Physics 111. Optional Extra Credit Assignment. Assigned Wednesday, March 9, 9:00 am. Due on or before Friday, March 11, 8:00 am

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1 Physics 111 Optional Extra Credit Assignment Assigned Wednesday, March 9, 9:00 am Due on or before Friday, March 11, 8:00 am Name Problem #1 /20 Problem #2 /20 Problem #3 /20 Problem #4 /20 Problem #5 /20 Total /100 Extra Points on Final Total /5

2 RulesofEngagement: 1. This optional extra credit assignment is assigned on Wednesday, March 9, This optional extra credit assignment is due on or before Friday, March 11, 2011 at 8 am. 3. Late extra credit assignments will not be looked at or graded and no extra credit points will be assigned. 4. This optional extra credit assignment will award you up to 5 additional points on top of your final class score. You are under no obligation to do any of this assignment. 5. You may do some or all of the assignment. Any amount you do will earn you some number of extra points on your final class score. 6. I will grade this optional assignment and the percent you earn on this assignment will be multiplied by 5 to determine the number of extra credit points you will earn on your final score. 7. I will determine grades for the class as if this assignment was not present and after assigning grade ranges will then add your extra credit points onto your final score. This has the potential to raise your grade by one letter grade. For example, if your final grade were at the class average of a B-, these extra points could potentially bring your grade up to a B. 8. Since this assignment is optional, you do not have to do it and your final grade will be whatever you earn. 9. This optional assignment can only help you, not hurt you. 10. You must work independently and consult no one on the preparation of this optional extra credit assignment. 11. You may consult only your textbook, class notes, or class homework assignment solutions for guidance. 12. You may not consult tutors of any kind, other faculty, other students, or the Internet for solutions. 13. I will treat this optional assignment as if it were an exam. I will answer questions, but will not help you with the solutions. 14. This optional assignment is also good preparation for the final exam, as many of these same topics will be covered. FreeResponseProblems:Thefiveproblemsbelowareworth100pointstotaland eachseparateproblemisworth20points.pleaseshowallworkinordertoreceive partialcredit.ifyoursolutionsareillegibleorillogicalnocreditwillbegiven.a numberwithnoworkshown(evenifcorrect)willbegivennocredit.pleaseusethe backofthepageifnecessary,butnumbertheproblemyouareworkingon.

3 1. A particular kind of mass spectrometer is shown below. Carbon from a sample is ionized in the ion source at the left. The resulting singly ionized 12 6 C + and 14 6 C + ions have negligibly small initial velocities and can be considered at rest. They are accelerated through the potential difference ΔV 1. They then enter a region where the magnetic field has a fixed magnitude of B = 0.2T. The ions pass through electric deflection plates that are 1cm apart and have a potential difference ΔV 2 that is adjusted so that the electric deflection and magnetic deflection cancel each other for a particular isotope: one isotope goes straight through, and the other isotope is deflected and misses the entrance to the next section of the spectrometer. The distance from the entrance to the fixed ion detector is a distance of 20cm. There are controls that let you vary the accelerating potential ΔV 1 and the deflection potential ΔV 2 in order that only 12 6 C + or 14 6 C + ions go all the way through the system and reach the detector. You count each kind of ion for fixed times and thus determine relative abundances. The various deflections insure that you count only the desired type of ion for a particular setting of the two voltages.

4 1a. Which accelerating plate is positive (left or right), which deflection plate is positive (upper or lower) and what is the direction of the magnetic field? Be sure to explain your reasoning for the choices you make. Accelerating plates: Left is + and right is so that the charges are accelerated across the gap. Deflection plates: Upper is + and lower is so that the electric field (and the electric force on a positively charged particle) points vertically down and points opposite to the magnetic force. The positively charged particle bends upwards in the magnetic field, by the RHR the magnetic field has to point into the page. 1b. Derive expressions for ΔV 1 and ΔV 2 in terms of the given quantities and the masses and charges of the ions. To determine ΔV 1, work is done accelerating charge q across the accelerating region. qδv 1 = 1 mv 2 2 ΔV 1 = mv 2 where the velocity is determined from the 2q bending of the charged particle in the far right external magnetic field. The velocity is thus F B = qvb = m v 2 R v = qrb. Thus we get m ΔV 1 = mv 2 2q = m 2q qrb m 2 = qr2 B 2 2m. To determine ΔV 2 we equate the electric and magnetic forces in the deflection region. We have F B F E = 0 F B = F E qvb = qe = q ΔV 2 ΔV 2 = Blv = Bl qrb = qrlb2 l m m 1c. What are the appropriate values for ΔV 1 and ΔV 2 for 12 6 C + and 14 6 C + respectively? (The spectroscopic masses for 12 6 C + and 14 6 C + are u and u respectively.) For 12 C ( ) 2 ( 0.2T) 2 ΔV 1 = qr2 B 2 2m = C 0.1m u 1.66 =1606V =1.6kV kg 1u ( ) 2 ΔV 2 = qrlb2 m = C 0.1m 0.01m 0.2T u kg 1u = 321.3V = 0.32kV

