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 :

Options

A. 0
B. 13 m/s down
C. 20 m/s, down
D. 27 m/s, down

Answer

B

Explanation

Initial velocity of C.M, `v_(cm) = (0.5(0) + (0.25)(19.6))/(0.5 + 0.25) = (19.6)/(3) m//s`
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`.