#$&* course phy122 3/15 6 *********************************************
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Given Solution: If the pieces attract, then the tape at point A is pulled toward point B. • The vectors AB_v and AB_u point from A to B. • Of these the vector AB_u is the unit vector. • So the tape at A experiences a force in the direction of the vector AB_u. If the pieces repel, then the tape at point B is pushed away from point A. • The direction of the force is therefore from A towards B. • The direction is therefore that of the vector AB_u. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qUsing the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The distance from A to B is the same as the distance from B to A. So the vectors AB_v and BA_v have the same length. Each vector therefore has magnitude equal to the distance between A and B. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The distance from A to B is the same as the distance from B to A. So the vectors AB_v and BA_v have the same length. Each vector therefore has magnitude equal to the distance between A and B. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qUsing the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Copied and pasted the answer from below. INSTRUCTOR RESPONSE: As noted in the preceding solution, AB_v stands for the vector whose initial point is A and whose terminal point is B, and AB_u stands for a vector of magnitude 1 whose direction is the same as that of AB_u. To get a unit vector in the direction of a given vector, you divide that vector by its magnitude. So AB_u = AB_v / | AB_v |. The purpose of a unit vector is to represent a direction. If you multiply a unit vector by a number, you get a vector in the same direction as the unit vector, whose magnitude is the number by which you multiplied it. The problem does not at this point ask you to actually calculate these vectors. However, as an example: Suppose point A is (4, 5) and point B is (7, 3). Then AB_v is the vector whose initial point is A and whose terminal point is B, so that AB_v = <3, -2>, the vector with x component 3 and y component -2. The magnitude of this vector is sqrt( 3^2 + (-2)^2 ) = sqrt(11). Therefore AB_u = <3, -2> / sqrt(11) = (3 / sqrt(11), -2 / sqrt(11) ). That is, AB_u is the vector with x component 3 / sqrt(11) and y component -2 / sqrt(11). (Note that these components should be written with rationalized denominators as 3 sqrt(11) / 11 and -2 sqrt(11) / 11, so that AB_u = <3 sqrt(11) / 11, -2 sqrt(11) / 11 > ). The vector AB_u is a unit vector: it has magnitude 1. The unit vector has the same direction as AB_v. If we want a vector of magnitude, say, 20 in the direction of this vector, we simply multiply the unit vector AB_u by 20 (we would obtain 20 * <3 sqrt(11) / 11, -2 sqrt(11) / 11 > = <60 sqrt(11) / 11, -40 sqrt(11) / 11 >). You don't need to understand this example at this point, but if you wish to understand it you should probably sketch this situation and identify all these quantities in your sketch. STUDENT RESPONSE AB_v/(AB_v)absolute value So AB_u would either be 1 or -1. INSTRUCTOR COMMENT AB_u and BA_v are vectors, so they have both magnitude and direction. Your answer is correct if everything is in one dimension (e.g., if all charges are on the x axis and all forces directed along the x axis). In one dimension direction can be specified by + and - signs. In two dimensions + and - signs are not sufficient. In two dimensions the direction of a vector is generally specified by its angle as measured counterclockwise from the positive x axis. In two dimensions, the magnitude of AB_u would be the same as that of BA_u, but the vectors would be in the opposite direction. confidence rating #$&*:l keep reading. Doesn’t make much sense at the moment. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the distance between A and B (i.e., to the separation between the two pieces of tape), increases the force decreases with the square of the distance (similarly if the distance decreases the force increases in the same proportion). This answer isn't exactly correct, since the two pieces of tape are not point charges. Since some parts of tape A are closer to B than other parts of tape A, and vice versa, the inverse square relationship applies only in approximation for actual pieces of tape. STUDENT COMMENT: Still not sure how to answer this question. I dont know what sort of difference there is between the magnitude vector and the unit vector. INSTRUCTOR RESPONSE: As noted in the preceding solution, AB_v stands for the vector whose initial point is A and whose terminal point is B, and AB_u stands for a vector of magnitude 1 whose direction is the same as that of AB_u. To get a unit vector in the direction of a given