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Unformatted text preview: j m v m t F A A A B A ˆ 20 2= ∆i m v m t F B B B A B ˆ ) 30 ( 1= ∆What is true about F AB and F BA ? They are equal and opposite. Now, substitute F BA = F AB and add the two equations above together. j m v m t F A A A B A ˆ 20 2= ∆i m v m t F B B B B A ˆ ) 30 ( 1= ∆j m i m v m v m A B B B A A ˆ 20 ˆ ) 30 ( 2 2+ = This is essentially the derivation for Eq 3/28 What is true about v A2 and v B2 ? They are the same so call it v 2 . Substitute and solve j m i m v m m A B B A ˆ 20 ˆ ) 30 ( ) ( 2+ = ) ( ˆ 3 . 10 ˆ 55 . 14 ) ( ˆ 20 ˆ ) 30 ( 2 mph j i m m j m i m v B A A B += + += Careful with units you would need to divide the weights by g, but they would end up cancelling in the equation. Same for the mph conversion....
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 Spring '05
 staff
 Derivative, Force, Momentum, Trigraph, mB vB1

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