Practice Exam 3 Solutions

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Physics

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Dec 6, 2023

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Short problems. Remember to show work for full credit. Problem 1: Magne5c force [3 pts.] The situa6ons depicted, labeled A through D, show the circular orbits of par6cles moving in a magne6c field. Circle all the labels that depict mo6on of posi6ve par6cles. Draw the forces on those charges. Problem 2: Spherical mirror [5 pts.] An object is placed in front of a mirror as shown. A. (3 pts.) Using ray tracing determine the loca6on of the image. Draw the image. B. (2 pts.) Is the image real or virtual? Explain. Problem 3: Energy released [3 pts.] In the nuclear reac6on below, determine how much energy is released, in MeV? H ! + He ! → H " + He # Atomic Mass (u) 2 H 2.014102 3 H 3.016049 3 He 3.016029 4 He 4.002603 f A B C D The image is virtual: rays only seem to originate from it. The energy released equals the decrease in rest energy: ! = −Δ% ' ! = (%( H " ) + %( He " ) − %( H ! ) − %( H # ).' ! = (3.016049u + 3.016029u − 2.014102u − 4.002603u )' ! = +0.015373 u ⋅ c ! = +0.015373 ⋅ 931.49 MeV = +14.3 MeV

Problem 4: Image of a lens [6 pts.] An object is located 20.0 cm from an unknown lens. The lens forms a real image that is twice the size of the object. A. (2 pts.) Is the image upright or inverted? Explain. B. [4 pts] Calculate the image posi6on and draw the image on the figure. Problem 5: Free neutrons [6 pts.] Free neutrons are unstable. They undergo beta decay into a proton with a half-life of about 3 minutes. A. (3 pts) In a sample with 1 × 10 $ free neutrons, how many protons would have been created aVer 12 minutes? B. (3 pts) Determine the number of decays per second in your ini6al sample. object Unknown lens here The sign of the magnifica9on % will tell us if the image is upright or inverted. Since the image is REAL then its image distance > $ must be POSITIVE. The object is also REAL and has a POSITIVE distance > % , giving: % = − > $ > % = − positive positive = − The magnifica9on is NEGATIVE; hence the image must be INVERTED. Since the image is twice as big as the object and inverted then % = −2 . Then % = − > $ > % −2 = − > $ 20 cm > $ = +40 cm. The 12 minutes are equivalent to about 4 half-lives. The number of neutrons leQ would be: F & = F ' G 1 2 H # = 10 ( 16 = 62,500 free neutrons Each decayed neutron creates a proton. Then the number of decayed neutrons is the difference F ' − F & . Then: F ) = F ' − F & = 1 × 10 ( − 62,500 = 937500 We need the 9me constant of the decay which can be calculated from the half-life as N = O */! ln 2 = 3 × 60 s 0.693 = 259.7 s The ac9vity of the sample is then: Q = F N = 10 ( 259.7 s = 3850.8 decays s Object distance: +20cm Image distance: +40cm Image is -2x the size of the object

Long problems. Remember to show work for full credit. Problem 6: Magne5c Fields by Currents [10 pts.] Two long wires carry the currents I 1 = 5A and I 2 = 10A in the direc6ons shown in figure. A. (3 pts.) What is the direc6on and strength of the magne6c field at point A in the figure? B. (3 pts) What is the direc6on and strength of the magne6c field at point B in the figure? C. (4 pts) What is the magnitude and direc6on of the force on an electron at point B moving to the right, parallel to the wires, with a velocity v=4.5 m/s? At point A: According to the right-hand-rule, current V * creates a magne9c field out of the page of magnitude: W * = X ' V * 2YZ = (1.26 × 10 ,- )(5 A) 2Y(0.02 m) = 5 × 10 ,. T . According to the right-hand-rule, current V ! creates a magne9c field into of the page of magnitude: W ! = X ' V ! 2YZ = (1.26 × 10 ,- )(10 A) 2Y(0.05 m) = 4 × 10 ,. T . The net field at point A is then W /01 = W * − W ! = 1 × 10 ,. T out of the page At point B: According to the right-hand-rule, current V * creates a magne9c field into of the page of magnitude: W * = X ' V * 2YZ = (1.26 × 10 ,- )(5 A) 2Y(0.02 m) = 5 × 10 ,. T . According to the right-hand-rule, current V ! creates a magne9c field also into of the page of magnitude: W ! = X ' V ! 2YZ = (1.26 × 10 ,- )(10 A) 2Y(0.01 m) = 20 × 10 ,. T . Since both point in the same direc9on, the net field at point A is then W /01 = W * + W ! = 25 × 10 ,. T into the page Since at point B the magne9c field points into the page and the par9cle moves right, the right-hand rule gives force on a POSITIVE charge as poin9ng up. For an electron instead the force will point downwards . The magnitude of the force would be: ] = |_|` 2 W = (1.6 × 10 ,*3 C) b4.5 4 5 c(25 × 10 ,. T) = 1.8 × 10 ,!! N . ! "⃗ for a posi)ve par)cle $ ""⃗ ⊗ & "⃗

