final weight of balloon
= weight of deflated balloon + weight of air filled in - weight of air displaced
=> 1.30g + mg - 6.74g = 1.55g
=> m = 6.74 + 1.55 - 1.30 g = 6.99 g
The air filled in and air displaced have the same volume, but the air filled in has higher pressure than the pressure of air outside which is at atmospheric pressure. Hence, inside air has higher density and higher mass.

iii) a)
Apparent weight of the balloon + nut
= (1.55 + 2.1) * 10^-3 * 9.81 N = 0.0358065 N
Mass of balloon + air + nut
= 1.3 + 6.99 + 2.1 = 10.39 * 10^-3 kg
Acceleration at the start
= (0.0358065) /(10.39 x 10^-3) m/s^2
= 3.45 m/s^2

iii) b)
As the balloon descends, three forces are acting on it.
1) its weight downwards,
2) force of buoyancy upwards and
3) force of air resistance upwards.
Of these, the first two forces are constant.
Air resistance is proportioanl to the downward velocity of the balloon. As the initial velocity at the start, downward velocity is zero, there is no air resistance and the balloon has the maximum downward acceleration. As it gains velocity due to this acceleration, the upward air resistance increases with velocity till it balances the first two constant forces. At that instant as all the three forces balance each other (taken with appropriate signs), there is no net force on the balloon and it continues to move down with whatever velocity it has gained by that time called the terminal velocity and has zero acceleration due to zero net force on it.
From the instant velocity reaches terminal velocity,
Buoyant force - Weight of balloon = Downward resistive force

[Note: Please note that this discussion is valid near the surface of the earth, where we can consider acceleration due to gravity "g" and air density to remain almost constant. At great heights, both "g" and air density vary and hence terminal velocity will change and not remain constant. It will reduce and ultimately become zero.]

iii) c)
Drag force
= apparent weight
= (1.55 + 2.1) x 10^-3 * 9.81 N
= 3.58 x 10^-2 N.

iii) d)
Fd=(CdrouAv^2)/2
=> 3.58 x 10^-2 = Cd x 1.21 x π(0.11)^2 * 2^2/2
=> Cd
= (0.0358) / (2.42 π * 0.0121)
= 0.389.

iv) a)
Mass of He filled
= volume x density
= 0.0055753 m^3 x 0.166 kg/m^2
= 0.0009255 kg
If F = downward force to be applied, then
F = weight of air displaced - weight of balloon
= (0.00674 - 0.0013 - 0.0009255) * 9.81 N
= 0.0443 N.

iv) b)
The He balloon on the surface of the earth has positive potential energy stored in it due to the force of buoyancy. As it rises, this energy is getting converted into gravitational potential energy and kinetic energy and a part is lost to overcome the resisitive force. Thus,

ΔPE(buoyancy) = ΔPE(gravitational) + ΔKE + work done against the resistive force.

In the process, total energy of the balloon and environment is conserved, but the mechanical energy of the balloon is reducing. This process of interconversion of the energies comes to a stop when the total
mechanical energy of the balloon (= PE due to buoyancy + PE due to gravity + KE) becomes zero.

Collision Problem

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:12:00 +0000

Q. Collision Problem.

Question
A car moving at speed v undergoes a one-dimensional collision with an identical car initially at rest. The collision is neither elastic nor fully inelastic; 2/17 of the initial kinetic energy is lost.
Find the velocities of the two cars after the collision. Express your answer in units of v.

Answer

Let equal mass of both cars = m
Initial velocity of 2nd car, u = 0
Let v' and u' be the final velocities of the 1st and 2nd car respectively.

Momentum is always conserved.
So, mv + 0 = mv' + mu'
=> v = v' + u' ... ( 1 )

2/17 of initial K.E. is lost => 15/17 of initial K.E. = Final K.E.
=> (15/17)(1/2)mv^2 = (1/2)mv'^2 + (1/2)mu'^2
=> (15/17) v^2 = v'^2 + u'^2 ... ( 2 )

From equations ( 1 ) and ( 2 ),
(v' + u')^2 - (v'^2 + u'^2) = v^2 - (15/17)v^2
=> 2v'u' = (2/17) v^2
=> (v' - u')^2 = (v' + u')^2 - 4v'u' = v^2 - (4/17)v^2 = (13/17)v^2
=> v' - u' = v √(13/17) = 0.874 v ... ( 3 )

Solving equations ( 1 ) and ( 3 ),
v' = 0.987 v and u' = 0.063v

Magnetic effects of electric current

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:09:00 +0000

Q. Magnetic effects of electric current

Question

A particle of mass m and charge q enters a uniform Magnetic field B (perpendicular to the paper in wards) at P with velocity V at an angle α and leaves the field at Q at an angle β as shown. Explain how these options are correct?

1) α must be equal to β,

2) Length PQ=(2mv*Sin α)/qB,

3) Particle remains in the field for time T=2mα/(qB).

Answer

1)

Since velocity is perpendicular to the direction of magnetic field, the charge experiences force in a direction perpendicular to both. As the force is perpendicular to the velocity, it does no work and results in uniform circular motion of the charge. If the angles α and β are the same, it means that the velcocity of the charge at both these points P and Q is tangential to the circular arc between P and Q inside the magnetic field which is the path of motion of the charge.
2)
The force acting on the charge
= qvB
which equals the centripetal force, making it move on the circular path of radius R given by
qvB = mv^2/R
=> R = mv/qB ... ( 1 )
A perpendicular from the center of the circular path on the chord PQ bisects it at M
PQ
= 2 PM
= 2 Rsinα
Plugging the value of R from ( 1 ) above,
PQ = (2mvsinα)/qB

3)
Arc length PQ
= 2αR
= 2α * mv/(qb)
Time taken to travel this distance with constant velocity, v is
T = distance /velocity
= [2αmv/(qB)]/v
= 2mα/(qB).

