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At the same instant that a 0.50-kg ball is dropped from a high building, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth’s surface with an initial velocity of 19.6 m/s. They move along nearby lines and pass without colliding. When the second ball is at its highest point the velocity of the center of mass of the two-ball system is : |
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Answer» 0 Acceleration of C.M.,`a_(cm) = (0.5 (-9.8) + 0.25(-9.8))/(0.5 + 0.25) = - 9.8 m//s^2` Time after which second BALL is at HIGHEST point`t = U/g= (19.6)/(9.8) = 2 "sec"` . After 2sec,`v_(cm) = u_(cm) + a_(cm) t = (19.6)/(3) + (-9.8) 2 = - 9.8 XX 4/3 = -13 m//s`. |
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