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Solution #1 f2007

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

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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 . 0 m shown. At time t = 0 s ball #1 at position x = 0 m (top of plane) is released. At the same time ball #2 is given a kick up the plane so its initial speed is 4.0 m/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 Δ t + 1 2 a 1 t ) 2 and x 2 = x 20 + v 20 Δ 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 Δ t + 1 2 a 1 t ) 2 = x 20 + v 20 Δ 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 Δ t = x 20 + v 20 Δ 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 -4.0 m/s, as its initial speed is 4.0 m/s in the negative direction. 0 + 0 = L + v 20 Δ t Δ t = - L v 20 = - 2 . 0 m - 4 . 0 m / s = 0 . 50 s 1. (6 points) Which direction is ball #2 travelling at the instant the balls meet? . . . . . . . . . . . . . . . . . . . . . . . The direction of travel for ball #2 would be determined by the sign of its velocity at the time Δ t found above.
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Solution #1 f2007 - II(16 points(This is the same situation...

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