A 12-Kg ball is connected to a rope and release from rest from an unknown height

Question

A 12-Kg ball is connected to a rope and release from rest from an unknown height. When the ball reaches its lowest point, it strikes a 24-Kg block, which is initially at rest. The 12-Kg ball stops after the collision and the 24-Kg block proceeds across a the floor with a speed of 4.2 m/s. Is the collision as described above elastic or inelastic? Explain From what height was the ball dropped? If the coefficient of kinetic friction between the block and the floor is 0.25, what is the displacement of the block?

Explanation / Answer

a)

Since ball and block does not move together after the collision , the collision is not inelastic . so the collision is elastic

b)

v = speed of ball just before hitting the block

V = speed of block after collision = 4.2 m/s

m = mass of ball = 12 kg

M = mass of block = 24 kg

using conservation of momentum

mv =MV

12 v = 24 x 4.2

v = 8.4 m/s

using conservation of energy

mgh = (0.5) m v2

h = v2/2g = 8.42/(2 x 9.8) = 3.6 m

c)

using the equation

Vf2 = Vi2 + 2 a d

02 = 4.22 - 2 (0.25 x 9.8) d

d = 3.6 m

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