how do you solve this question? Two balls, of masses m(a) = 40 g and m(b) = 60 g

Question

how do you solve this question? Two balls, of masses m(a) = 40 g and m(b) = 60 g, aresuspended. The lighter ball is pulled away to a 60 degreeangle with the vertical and released.. (a) what is the velocity ofthe lighter ball before impact? (b) What is the velocity of eachball after the elastic collisoin? (c) What will be the maxiumumheight of each ball after the elastic collision? The balls are suspended 30 cm from the ceiling how do you solve this question? Two balls, of masses m(a) = 40 g and m(b) = 60 g, aresuspended. The lighter ball is pulled away to a 60 degreeangle with the vertical and released.. (a) what is the velocity ofthe lighter ball before impact? (b) What is the velocity of eachball after the elastic collisoin? (c) What will be the maxiumumheight of each ball after the elastic collision? The balls are suspended 30 cm from the ceiling

Explanation / Answer

(a)
   according to the law of conservation of energy the KEat the lowest point is equal to the PE at
   the heighest point so we can write
   (1 / 2) m vbottom2 = m gh
   so the speed of the lighter ball at the bottom of theswing will be
   vA = (2 g hA)
         = [2(9.80 m / s2)(0.30 m - 0.30 m cos60o)]
         = ......... m /s
(b)
   as the collision is elastic we get the velocities ofeach ball as
   vA' = [(mA - mB)/ (mA + mB)] vA
          = .........m / s
   vB' = [2 mA /(mA + mB)] vA
          = .........m / s
(c)
   after the collision their maximum heights willbe
   hA' =vA'2 / 2 g
   hB' =vB'2 / 2 g

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