ch09-p065 - collision (at x = 0 and t = 0): v 1 f = m 1 m 2...

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and particle 2 is “gaining” at a rate of (10/7) m/s leftward; this is their relative velocity at that time. Thus, this “gap” of 23 cm between them will be closed after an additional time of (0.23 m)/(10/7 m/s) = 0.16 s has passed. At this time ( t = 0.82 + 0.16 = 0.98 s) the two particles are at x = (–2/7)(0.98) = –28 cm. 65. We use Eq 9-67 and 9-68 to find the velocities of the particles after their first
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Unformatted text preview: collision (at x = 0 and t = 0): v 1 f = m 1 m 2 m 1 + m 2 v 1 i = 0.1 kg 0.7 kg (2.0 m/s) = 2 7 m/s v 2 f = 2 m 1 m 1 + m 2 v 1 i = 0.6 kg 0.7 kg (2.0 m/s) = 12 7 m/s 1.7 m/s. At a rate of motion of 1.7 m/s, 2 x w = 140 cm (the distance to the wall and back to x = 0) will be traversed by particle 2 in 0.82 s. At t = 0.82 s, particle 1 is located at x = (2/7)(0.82) = 23 cm,...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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