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# 101e304 - Physics 171.101 Exam 3 1 SOLUTIONS Prof Barnett...

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Unformatted text preview: Physics 171.101 Exam 3 April 15, 20071 SOLUTIONS Prof. Barnett You may use anything written on the front and back of ONE 3”x 5” index card. Everything should be written in ink and no “snow paint” should be used. ‘7 - '9 '7 Use g = 10 Ill/86C“, 7T : 3, [hoop : mr‘, [dz-5C : £1717". 1. A 5 kg, mass with velocity 171 : 10% m/sec collides head on with a stationary 10 kg mass. After the collision the velocity of the 5 kg mass is VF : (2.0% + 1.03) m/sec. A) (9 pts) What is the velocity of the 10 kg mass, 1710, after the collision? Use momentum conservation. V ' ' :_-..,.._.(-_i, W i t " _ _" 't‘ 1 _ ﬁfinal ﬁfinal Ptotal—PfinZMZ—PS +1310 “final _ “initial “final P10 0 _ P5 ifrmr/ ::(5 kg) X (vgnnuu _ L5 ) : (50 i — 10 i — 5 .5) kg m/sec : (40 i — 5 j) kzg III/511' 1716mm : ﬁﬁml/m kzg : (4 / — 0.5 j) vii/sw- B) (8 pts) What is the velocity of the center of mass of this system before and after the collision? 1757‘?“ : Einitial/Mtoml = (50 kg m/sec)/(15 kg) : 3%; m/sec, total *final __ Vinitial m 31‘- CM — CM — 32 771/360. C) (8 pts) 1s mechanical energy conserved during this collision? If not, by how much does the mechanical energy change? lli/[Einitial I éjllgﬂ/sg Z %(5)(10)2 : 250 JOUlGS xiii/[Efinal : é‘A/[sl/SQ ‘i’ %A[101/120 ll [Oh—I (5 mm2 +1‘2)+§(10)(42 + (0.5)?) : 93.75 J MECHANICAL ENERGY IS NOT CONSERVED. AAJE = 93.75 J— 250 J 2 ~15625 J 4. A transverse sinusoidal wave on a string has the form: y(:r,t) : 0.3 sin(47mr + 807W) cm with the dimensions of :r and 75 being meters and secon ls. A piece of the string that is 50 cm long has a mass of 0.25 grams. A) (5 pts) What is the phase velocity of this wave? Specifically indicate the tlll‘t‘t‘ilUU the wave is trawling: toward Lt or toward '7!“ ”0pm“ 2 (80 7r)/(4 7r) : ‘20 772/566 toward —x direction. B) (5 pts) What is the maximum displacement of the string? ymax : 0.3 cm. C) (5 pts) What is the wavelength of the wave? 47r = 27r/A -+ A = 0.5 meters. D) (5 pts) What is the maximum speed of an atom in the string as it vibrates back and forth due to the passage of the wave? %% : (0.3 cm)(80 7r) cos(' . -) lily? max = (03 cm)(80 7r) 2 72 771/560 (using 7? : 3) E) (5 pts) What is the tension of the string? ”phase = T/h ———> T : M 400(m/s)2 ,u = (0.25 ><10ﬁ3 kg)/(0.5 m) = 0.5 ><10‘3 kg/m. T = 0.2 Newtons. ...
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101e304 - Physics 171.101 Exam 3 1 SOLUTIONS Prof Barnett...

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