Exam 2 Spring 2006 Solutions

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

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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

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Physics 221 2006 S Exam 2 Solutions - 2 - [31] (C) The total mechanical energy is the sum of potential and kinetic energy so 22 11 (100 )(0.3 ) (2 )(3 ) 4.5 9 13.5 Nm ms EUK k x m v mk g JJ J =+= + =+ = [32](C) Since p mv = and 2 1 2 Km v = so that 2 K p v = . Going through each of the four particles: 00 24 2 K K PQ pp KK RS vv == Thus QPSR vvvv >=> [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 =− . Another way to do it is to note that if we plug t=0 into the general expression where x=1m we get () 1 23 1 cos( ) (2 )cos( ) arccos mA m ∴ =± to resolve the trig ambiguity note that the velocity is positive and since (0 ) s i n ( ) vt A only the 3 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
Physics 221 2006 S Exam 2 Solutions - 3 - 22 11 0 (10 )(2 ) 20J N m EKU k x 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

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Physics 221 2006 S Exam 2 Solutions - 4 - [38](A) Taking the zero of potential at equilibrium, at that position the mechanical energy is 01 0 J1 0 J EUK =+=+ = . At the maximum of the oscillation the kinetic energy is zero. Thus 2 1 2 22 ( 1 0 ) 45cm 100 N m Ek x EJ 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 12 FFW + = . Taking the torque condition about the left end of the plank, 21 2 32 3 2 4 0 45 net LF LW FW N τ By the force condition 1 15 FN = . [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 mm ss m cm s kg i kg vi kg kg + + [41](A) The moment of inertia of the horizontal rod is 2 11 33 (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 (5 )(2 ) 20 v I mL kg m kg m = thus the total moment of inertia is 2 26.7 vh I II k g m =+= . The rotational kinetic energy is therefore: 1 2 (26.7 )(3 ) 120 KI k g ms J ω = [42](B) The angular momentum is ˆˆ ˆ ˆ (3 2 ) ( ) Js ˆ [(2 1 ( 1 1)) (1 ( 1) 3 1) (3 ( 1) 2 ( 1)) ]Js ˆ (3 4 )Js Lrp i jk i jk ij k k =× = + + ×−−+ = ⋅−−⋅ + ⋅− − ⋅ + ⋅− − ⋅− =− G G G
Physics 221 2006 S Exam 2 Solutions - 5 - [43](C)

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## This homework help was uploaded on 03/27/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Fall '08 term at Iowa State.

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

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