Momentum and CollisionHard
Question
A ball of mass 2 kg dropped from a height h above a horizontal surface redounds to a height h after one bounce. The graph that relates H to h is shown in figure. If the ball was dropped from an initial height of 81 m and made ten bounces, the kinetic energy of the ball immediately after the second impact with the surface was
Options
A.320 J
B.480 J
C.640 J
D.Can′t be determined
Solution
From graph e = 
Kinetic energy of ball just after second bounce
=
m (e2u)2 = 
me4u2 = (e4) (mgH) = 
(2)(10)(81) = 320 J
Kinetic energy of ball just after second bounce
=
m (e2u)2 = me4u2 = (e4) (mgH) = (2)(10)(81) = 320 JCreate a free account to view solution
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