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Ball “A” and Ball “B” have the same mass- 1.0 kg. Ball “A” is moving toward Ball “B”. Ball “B” is stationary. Ball “B”. Ball “A”. When “A” travels at 2.0 m/s and strikes “B”, Ball “B” moves upward at 35 o and Ball “A” moves downward at 55 o.

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## Ball “A”

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**Ball “A” and Ball “B” have the same mass- 1.0 kg**Ball “A” is moving toward Ball “B”. Ball “B” is stationary Ball “B” Ball “A”**When “A” travels at 2.0 m/s and strikes “B”, Ball**“B” moves upward at 35o and Ball “A” moves downward at 55o . Whatever momentum “A” has before the collision A’ and B’ share after they collide rb’ ra 35o 55o ra’ mva + mvb = mva’ + mv2’ 1.0 kg x 2.0 m/s + 0 = mva’ + mvb’ 2.0 kg m/s = ra’ + rb’**rB’**rB’ rA 35o 35o rA’ rA rA’ rb’ = Cos35 x 2.0 kg m/s rb’ = 1.6 kg m/s rA’ ra’ = Sin 35 x 2.0 kg m/s rB’ ra’’ = 1.1 kg m/s rA 35o**rB’**rA 35o rA’ rB’ 35o rA rb’ = Cos35 x 2.0 kg m/s rA’ rb’ = 1.6 kg m/s vb’ = rb/mb vb’ = 1.6 kg m/s / 1 kg = 1.6 m/s rA’ rB’ ra’ = Sin 35 x 2.0 kg m/s rA ra’’ = 1.1 kg m/s 35o va’ = ra/ma va’ = 1.1 kg m/s / 1 kg = 1.1 m/s**rBx’ = Cosq1rB’**rA = rBx’ + rAx’ rBy’ = Sinq1rB’ rBy’ rB’ rBx’ q1 rA q2 rAx’ rA’ rAy’ rAx’ = Cosq2rA’ rAy’ = Sinq2rA’ 0 = rBy’ + rAy’**p = p’**mvA = mvA’Cosq2 + mvB’Cosq1 vBy’ vB’ vBx’ q1 vA q2 vAx’ vA’ vAy’**Two balls of equal mass strike each other. Ball A moves at**5.20 m/s and strikes ball B, which is stationary. After the collision ball A travels above the original path at an angle of 21.00. What is the “p” and “v”s of each ball after the collision? The mass is 2.0 kg. • rA’ = Cosq1rA rA’ = Cos 21.00 x 10.40 m/s rA’ = 9.7 kg m/s • rA’ = mvA’ vA’ = 9.7 kg m/s / 2.0 kg = vA’ = 4.85 m/s 2. rB’ = Sinq1rB rB’ = Sin21.00 x 10.40 m/s rB’ = 3.73 kg m/s rB’ = mvB’ vB’ = 3.73 kg m/s / 2.0 kg = vB’ = 1.86 m/s

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