ࡱ> %'$5@ }bjbj22 36XXY #6666j j j j \ G qqq$ROTqTT667,,,T6l,T,,J,J pb=j |J0GJJ6666J|q",qqq  Mechanics AP Physics C 1995 1.  EMBED Equation.DSMT4  (a) Area = 4 + 4 + 4 = 12 Ns (b)  EMBED Equation.DSMT4  (Use J=12 on the 5kg ball...)  EMBED Equation.DSMT4  (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)  EMBED Equation.DSMT4  EMBED Equation.DSMT4  (d)  EMBED Equation.DSMT4  = 169  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  So the kinetic energy lost was:  EMBED Equation.DSMT4  (e) Time to floor (using d = at2 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:  EMBED Equation.DSMT4  2. The key to these potential energy graph/function problems is:  EMBED Equation.DSMT4  (a biggie!) ex/ Recall Uspring = kx2 and notice:  EMBED Equation.DSMT4 . Well, getting back to our problem (where r = x)...  EMBED Equation.DSMT4  (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... (i)  EMBED Equation.DSMT4  (Graph shows us the minimum.) (ii)  EMBED Equation.DSMT4  (b) Now Fnet =  EMBED Equation.DSMT4  (Now graph!) This is a rational function such as y = 1/x2 shifted down say 3 as in y =  EMBED Equation.DSMT4    #2 continued on the next page... Mechanics AP Physics C 1995 p. 2 2. (continued) (c) We'll use the Work-Energy Theorem:  EMBED Equation.DSMT4  where the work done will equal the decrease in potential energy or  EMBED Equation.DSMT4  with:  EMBED Equation.DSMT4  (where above, we found r0 = b)  EMBED Equation.DSMT4  (where above, we found U0 = 2a with U0 is Uf here) So the work done is:  EMBED Equation.DSMT4  (Wait! in terms of U0???) Oh yeah, U0 = 2a, so a = U0 so W =  EMBED Equation.DSMT4  (d) Using the Work-Energy Thm again, we notice that since we started from rest... To come to rest again we'll need  EMBED Equation.DSMT4 = 0J or Uf = Ui . Ui =  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , so we can equate these to get:  EMBED Equation.DSMT4  (e) The particle accelerate to the right until r = r0. Then acceleration will be back to the left bringing the particle rest when U = EMBED Equation.DSMT4  again. It returns and repeats the process forever. 3. In terms of Ra and Rb and T (the same for both masses)... (a) Using  EMBED Equation.DSMT4 ... and Newton's Universal Law of Gravity on mass 'A'... We get:  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , we get:  EMBED Equation.DSMT4  and using the boxed result above...  EMBED Equation.DSMT4  Notice that Ra + Rb is the force distance and Ra is the orbital radius or distance for Ma Solving for Mb ...  EMBED Equation.DSMT4  (c) Similar to (b) but with Mb's orbital radius Rb (not Ra), we'll get:  EMBED Equation.DSMT4  and solve for Ma ...  EMBED Equation.DSMT4  = 2Mb (d)  EMBED Equation.DSMT4  (e)  EMBED Equation.DSMT4  26 m/s 0.5kg 5kg | 2m 1.2m r0 y = -a/b Notice the vertical asymptote: x = 0 and the horizontal asymptote: y = -a/b and the x-intercept of r0  "+./FGHIJkmn    ) > A B Z վխviթdd```h hFAH*jhFAhFAEHUj#1B hFACJUVaJj4hFAhFAEHUj1B hFACJUVaJjhFAUhFAhe,Fjh@h@EHUj1B h@CJUVaJjh@Uh@jh@UmHnHujhe,FUmHnHu h615 h@5# KhC @ G H Q % K m|gd$a$gdgd@$a$gd@Y|Z \ g h  $ % < 꿲꣖zk^j7hZhZEHUj#1B hZCJUVaJjBhZhZEHUj1B hZCJUVaJj'hZhZEHUj1B hZCJUVaJj6 hZhZEHUj1B hZCJUVaJj hZhZEHUj1B hZCJUVaJhZjhZUh hH* < = > ? 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