A 450g rubber ball is dropped from 2.0m above the floor. After the bounce, the b
ID: 2294617 • Letter: A
Question
A 450g rubber ball is dropped from 2.0m above the floor. After the bounce, the ball only rises to a height of 1.5m above the floor.
A. Determine the change in momentum of the ball when it hits the floor and bounces up.
B. Calculate the amount of heat generated when the ball hits the floor.
C. Explain how momentum is conserved when the ball collides with the floor.
(hint: either use motion equations or conservation of energy to determine the speed of the ball when it reaches the floor.)
Please show all work
Explanation / Answer
v1 = -sqrt(2*g*h1)
= -sqrt(2*9.8*2)
= -6.26 m/s
v2 = sqrt(2*g*h2)
= sqrt(2*9.8*1.5)
= 5.42 m/s
A) delta P = m*(v2-v1)
= 0.45*(5.42 - (-6.26))
= 5.26 kg.m/s
B) heat generated = chnage in mechanical enrgy
= m*g*(h1-h2)
= 0.45*9.8*(2-1.5)
= 2.205 J
C) here momentum is not conserved. because the ground exert force on the ball. Beacuse of that force
the momentum of the ball changes.
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