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Mechanics – AP Physics C – 1995 1. J F t F (t ) dt (a) Area = 4 + 4 + 4 = 12 Ns (b) J p (Use J=12 on the 5kg ball...) 26 m/s 0.5kg | 5kg 2m 12 5v f 5(0) v f 2.4 m s 1.2m (c) For the 0.5kg cube in collision with... p0 = pf (Using conservation of momentum for the 2-mass system...) .5(26) + 5(0) = .5vcf + 5(2.4) (i ) 13 .5vcf 12 vcf 2 m s (ii ) right (d) K K f Ki 1 (.5)262 2 1 (.5)22 2 12 5(2.4)2 = 169 1 14.4 = 153.6 J So the kinetic energy lost was: 153.6 J (e) Time to floor (using d = ½at2 or 1.2 = 4.9t2 or 5t2 with g = 10m/s2...): t = .495 or .490 s. Now (using d = rt): dcube = 2(.49) =0.98m vs dball = 2.4(.49) = 1.176m. Subtracting we get a difference in landing points of: .196m 2. The key to these potential energy graph/function problems is: dU F ( x) (a biggie!) dx dU 12 k 2 x kx Fs ( Hooke ' s Law) . dx ab a Well, getting back to our problem (where r = x)... U (r ) ba r b r abr 1 r (a) Potential Energy (U) is at a minimum when r = r0 but in terms of 'a' and 'b'... ??? Well, let's do some calculus and see what we get... dU a a ab (i) ab r 2 0 2 r 2 b2 r0 b (Graph shows us the minimum.) dr b b r a ab (ii) U (r0 ) U (b) (b) a a 2a b (b) dU a a ab ab a ab r 2 2 2 F (r ) (Now graph!) (b) Now Fnet = dr b b b r r 1 This is a rational function such as y = 1/x2 shifted down say 3 as in y = 2 3 x ex/ Recall Uspring = ½ kx2 and notice: Notice the vertical asymptote: x = 0 and the horizontal asymptote: y = -a/b and the x-intercept of r0 r0 y = -a/b #2 continued on the next page... Mechanics – AP Physics C – 1995 p. 2 2. (continued) (c) We'll use the Work-Energy Theorem: Wnet K where the work done will equal the decrease in potential energy or W U (U f Ui ) Ui U f with: Ui U U ab a2 2a 52a r0 2 b 2 a b b 2 b 2 (where above, we found r0 = b) U f U r0 2a (where above, we found U0 = 2a with U0 is Uf here) 5 1 So the work done is: W U i U f a 2a a (Wait! in terms of U0???) 2 2 Oh yeah, U0 = 2a, so a = ½U0 so W = 14 U 0 (d) Using the Work-Energy Thm again, we notice that since we started from rest... To come to rest again we'll need Wnet K = 0J or Uf = Ui . ab ab , so we can equate these to get: 52 a ba r r r (e) The particle accelerate to the right until r = r0. Then acceleration will be back to the left bringing the particle rest when U = 52 a again. It returns and repeats the process forever. 3. In terms of Ra and Rb and T (the same for both masses)... v2 (a) Using Fc m mac ... and Newton's Universal Law of Gravity on mass 'A'... R M bM a Mb M a a G We get: G a c c Ra Rb 2 Ra Rb 2 Ui = 5 a 2 and U (r ) ba r 4 2 2 v2 2 2 (b) ac R and , we get: ac R R 2 and R T T T 2 Mb 4 R using the boxed result above... G a T2 Ra Rb 2 Notice that Ra + Rb is the force distance and Ra is the orbital radius or distance for Ma 2 M b Ra Solving for Mb ... 4 2 Ra Rb 2 GT 2 (c) Similar to (b) but with Mb's orbital radius Rb (not Ra), we'll get: G Ma Ra Rb 2 4 4 2 Rb 2 and solve for Ma ... M a Rb T (d) I M a Ra2 M b Rb2 (e) L I M a Ra2 M R 2 T 2 b b 2 T 2 Ra Rb 2 GT 2 = 2Mb