Two identical steel balls, each of mass 2.9 kg, are suspended from strings of le

Question

Two identical steel balls, each of mass 2.9 kg, are suspended from strings of length 30 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle of ? = 37° with the vertical and let it go. It collides elastically with the other ball.
a) How high will the other ball rise?
b) Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision?
Answers= a) 6.06 cm, b) 1.52 cm

Explanation / Answer

m = 2.9 kg, L = 30 cm, = 37°,
a) How high will the other ball rise?

initial energy = mgL(1 - cos)

final energy = mgh

mgL(1 - cos) = mgh

h = L(1 - cos) = 0.0604 m = 6.04 cm

b) Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision?

mgL(1 - cos) = mv2/2, so initial speed before collision v = [2gL(1 - cos)]

during collision: mv = (m + m)v', so v' = v/2

2m*v'2/2 = 2m*gh,

h = v'2/2g = (v/2)2/2g = (1/4)*v2/2g = (1/4)*L(1 - cos) = 0.0151 m = 1.51 cm

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