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Published byRebecca Ritchie Modified over 3 years ago

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Ball A Ball B Ball A is moving toward Ball B. Ball B is stationary Ball A and Ball B have the same mass!

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When A strikes B, Ball B moves upward and Ball A moves downward. Whatever momentum A has before the collision A and B share after they collide vAvA v B v A KE = KE v 1 2 + v 2 2 = v 1 2 + v 2 2 If v 2 = 0 then, v 1 = v 1 2 + v 2 2

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vAvA v B v A vAvA v B v A vAvA v B v A v B = Cos 1 v A v A = Sin 1 v A

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vAvA v B v A v Bx v By v Ax v Ay v Bx = Cos v B v By = Sin v B v Ax = Cos v A v Ay = Sin v A v A = v Bx + v Ax 0 = v By + v Ay

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vAvA v B v A v Bx v By v Ax v Ay p = p mv A = mv A Cos 2 + mv B Cos 1 m 1 = m 2 v A = v A Cos 2 + v B Cos 1

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1 2 vAvA v B p = p On y axis 0 = Sin 1 v B + Sin 2 v A Sin 2 = -Sin 1 v B /v A v a vAvA v B v A v Bx v By v Ax v Ay

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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 21.0 0. What is the p vs of each ball after the collision? 1. v A = Cos 1 v A v A = Cos 21.0 0 x 5.20 m/s v A = 4.85 m/s 2. v A = v A 2 + v B 2 v B = v A 2 - v B 2 v B = 1.88 m/s 3. Sin 2 = -Sin 1 v A /v B 2 = -Sin21 0 x 4.85m/s 1.88m/s = 68 0

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ax2 + bx + c 2x2 + 11x + 5 ac 2(5) = b =11 ( ) 2x2 + 10x + 1x

ax2 + bx + c 2x2 + 11x + 5 ac 2(5) = b =11 ( ) 2x2 + 10x + 1x

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