Solution #1 f2007 - II . (16 points) ( This is the same...

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Unformatted text preview: II . (16 points) ( This is the same situation as problem I. ) Two balls are placed at opposite ends of the inclined plane of length L = 2 . 0m shown. At time t = 0s ball #1 at position x = 0m (top of plane) is released. At the same time ball #2 is given a kick up the plane so its initial speed is 4.0m/s. At what time do the two balls meet? ( On Earth. ) . . . . . . . . . . . . . . . . . . . . . . . Since both balls undergo constant acceleration, the constant accelera- tion kinematics equations can be used. x 1 = x 10 + v 10 t + 1 2 a 1 ( t ) 2 and x 2 = x 20 + v 20 t + 1 2 a 2 ( t ) 2 When the balls meet, they are in the same position at the same time. x 1 = x 2 x 10 + v 10 t + 1 2 a 1 ( t ) 2 = x 20 + v 20 t + 1 2 a 2 ( t ) 2 Although the accelerations are not known, they are the same ( a 1 = a 2 ), so the 1 2 a ( t ) 2 terms can be subtracted from each side. x 10 + v 10 t = x 20 + v 20 t The initial position of ball #1 is zero, and the initial velocity of ball #1 is zero. The initial position of ball #2 is L . Note that the initial velocity of ball #2 is -4.0m/s, as its initial speed is 4.0m/s in the negative direction. 0 + 0 = L + v 20 t t =- L v 20 =- 2 . 0m- 4 . 0m / s = 0 . 50s 1. (6 points) Which direction is ball #2 travelling at the instant the balls meet?...
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This note was uploaded on 04/09/2008 for the course PHYSICS 2211 taught by Professor Uzer during the Spring '08 term at Georgia Institute of Technology.

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Solution #1 f2007 - II . (16 points) ( This is the same...

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