I am doing an energy analysis of a falling ball that has been cut in half. the f

Question

I am doing an energy analysis of a falling ball that has been cut in half. the focus of the lab is elastic potential energy because the rubber ball is flipped inside out then dropped on the flipped side therefore creating a bounce when dropped because the flipped side pops out at impact to the ground. The mass of the chunk is .0154kg. the height the chunk bounces to is 1.3339 m. I made my starting point at the floor meaning h = 0. The gravitational potential energy of the inside out chunk at 1.0m was Ug=mgy; Ug=(.0154)(9.8)(1.0)=.15092 (units?) so the Ug at max height after the bounce is mgy=(.0154)(9.8)(1.3339)=.2013 (units?) My question is a bit of a long one, how do I use the conservation of energy to find the speed of the chunk just before it hits the ground? Also what is the difference between the two potential energies I calculated and what does this amount of energy mean? And is energy conserved? explain. sorry for the long problem guys.

Explanation / Answer

When you flip the rubber ball some amount of energy is stored in it in form of elastic potential energy, lets call it Ue.

When the ball is at height 1m it has a gravitational potential energy equal to mgy = 0.1509, lets call it Ug.

So when you leave the ball from that height it has both elastic as well as gravitational potential energy U=Ue+Ug.

When the falls under gravity before it hits the ground, only its gravitational potential energy is getting converted into kinetic energy while the elastic energy is still stored in it, Hence the speed at the point just before impact can be calculated using,

K.E. = Ug , 0.5*m*v^2 = 0.1509 . which gives you v = 1.399 units

Now when the ball bounces off the ground the elastic potential energy is released and adds into the kinetic energy.Assuming a perfectly elastic collision, the kinetic energy with the ball after the impact

K.E = Ug + Ue

This kinetic energy is finally converted into potential energy of the ball at the height of 1.3339m. Hence the difference you see in the two potential energy (before and after the impact) is the the elastic potenatial energy stored in the ball.

Which tells you, Ue = .2013 - 0.1509 = 0.0504 units

Hope it helps

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