Two identical balls of clay are positioned such that one piece is located 4.80m

Question

Two identical balls of clay are positioned such that one piece is located 4.80m directly above the other, which is on the ground. The upper piece of clay is released from rest while the lower one is shot straight up from the ground at a speed of 6.00m/s. When the clay balls collide, they stick together. Find the speed of the balls when they strike the ground together. Treat the balls as point
particles (i.e. ignore their extension in space). [Hint: You can do this as a collision problem, but it is much easier to do using the
center of mass frame.]

Explanation / Answer

the two balls collide after t = H/u = 4.8/6 = 0.8 s

the two balls meet at = h = (6*0.8)-(0.5*9.8*0.8^2) = 1.664 m

m1 = m         u1 = g*t = -9.8*0.8 = -7.84 m/s

m2 = m         u2 = 6-(9.8*0.8) = -1.84 m/s

momentum before collision = Pi = m1*u1 + m2*u2

immediately after collision momentum Pf = (m1+m2)*V

Pf = Pi

V = (-7.84-1.84)/2 = -4.84 m/s

after they reach the ground Vf = -(V + sqrt(2*g*h))

Vf = (4.84 +sqrt(2*9.8*1.664))

Vf = 10.55 m/s

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