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Introductory Physics Homework Help
Conservation of Momentum of a box of mass
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[QUOTE="blue_lilly, post: 4721073, member: 458140"] [h2]Homework Statement [/h2] Box A of mass 1.20 kg is sliding to the right across a frictionless table at a speed of 2.52 m/s. Box A collides with Box B which has a mass of 2.56 kg, and Box A bounces straight back to the left with a speed of 0.665 m/s. A) What is the momentum of Box A before the collision? [INDENT][B]3.02 kg*m/s To the Right [/B][COLOR="DarkOrchid"]Correct Answer[/COLOR][/INDENT] B) What is the momentum of Box A after the collision? [INDENT][B]0.798 kg*m/s Correct: To the Left [/B] [COLOR="DarkOrchid"]Correct Answer[/COLOR][/INDENT] C) What is the momentum of Box B after the collision? [COLOR="Red"]This is the one I need help with![/COLOR] [h2]Homework Equations[/h2] Momentum(P)=mass(kg)*Velocity(kg*m/s) Momentum(P-before)=Momentum(P-after) [h2]The Attempt at a Solution[/h2] A) P=m*V=(1.20 kg)(2.52 m/s)=3.02 kg*m/s To the right CORRECT B) P=m*V=(1.20 kg)(.665 m/s)=.798 kg*m/s To the Left CORRECT This is the part I am having trouble with C) Momentum(before)=Momentum(after); P=m*V P[SUB]Before[/SUB]A+P[SUB]Before[/SUB]B=P[SUB]After[/SUB]A+P[SUB]After[/SUB]B [(1.20 kg)(2.52 m/s)]+[(2.56 kg)(0 m/s)] = [(1.20 kg)(.665 m/s)]+[] [3.02 kg*m/s]+[0 kg*m/s] = [.798 kg*m/s]+[P[SUB]After[/SUB]B] [(3.02 kg*m/s)/(.798 kg*m/s)] = P[SUB]After[/SUB]B P[SUB]After[/SUB]B = [(3.02 kg*m/s)/(.798 kg*m/s)] P[SUB]After[/SUB]B = 3.78 kg*m/s : To the Right The answer is incorrect. I am assuming the [To the right] portion is correct because BoxA and BoxB collided; this means that they are asserting equal and opposite force on each other and because BoxA went to the Left, BoxB must go to the Right. This leaves the incorrect portion to be the Momentum of BoxB after the collision. I double checked my math, a numerous amount of time, so I don't know where I am going wrong. Help is appreciated! [/QUOTE]
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Conservation of Momentum of a box of mass
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