5 For 14 C ΔV 1 = qr2 B 2 2m = C ( 0.1m) 2 ( 0.2T) u 1.66 =1377V =1.4kV kg 1u ( ) 2 ΔV 2 = qrlb2 m = C 0.1m 0.01m 0.2T u kg 1u = 275.3V = 0.28kV 1d. Suppose that your sample is actually a piece of cloth from an old artifact. If is found to contain only 6.0% of 14 6C that a sample of newly made cloth contains, how old is your sample? N( t) = 0.06N o = N o e λt t = 1 ln 0.06 λ ( ) = 5730yr ln( 0.06) = 23,262yrs ~ 23000yrs

6 2. The Bell helicopter shown below has two blades extending out from a hub at the center of the aircraft. Each blade has a length from the hub to the tip of the rotor of 6.7m. The hub and blade assembly rotates at 6 rev/sec. blade tip hub 2a. What is the angular velocity of one of the blades and what is the translational velocity of the tip of the blade? ω = 2π rad 6 rev rev s = 37.7 rad s v tip = Rω = 6.7m 37.7 rad s = m s 2b. What is the induced potential difference between the blade tip and the center hub if the Earth has a component of its magnetic field 50µT and is pointing vertically? There are at least two ways that you can do this problem. One way: Since all parts of the blade do not have the same translational velocity, we need to determine the average velocity to use in the equation for the motional emfε = Blv average = Bl v tip + v hub = Bl v tip = T 6.7m m s = 0.042V Second way: The blade sweeps out some fraction of a full circle in a time Δt. The fraction of the circle that is swept out in this time Δt is given as θ swept out 2π = ωδt. Using Faraday s law we have 2π ωδt ε = Δ( BA) = B ΔA Δt πl 2 Δt = B 2π = Δt ε = 1 2 Bl2 ω = T 6.7m ( ) 2 ( 37.7 rad ) = 0.042V 2c. What is the magnitude of the electric field that is induced over the blade? E = ΔV Δx = 0.042V 6.7m = V m s

7 3. The Physics of Rainbows: Suppose that you are standing outside with your back to the sun just after a rainstorm and you are looking at the rain clouds off in the distance. Consider white light (white light is composed of the colors red, orange, yellow, green, blue, and violet) incident on a spherical raindrop located in the storm clouds in front of you and that the light is incident on the raindrop at an angle of 30 o as shown in the figure below. Each of the colors has its own frequency of oscillation and the index of refraction of a material varies slightly with frequency. This is known as dispersion. When white light passes from air into a material with a higher refractive index, the material of higher refractive index disperses the white light into its component colors. Thus the different colors of light get bent by varying amounts in the material and we have a rainbow of color produced, in this case in the raindrop, and this color separation is maintained as the colors exit the raindrop and we see a rainbow of color in the sky. Assume that the indices of refraction for water for red and blue light are n red = and n blue = respectively and that a typical light ray (orange) is also shown below.

8 3a. Suppose that red and blue light enter the drop at the 30 o angle shown. Determine the angles of refraction for red and blue light as well as which color (red or blue) gets bent more. On the raindrop above draw and label the rays corresponding to red and blue light in relation to the orange ray shown. Sincetheseanglesaremeasuredwithrespecttothenormal,bluelightbendsmorethanred. 3b. What are the critical angles (θ c,r and θ c,b ) for red and blue light striking the back surface of the raindrop? Draw and label on the raindrop above the rays corresponding to red and blue light in relation to the orange ray shown reflecting off of the back surface. (Note: A rainbow of light is formed from many such drops with the red light emerging at the same angle from each drop and adding together to form a bright red band, with similar effects for the other colors.) 3c. Draw the distribution of colors when the light leaves the bottom of the water droplet? Try to be as accurate as possible with your sketch and make sure you show the ordering of the colors and explain your choice for that ordering. On the drawing.