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Problem 7: Refrac5on [12 pts.] A tank holds a layer of oil, 1.20 cm thick, ﬂoa6ng on a layer of syrup that is 0.83 cm thick. A light ray originates at point A at the bo^om of the tank and refracts into the oil at a point 0.90 cm to the right of A. The ray then arrives at the oil-air interface at the cri6cal angle, 2.00 cm to the right of point A. A. (3 pts.) According to the figure, what is the angle of incidence at the syrup-oil boundary? B. (3 pts.) According to the figure, what is the angle of refrac6on of the light into the oil? C. (3 pts.) Find the index of refrac6on of the oil ( ) %&' = 1.0 ). D. (3 pts.) Find the index of refrac6on of the syrup. 0.83 cm A 0.9 cm 2.0 cm 1.20 cm air oil syrup tan e * = 0.9 0.83 e * = tan ,* G 0.9 0.83 H = 47.3° tan e ! = 2 − 0.9 1.2 = 1.1 1.2 e ! = tan ,* G 1.1 1.2 H = 42.5° The angle of incidence at the air-oil interface is e * = 42.5° . Since this is the cri9cal angle then, according to Snell’s law: g 678 ⋅ sin(42.5°) = 1 ⋅ sin 90 = 1 g 678 = 1 sin(42.5°) = 1.48 At the syrup-oil interface the angle of refrac9on is 42.5° . Using Snell’s law: g 9:;<= sin(47.3°) = g 678 sin(42.5°) = 1 g 678 = 1 sin(47.3°) = 1.36 ' ! 0.9 cm 0.83 cm ' " 1.1 cm 1.20 cm

Problem 8: Imaging system [12 pts.] A. (2 pts.) First consinder a converging lens (L2). To determine its focal length, you place “Object A” 10cm in front of the lens. You find, that the magnifica6on factor for the image is -1/4. Determine the image distance and focal length of the lens. Also draw the image in the sketch below. B. (3 pts.) For the same lens, you find that the image of another “Object B” forms 3cm behind the lens. Calculate the object distance for Object B and indicate the object loca6on in the drawing below The magnifica9on is % = − > # > $ = − * # . So > $ = > $ # = 2.5'% . Find the focal length of the lens using the thin lens equa9on: * ? = * > $ + * > # with > % = 10'% and > $ = 2.5'% to find h = b * *'@4 + * !..@4 c ,* = b * *'@4 + # *'@4 c ,* = 2'% . Image distance: 2.5cm Focal length 2cm The prompt is then asking to find the object loca9on when the image distance is 3cm. It’s the same lens, so the focal length does not change. (The magnifica9on changes, since it depends on object and image distance.) Use the thin lens equa9on again: * ? = * >B $ + * >B # to find >′ % = b * ? − * > % # c ,* = b * !@4 − * "@4 c ,* = 6'% Object B distance: 6cm Focal length 2cm

Mark the posi%on you found for Object B in the drawing below. We will now insert an unknown lens (L1) 2cm in front of L2. L1 is chosen such that it creates a virtual image of Object A in the posi6on of Object B. C. (4 pts.) Write down the object and image distances for this lens. Then calculate the focal length of L1. What type of lens is this – converging or diverging? D. (3 pts.) Calculate the overall magnifica6on from Object A to the final image of the en6re imaging system. Object B loca9on = image loca9on for L1 4cm 8cm First, determine the object and image distance for the unknown lens L1: >" % = 10'% − 2'% = 8'% is the distance of Object A to lens L1. >" $ = −(6'% − 2'%) = −4'% is the distance of the image that the lens is supposed to form to L1. Note that since it is a virtual image, the image distance is nega9ve. Then find the focal length of the lens L1 using the thin lens equa9on: * ? &' = * >" $ + * >" # to find h D* = −8'% . A nega9ve focal length indicates a diverging lens. The total magnifica9on is the product of both magnifica9ons. For each lens, the magnifica9on is % = − > # > $ . For L1, > E,D* = −4cm and > G,D* = 8'%. Object A located 8cm from L1 forms a virtual image 4cm in front of L1. This becomes the object for L2 with at an object distance of 6cm and with an image distance of 3cm. % I%IJK = % D* × % D! = −> E,D* > G,D* × −> E,D! > G,D! = − −4'% 8'% × − 3'% 6'% = 1 2 × − 1 2 = − 1 4

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Problem 9: Nuclear decay [10 pts.] A Co !" #$ isotope undergoes beta minus decay with a half-life of 5.27 years. A. (2 pts.) In the following decay reac6on determine A and Z. Co "( $) → Ni * + , + 0 - + Energy B. (2 pts.) In a sample of 60 Co, what percentage of 60 Co remains aVer 2 years? C. (3 pts.) In 1 gram of pure 60 Co, there are 10 22 radioac6ve nuclei. Determine the ac6vity R (in decays per second) of this sample. D. (3 pts.) What is the ac6vity of this sample aVer 2 years? Under k , decay: l LM<N1O0; = l =M;0/1 + 1 and m LM<NO10; = m =M;0/1 . Then l = 27 + 1 = 28 m = 60 The number of unstable nuclei at any 9me is F(n) = F ' o ,I/P . The 9me constant N is N = O * ! ln 2 = 5.27 years ln 2 = 7.60 years. AQer 2 years then F(2 years ) = F ' o , ! :0M;9 -.(' :0M;9 = F ' 0.768. Then the percentage of remaining Co !- (' is F(2 years) F ' = 0.768 = 76.8% For a given number of unstable nuclei the ac9vity is simply Q = F/N : Q = F N = 1 × 10 !! 7.60 years = 1 × 10 !! 7.60 × 365 × 24 × 60 × 60 s = 4.17 × 10 *" decays s The 9me dependency of the ac9vity is similar to that of the number of unstable par9cles: Q(n) = Q ' o ,I/P Q(2 years) = 4.17 × 10 *" decays s o ,!/-.( = 3.2 × 10 *" decays s

Practice Exam 3 Solutions

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