problem23_91

University Physics with Modern Physics with Mastering Physics (11th Edition)

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23.91: a) s m 700 kg 10 3.0 kg 10 0 . 6 s) m kg)(1300 10 3 ( ) s m 400 )( kg 10 6 ( 5 5 5 5 2 1 2 2 1 1 = × + × × + × = + + = - - - - m m v m v m v cm b) . ) ( 2 1 2 1 2 1 2 2 1 2 1 2 2 2 2 1 1 cm rel v m m r q kq v m v m E + - + + = After expanding the center of mass velocity and collecting like terms: . ) ( 2 1 ] 2 [ 2 1 2 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 1 r q kq v v μ r q kq v v v v m m m m E rel + - = + - + + = c) J. 9 . 1 m 0.0090 C) 10 5.0 C)( 10 0 . 2 ( s) m kg)(900 10 0 . 2 ( 2 1 6 6 2 5 - = × - × + × = - - - k E rel d) Since the energy is less than zero, the system is “bound.” e) The maximum separation is when the velocity is zero:
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Unformatted text preview: . m 047 . J 9 . 1 ) C 10 . 5 )( C 10 . 2 ( J 9 . 1 6 6 2 1 =-×-× = ⇒ =---k r r q kq f) Now using particles the so J. 6 . 9 : find we , s m 1800 and s m 400 2 1 + = = = rel E v v s. m 980 kg 10 . 2 ) J 6 . 9 ( 2 2 5 2 1 = × = =--μ E v v rel...
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This document was uploaded on 02/06/2008.

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