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Unformatted text preview: Taylor 3.4 For the sake of definiteness let’s say that the cart moves in the the positive ˆ x direction, and the hobos in the negative ˆ x direction. We are told the magnitude of the hobos’ speed relative to the car so, using the notation where v A/B is the velocity of A relative to B , we have  v H/C  = u ⇒ v H/C = u ˆ x . (1) Use a couple of basic facts about relative velocities in order to find the velocity of the cart relative to the earth and the velocity of the hobo relative to the earth, ˆ v H/C = ˆ v H/E + ˆ v E/C ˆ v E/C = ˆ v C/E ⇒ ˆ v H/C = ˆ v H/E ˆ v C/E If I denote the hobo velocity relative to the ground as v h ˆ x ( v h is the speed) and the cart velocity relative to the ground as v c ˆ x ( v c is the speed), then the last equation above says u ˆ x = v h ˆ x v c ˆ x ⇒ v h = u v c (a) Conservation of momentum initial p = 0ˆ x (2) final p = 2 m h ( u v c )ˆ x + m c ( v c )ˆ x (3) Setting initial p equal to final p and rearranging v c = 2 m h 2 m h + m c u (4) (b) Break this answer into two parts. First solve for when just one hobo has jumped. Then using that information, solve for whenone hobo has jumped....
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 Fall '08
 Van,D
 Angular Momentum, Momentum, Velocity, MH

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