ch09-p104 - 104. (a) We use Fig. 9-22 of the text (which...

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Km v m m ii == = 1 2 1 2 475 113 22 (. ) . and the final kinetic energy is vm v m m fff =+= + = 1 2 1 2 1 2 350 200 81 1 2 2 2 (. ) (. ) . . c h Kinetic energy is not conserved. 104. (a) We use Fig. 9-22 of the text (which treats both angles as positive-valued, even though one of them is in the fourth quadrant; this is why there is an explicit minus sign in Eq. 9-80 as opposed to it being implicitly in the angle). We take the cue ball to be body 1 and the other ball to be body 2. Conservation of the x and the components of the total momentum of the two-ball system leads to: mv 1 i = mv 1 f cos θ 1 + mv 2 f cos 2 0 = – mv 1 f sin 1 + mv 2 f sin 2 . The masses are the same and cancel from the equations. We solve the second equation for
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