{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam 2 Spring 2006 Solutions

Exam 2 Spring 2006 Solutions - Physics 221 2006 S Exam 2...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 221 2006 S Exam 2 Solutions - 1 - Answer key 1 Solutions 2 Questions 9 31 C 41 A 51 A 32 C 42 B 52 B 33 D 43 C 53 D 34 C 44 A 54 E 35 B 45 C 55 E 36 B 46 A 56 E 37 E 47 D 57 C 38 A 48 C 58 E 39 E 49 B 40 B 50 D
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Physics 221 2006 S Exam 2 Solutions - 2 - [31] (C) The total mechanical energy is the sum of potential and kinetic energy so 2 2 1 1 2 2 2 2 1 1 2 2 (100 )(0.3 ) (2 )(3 ) 4.5 9 13.5 N m m s E U K kx mv m kg J J J = + = + = + = + = [32](C) Since p mv = and 2 1 2 K mv = so that 2 K p v = . Going through each of the four particles: 0 0 0 0 0 0 0 0 2 4 2 K K P Q p p K K R S p p v v v v = = = = Thus Q P S R v v v v > = > [33](D) The total kinetic energy is the same before and after an elastic collision so (D) is false. [34](C) The interval between t=2s and t=8s goes from a maximum to a minimum and so the 6s interval is half a period. The period is therefore T=12s. [35](B) Take the peak at t=2s as the “first peak”. It is 2s to the right of the origin which is 1/6 of a period so φ is negative and 3 (2 )(1/ 6) π φ π = − = − . Another way to do it is to note that if we plug t=0 into the general expression where x=1m we get ( ) 1 2 3 1 cos( ) (2 )cos( ) arccos m A m π φ φ φ = = = ± = ± to resolve the trig ambiguity note that the velocity is positive and since ( 0) sin( ) v t A φ = = − only the 3 π solution gives the correct velocity sign. (of course these solutions are modulo 2 π ) [36](B) When the mass is at a maximum displacement, then the kinetic energy is 0 so the mechanical energy is equal to the potential energy. Such a case is t =2s where x =2m so
Image of page 2
Physics 221 2006 S Exam 2 Solutions - 3 - 2 2 1 1 2 2 0 (10 )(2 ) 20J N m E K U kx m = + = + = = This mechanical energy does not change with time because in a simple harmonic oscillator the force is conservative. [37](E) This is an inelastic collision so momentum is conserved. The final momentum of the system is (3000 )(10 ) 30000 m s p kg Ns = = . All of this momentum must be carried by the initial car so its velocity is 30000 30 1000 m s Ns v kg = =
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Physics 221 2006 S Exam 2 Solutions - 4 - [38](A) Taking the zero of potential at equilibrium, at that position the mechanical energy is 0 10J 10J E U K = + = + = . At the maximum of the oscillation the kinetic energy is zero. Thus 2 1 2 2 2(10 ) 45cm 100 N m E kx E J x k = = = = [39](E) Let 1 F be the force of scale 1 and 2 F be the force of scale 2 with W the weight of the plank. The force condition in the vertical direction is 1 2 F F W + = . Taking the torque condition about the left end of the plank, 2 1 2 3 2 3 2 4 0 45 net LF LW F W N τ = = = = By the force condition 1 15 F N = . [40](B) The velocity of the center of mass does not change during the collision. We can find this velocity from the initial state: ˆ (200 )(10 ) (400 )(0 ) ˆ (3.3 ) 400 200 m m s s m cm s kg i kg v i kg kg + = = + [41](A) The moment of inertia of the horizontal rod is 2 2 2 1 1 3 3 (5 )(2 ) 6.6 h I mL kg m kg m = = = . All of the mass of the vertical rod is at a distance of 2m from the axis so 2 2 2 (5 )(2 ) 20 v I mL kg m kg m = = = thus the total moment of inertia is 2 26.7 v h I I I kg m = + = .
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern