266_Physics ProblemsTechnical Physics

266_Physics ProblemsTechnical Physics - 268 Linear Momentum...

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268 Linear Momentum and Collisions Section 9.6 Motion of a System of Particles P9.45 (a) v v vv ij CM kg ms kg == + = −+ + m M mm M ii 11 22 2 00 2 00 3 00 3 00 1 00 6 00 500 .. ± . ± ± . ± . bg ej vi j CM =+ 140 240 . ± . ± (b) pv i j i j + M CM kg m s kg m s 5 00 1 40 2 40 7 00 12 0 ± . ± . ± . ± P9.46 (a) See figure to the right. (b) Using the definition of the position vector at the center of mass, r rr r ri j CM CM CM kg m 2.00 m kg m, m kg kg m = + + = +− + =− 12 200 100 300 400 , . . . . ± . ± af FIG. P9.46 (c) The velocity of the center of mass is v P j CM CM kg m s m s kg m s m s kg kg + + = + M 050 , . , . . ± . ± b g b g (d) The total linear momentum of the system can be calculated as Pv = M CM or as v Either gives Pi j 15 0 5 00 . ± . ± kg m s P9.47 Let x = distance from shore to center of boat A = length of boat ′= x distance boat moves as Juliet moves toward Romeo The center of mass stays fixed.
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