vector, you divide that vector by its magnitude. So AB_u = AB_v / | AB_v |. The purpose of a unit vector is to represent a direction. If you multiply a unit vector by a number, you get a vector in the same direction as the unit vector, whose magnitude is the number by which you multiplied it. The problem does not at this point ask you to actually calculate these vectors. However, as an example: Suppose point A is (4, 5) and point B is (7, 3). Then AB_v is the vector whose initial point is A and whose terminal point is B, so that AB_v = <3, -2>, the vector with x component 3 and y component -2. The magnitude of this vector is sqrt( 3^2 + (-2)^2 ) = sqrt(11). Therefore AB_u = <3, -2> / sqrt(11) = (3 / sqrt(11), -2 / sqrt(11) ). That is, AB_u is the vector with x component 3 / sqrt(11) and y component -2 / sqrt(11). (Note that these components should be written with rationalized denominators as 3 sqrt(11) / 11 and -2 sqrt(11) / 11, so that AB_u = <3 sqrt(11) / 11, -2 sqrt(11) / 11 > ). The vector AB_u is a unit vector: it has magnitude 1. The unit vector has the same direction as AB_v. If we want a vector of magnitude, say, 20 in the direction of this vector, we simply multiply the unit vector AB_u by 20 (we would obtain 20 * <3 sqrt(11) / 11, -2 sqrt(11) / 11 > = <60 sqrt(11) / 11, -40 sqrt(11) / 11 >). You don't need to understand this example at this point, but if you wish to understand it you should probably sketch this situation and identify all these quantities in your sketch. STUDENT RESPONSE AB_v/(AB_v)absolute value So AB_u would either be 1 or -1. INSTRUCTOR COMMENT AB_u and BA_v are vectors, so they have both magnitude and direction. Your answer is correct if everything is in one dimension (e.g., if all charges are on the x axis and all forces directed along the x axis). In one dimension direction can be specified by + and - signs. In two dimensions + and - signs are not sufficient. In two dimensions the direction of a vector is generally specified by its angle as measured counterclockwise from the positive x axis. In two dimensions, the magnitude of AB_u would be the same as that of BA_u, but the vectors would be in the opposite direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery introductory set #1, 1-5 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Introset 1: Using a diffraction grating we determine that the wavelength of light from certain laser is 610nm. What is the energy in a photon of this light? If an electron is ejected from a photoelectric metal by light of this wavelength, what is the magnitude of the maximum potential difference through which the electron might travel, provided the metal has work function of 1.7 electron volts? Solution: The energy of a photon is h x f, where F is the frequency of the electromagnetic wave and h is plancks constant (6.63 x 10 ^-34 J s). The frequency of the wave is f= c/lambda, wehre c= 3*10^8m/s is the speed of light in a vaccum. We therefore have: Frequency of light f= 3 * 10 ^8/ 610 nm=3*10^8 m/s /(610x 10^-9m)= .4918 x 10^15 Hz Photon energy= h*f=6.63 x 10^-34J s *.4918*10^15 Hx= 3.265 x 10 ^-19J An electron volt is the energy gained by an electron as it travels through a potential difference of 0 volt, this energy is equal to the 1 ev= 1.6 x 10^-19J Expressing this energy in units of electron volts, we have Photon energy= 3.265 * 10 ^-19J/ 1.6*10^-19 J/ev= 2.04 When a photon of wavelength 610 nm gives up its total 2.04 of energy to an electron, and when the electron loses no more than the required 1.7 ev, the electron will escape with KE Escape KE= 2.07-1.7 ev= .34 ev Max PE overcome = .34 volts Introset #2 The intensity of 836 nm light falling on a photoelectric metal is 110w/m2 (max sunlight intensity is at the surface of the earth is on the order of 100w/m^2) How many photons fall on a 1 cm ^2 area in a second? What is the approximate average spacing of the photons that fall on this are in a second? Solution: Each photon of 836 nm light has energy E=hr, where h is Planck’s constant 6.62 x 10^-34J s. The frequency of 836 nm light is (3 x 10^8m/s)/(836nm)= .3588 x 10^15 Hz Photon energy: (6.62*10^-34J) * (.3588 * 10 ^15 Hz)= 2.375x 10 ^-19J At 110 w/m^2, energy falls on1cm^2 at a rate of: (110 w/m^2)* (10^-2m)^2= .011 watts The energy falling on this are in a second is: energy in one sec= .011 J/s x 1 s=.011 J Each photon carries 2.375 * 10 ^-19 J of energy, in one second will have: Number of photons in 1 sec=energy in 1 sec/energy per photon: .011J/(2.375 x 10^-19J) =.04631 x 10^16 The average area per photon is therefore 1 cm^2/ (.04631 x 10^16)= 21.59 x 10 ^-16 cm^2 First estimate of distance between photon strikes= distance= sqrt(21.59 x 10^-16cm^2)^2 4.646 * 10 ^-8 cm=4.646 x 10 ^-10 nm= 4.646 Angstroms Equilateral triangle model for spacing= sqrt (3)/4 x s ^2= 21.59 x 10^-16 cm^2 Solving for the side s, we obtain: Second estimate distance between photon strikes: Dist=s=sqrt [(21.59 8 10 ^-16 cm/ 4/ sqrt (3)]= 3.057 x 10 ^-8 cm= 3.057 Angstroms Introset #3 The intensity of light with a mean wavelength of 758 nm falling on a solar sail is 200 w/m^2, this sail is oriented perpendicular to the direction of incoming light, in order to intercept as much light as possible. What is the total momentum of the photons falling on a 18000 m^2 of the sail every second? If the sail is a perfect mirror, reflecting the photons without any absorption or change in their momentum, then what average force is exerted on the sail? How long would it take to accelerate a 1730 kg craft( mass indicates that of the sail) from rest to velocity .1 c using such a sail? Solution: The momentum of a photon is p = E/c The total momentum falling on the sail in one second is therefore equal to the total energy of all the photons falling in one second divided by c. The energy in one second, per square meter, at 200 w/ m^2= 200 J/s/m^2 is: Energy per square meter per second: e/A= 200J/m^2 The momentum of all the photons required to deliver 200 J of energy is: Ptotal= e total/c= 200j/(3*10^8m/s)= .6666 x 10^-6kg m/s We thus have a total momentum of (.6666 x 10^-6 kg m/s / m^2 x 18000 m^2)= 1.199 x 10^ -2 kg m/s Magnitude of total momentum change in one second: dp= 2 * 1.199 *10^-2 kg m/s= 2.398*10^-2kg m/s By impulse momentum theorem, the average force on the sail will be: Fave= dp/dt= (2.398 * 10^-2 kg m/s)/ 1 sec= 2.398* 10^-2N The acceleration of 1730 kg mass will therefore be: a= f/m=2.398 *10^-2N/(1730kg) = .1386 * 10^ -4 m/s^2 At this rate the time required for a velocity change of .0 c= 3 * 10 ^7 m/s Dt= (3 * 10 ^7m/s)/ (.1386 x 10 ^-4 m/s ^2)= 2.164 x 10 ^12 sec Introset #4 A beam consisting of 50 ev electrons (electron mass 9.11 x 10 ^-31 kg) is incident on a thin wafer of crystal with layer spacing 4.5 Angstroms. Surprisingly we find that the electrons which are particles, are scattered in such a way as to form an interference pattern identical to that of a wave. What will be the distance between central interference maxima at a distance of 11 cm from the wafer? Solution: deBroglue wavelength:lambda=h/p where h is Planck’s constant. To find the momentum of the electron from the given information we must find KE in Joules. KE of the given electron in Joules is 50 ev* (1.6 x 10^-19 J/ev)=.8 * 10 ^-17J We can find the momentum mv or the electron my multiplying its KE which is .5mv^2, by 2m to obtain m^2v^2 then taking the square root Momentum= sqrt [2 x 9.11 x 10^-31kg) * (.8 x 10^-17J)]=3.794733x 10^-24 kg m/s deBroglie wavelength of the electron is therefore: wavelength: lambda= h/p= 6.63 x 10^-34 J s/ (3.794733 x 10^-24 kg m/s)= 1.74158 x 10^ -10 m= 1.747158 x 10^-10 Angstroms sin (theta)= lambda/ a The angle at which the max occurs is theta= sin -1 (1.747158 Angstroms/ 4.5 Angstroms)= 22.8577 degrees At a distance of 11 cm on the separation of this angle from the central beam will be: Sep of adjacent maxima= 11 cm x sin (22.8577 degrees)= 4.634396 cm Introset #5 What is the approximate uncertainty in the velocity of an electron (mass 9.11 x 10^-31 kg) known to remain within a distance of 1.57 Angstroms of a proton? What kinetic energy would the electron have at this velocity? Approximately how many times would the electron orbit the proton in a circular orbit at this distance and velocity? Compare the centripetal acceleration at this distance and velocity to the acceleration of the electron due to electrostatic force between an electron and a proton at this distance. Solution: The product of uncertainty in the position (standard units in meters) and the uncertainty of momentum (standard units of kg m/s) is equal to Planck’s constant 6.63 x 10 ^-34 J x (units in kg m^2/ s^2 x s= kg m^2/s Uncertainty principle: dx x dp=h Position uncertainty : 2 * 1.57 Angstroms= 3.14 x 10^-10 m, the diameter of the circle on which the electron orbits. This applies an uncertainty of momentum uncertainty: Dp=h/dx (6.63* 10 ^-34 kg m^s/s) / (3.14 x 10^-10m)= 4.222 x 10^-34 kg m/s Uncertain velocity: Dv=dp/m= 4.222 x 10 ^-24 kg m/s/ 9.11 x 10^-34 kg)= .4634 x 10 ^6 m/s Electron KE: .5mv^2: .5(9.11 x 10 ^-31 km)x (.4634 x 10^6m/s)^2= .9781 x 10^-19 J Time to orbit= circumference / velocity= 2 pi x 1.57 x 10^-10m / (.4634 x 10^6 m/s)^2/ (1.57 x 10 ^-10m) = .1367 x 10 ^22 m/s ^2 Centripetal force= m x centripetal acceleration= (9.11 x 10 ^-31 kg) x (.1367 x 10 ^22 m/s ^2)= 1.245 x 10^-9 N Magnitude of the force of attraction between proton and electron is: Coulomb force= k q1q2/ r2= 9* 10^9 N m^2/ C2 x (1.6 x 10 ^-19 C)^2 / (1.57 x 10 ^-16m )^2= 9.347 x 10 ^-9 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!