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1 entonces 1 1 2 2 1 2 2 2 21 2 la velocidad del

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?𝑅= ?(1 +?𝑅).Entonces:?? (1 +?𝑅) =12????2+12?(???𝑅)2???2=2??(1 +?𝑅)? +?𝑅2.La velocidad del centro de masa:???=2??(1 +?𝑅)? +?𝑅2b)Utilizando la definición de Impulso y variación de cantidad de movimiento? = 𝑝?− 𝑝0?∆? = ????− ????0donde???0= 0 ?/?y? = ? − ?así tenemos:?∆? − ?∆? = ????Por definición:?????=𝑣0+𝑣𝑓2;?0= 0de esta forma tenemos???????=???2.También sabemos que???????=??− ?0∆?=? − 0∆?→ ∆? =????????finalmente:∆? =2????.De la ecuación del impulso:(? − ?)∆? = ????(? − ?)2????= ????
(? − ?) =????22?→ ? = ? −????22?= ? −?2?(2?? (1 +?𝑅)? +?𝑅2) = ? − (? (1 +?𝑅)1 +??𝑅2).Valor de la fuerza de fricciónf? = ? (1 −1 +?𝑅1 +??𝑅2).
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Term
Summer
Professor
LUIS ALBERTO SORIANO CARILLO
Tags
Cinem tica, Fricci n, Energ a potencial, Energ a cin tica, Cantidad de movimiento, Conservaci n de la energ a

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