Kmvmmii===121247511322(. ). and the final kinetic energy is vmvmmfff=+=+=121212350200811222(. ) (. ). .chKinetic 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 xand the components of the total momentum of the two-ball system leads to: mv1i= mv1fcos θ1+ mv2fcos 20 = –mv1fsin 1+ mv2fsin 2.The masses are the same and cancel from the equations. We solve the second equation for
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