9 4. Given the circuit shown below answer the following questions. 5Ω 3Ω 6Ω 7Ω 12V 4Ω 5Ω 4a. What is the equivalent resistance of the circuit? R right branch = 5Ω + 5Ω + 1 6Ω + 1 7Ω R left branch = 3Ω + 4Ω = 7Ω 1 R eq = + R right branch 1 R left branch 1 1 =13.2Ω 1 = 13.2Ω + 1 7Ω 4b. What is the total current supplied by the battery? I total = V R eq = Ω = 2.6A 1 = 4.6Ω 4c. What are the potential drop across and the current through the 6Ω resistor? I right branch = V right R right branch = 12V 13.2Ω = 0.91A V 6 = V 7 =12V 2V 5Ω =12V 2 I right branch R 5Ω V 6 = I 6 R 6 I 6 = V 6 R 6 = 2.9V 6Ω = 0.48A ( ) =12V A 5Ω ( ) = 2.9V

10 4d. What is the power dissipated across the 3Ω resistor? I left branch = I total I right branch = V left branch = 12V R left branch 7Ω =1.7A P 3 = I 2 3 R 3Ω = (1.7A) 2 ( 3Ω) = V 2 3 = ( 5.1V )2 R 3Ω 3Ω = I 3V 3 =1.7A 5.1V = 8.7W 4e. Suppose that we represent your body as a resistive circuit and that currents between 0.1A and 1.0A can cause ventricular fibrillation (uncontrolled beating of the heart which can lead to death.) Suppose that the resistance of a single foot on wet soil is 200Ω and the resistance of the body is 1400Ω, and you are standing on wet ground, what is the current that flows through the body if the potential difference between your feet was 1500V? Is this enough of a potential difference to cause your heart to stop? Yourfeetareinserieswithyourbodysotheequivalentresistanceisthesum ofthebody sresistanceandtwicetheresistanceofafoot,or1800ω.thus thecurrentthatflowsisgivenbyohm slawv = IR I = V R = 1500V 1800Ω = 0.83A, andthisisenoughtocauseaheartattack.

11 5. Age of the Iceman: Early one afternoon in the fall of 1991 a German couple were hiking in the Italian Alps and noticed something brown sticking out of the ice 8 10m ahead. At first they took the object to be a doll or some rubbish, but upon closer inspection discovered it to be the body of a person trapped in the ice, with only part of the body exposed above the snow. Subsequent investigation revealed the remarkably well-preserved body of a stone-aged man, shown below, who had died in the mountains and had become entombed in the ice. When 14 C-dating methods were applied to the remains of the iceman it was found that the 14 C activity was 0.121Bq per gram of carbon. 5a. What is the decay constant that characterizes the decay of 14 C given the half-life of 14 C is 5730 yrs? λ = = yr t yr = yr s = s 1 5b. What was the initial activity of the sample when the person died? ( ) 1g A o = λn o = s 1 1mol 12 6 C 12g atoms = 0.248Bq 1mol 5c. How long ago did the iceman die? ( ) = A o e λt t = 1 λ ln A( t) A t A 0 1yr = ln 4 = 5930yr

12 ElectricForces,FieldsandPotentials ElectricCircuits LightasaWave F = k Q Q 1 2 r ˆ 1 r 2 c = fλ = ε o µ o E = F q E Q = k Q r r ˆ 2 PE = k Q 1Q 2 r V ( r) = k Q r E x = ΔV B,A Δx W A,B = qδv A,B MagneticForcesandFields F = qvb sinθ F = IlB sinθ τ = NIAB sinθ = µb sinθ PE = µb cosθ B = µ 0I 2πr ε induced = N Δφ B Δt Constants g = 9.8 m s Nm G = kg 2 1e = C k = 1 = πε o C 2 Nm 2 = N Δ ( BA cosθ ) Δt ε o = Nm 2 C 2 1eV = J µ o = 4π 10 7 Tm A c = m s h = Js m e = kg = 0.511MeV c 2 m p = kg = 937.1MeV c 2 m n = kg = 948.3MeV c 2 1amu = kg = 931.5MeV c 2 N A = Ax 2 + Bx + C = 0 x = B ± B2 4AC 2A Physics111EquationSheet Geometry Circles : C = 2πr = πd A = πr 2 Triangles : A = 1 2 bh Spheres : A = 4πr 2 V = 4 3 πr3 S( t) = energy time area = cε E 2 o t I = S avg = 1 cε E 2 2 o max P = S c = Force Area S = S o cos 2 θ v = 1 εµ = c n θ inc = θ refl n 1 sinθ 1 = n 2 sinθ 2 1 f = d o d i M = h i h o = d i d o M total = N i=1 M i = c B 2 max 2µ 0 ( ( ) = c B2 t) µ 0 dsinθ m = d tanθ m = d y m D = mλ or ( m + 1 2)λ asinφ m' = atanφ m' = a y m' D = m'λ LightasaParticle&Relativity NuclearPhysics Misc.Physics110 Formulae F = Δ p Δt = Δ( mv) = ma Δt F = ky F C = m v 2 R r ˆ W = ΔKE = 1 m v f v i PE gravity = mgy PE spring = 1 2 ky 2 x f = x i + v ix t a xt 2 v fx = v ix + a x t v 2 vx = v 2 ix + 2a x Δx ( ) = ΔPE