Electromagnetic Induction

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 17:04:00 +0000

Q. Electromagnetic Induction.

Question

A conducting rod of length l is moved at constant velocity Vo on two parallel, conducting, smooth, fixed rails, that are placed in a uniform constant Magnetic field B perpendicular to the plane of rails as shown in the figure. A resistance R is connected between the two ends of the rail, then please explain how the following options are correct?

1) Thermal power dissipated in the resistor is equal to the rate of work done by the external person pulling the rod.

2) If applied external force is doubled, then a part of the external power increases the velocity of the rod.

3) If resistance R is doubled, then power required to maintain the constant velocity v becomes half.

Answer

1)
If "I" is the current flowing through the conducting rod and its length is "L", then the force acting on it due to magnetic field in the loop of intensity B is
F = IBL
=> mechanical power developed ,
P = FVo = IBLVo ... ( 1 )

Thermal power dissipated
= electric power expended in the wire
= emf developed x current
= BLVo * I
= IBLVo ... ( 2 )

From ( 1 ) and ( 2 ),
thermal power dissipated in the resistor
= work done by the external force.

2)
When force F is acting on the conducting rod, velocity is constant at Vo and according to Newton's first law, no net force is working on the rod. This is because, the force applied balances the force developed on the conducting rod in the opposite direction due to rate of change of magnetic flux. Now when F is increased to 2F, initially net force is 2F - F = F which accelerates the rod and increases its velocity due to the acceleration. With increase in velocity, the opposite magnetic force keeps increasing and when the velocity doubles to Vo becomes equal to 2F. At that instant, again the net force becomes zero and the rod continues to move with velocity 2Vo. During this period, net force changes from F to zero, and the varying net force increases the velocity of the rod from Vo to 2Vo.

3)
If R is doubled, current I becomes half and the force required to maintain velocity also becomes half. If a constant velocity Vo is maintained in this condition, then the power required also becomes half as the force is half and velocity is the same since power = force x velocity.

Newton's Laws of Motion

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 16:58:00 +0000

Q. Newton's Laws of Motion

Question

1) In the first figure, a crate of mass m = 210 kg is pushed atconstant speed up thefrictionless ramp (=26.0°) by a horizontal force F.

(i) Find the magnitude of F and

(ii) the magnitude of the force exerted by the ramp on the crate.

2) In earlier days, horses pulled barges down canals in the manner shown in the second figure. Suppose that the horse pulls on the rope with a force of 7900 N at an angle of 18° to the direction of motion of the barge, which is headed straight

along the canal. The mass of the barge is 10000 kg, and its acceleration is 0.10 m/s2.

What are the magnitude and the direction of the force on the barge from the water?

Answer

Three forces act on the crate:

i) its wt. (210)*(9.8) = 2058 N acting downwards,

ii) Horizontal force, F and

iii) Reactive force, N, from the ramp on the crate perpendicular to the inclined plane.

Since mass is pushed at constant speed, there is no net force acting on the crate which means that these three forces balance each other.

Balancing horizontal and vertical component of forces,
F = Nsin θ ... ( 1 ) and
mg = N cos θ ... ( 2 )

Dividing eqn. ( 1 ) by ( 2 ),
F = mg tan θ = 2058 tan (26.0°) = 1004 N

Putting this value of F in eqn. ( 1 ),
N = F / sin θ = 1004 / sin (26.0°) = 2290 N

2)
Two force act on the barge in the horizontal direction:
i) Horizontal component of tension, T in the rope = 7900 cos 18° = 7513 N and
ii) Opposing drag force from water on the barge = F

Under the effect of these two forces, the barge has acceleration of 0.10 m/s^2.
So, (10000) * (0.10) = 7513 - F
=> F = 6513 N

There is another vertical reactive force, F', on the barge from water which is in the downward direction opposite to the upward component of tension T in the rope
= T sin 18° = 7900 * sin 18° = 2441 N

Net force from water is the resultant of F and F'
Magnitude of net force from water
= √[ (6513)^2 + (2441)^2 ]
= 6955 N

If θ is the angle made by this force with the horizontal, then
θ = - arctan (2441) / (6513) = - 20.5°

Physics Laws of motion applied to rotational motion

noreply@blogger.com (Study Physics) — Thu, 04 Nov 2010 16:51:00 +0000

Q. Physics Laws of motion applied to rotational motion

Question
A pulley has a radius of 2.70 m and a moment of inertia of 39.0 kg.m^2. A hanging mass is 4.20 kg and it exerts a force tangent to the edge of the pulley. What is the angular acceleration of the pulley?

Answer

Let T = tension in the rope
and a = downward acceleration of the hanging mass

=>
mg - T = ma = mRα ... ( 1 ) for the hanging mass
T*R = Iα
=> T = Iα/R ... ( 2 ) for the pulley

Adding equations ( 1 ) and ( 2 ),
mg = mRα + Iα/R
=> α
= (mg) / (mR + I/R)
= (4.2 * 9.81) / (4.2*2.7 + 39/2.7)
= 1.6 rad/s^2.