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). For your alternate solution, use the center of mass frame of the two balls.

Explanation / Answer

suppose they meet after time t.

so [ gt^2/2 ] + [ 6t - gt^2/2] = 4.80

t = 0.8 sec

velocity of ball 1, v1 = - gt = - 9.8 * 0.8 = - 7.84 m/s

v2 = 6 - gt = -1.84 m/s

after collision, applying momentum conservation,

m * -7.84 + m * -1.84 = 2mv
v = 4.84 m/s

height from ground,

h = 4.80 - (9.8 x 0.8^2 / 2) = 1.664 m

applying

vf^2 - vi^2 = 2ad

vf^2 - 4.84^2 = 2(-9.8)(-1.664)

vf = 7.49 m/s

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