Asteel ball bearing of mass 0.06 kg at a height of 1.5m above a steel tables is released from rest and it is found to rebound to a height of 1.3m calculate the gravitational potential energy lost during the fall and the kinetic energy of the ball bearing jusr before impact and the speed of the ball bearing just before impact and the gravitational potential energy gained by the ball bearing when it rebounds to a height of 1.3m. assume that the gravitational field strength is g=10n/kg

Gravitational Potential Energy Lost During the Fall





Kinetic Energy Before Impact

As the ball falls, the gravitational energy is converted to kinetic energy. Just before impact, when the ball is just touching the ground, the gravitational potential energy is 0 (as the height above the ground is essentially 0), meaning that all of the gravitational potential energy has been converted. This means that the kinetic energy at this point is 0.9J.

Speed of the Ball Before Impact











Gravitational Potential Energy Gained After Rebound

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward