If the energy of a ball falling from a height of 10 meters is reduced by 40% how high will it rebound?
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Answer:
a bouncing ball loses 40% of its potential energy with each bounce how would it tend to rebound ? The Gravitational potential energy (PE) at 10m height is given by mgh where m is the mass of the ball, g is the acceleration due to gravity, and h is the height. Then when it strikes the ground it loses energy by sound and heat produced. If that is 40% reduction then 60% of the initial PE remains, which is 0.6mgh. This is the reduced total energy, neglecting air resistance, that will become the new PE at the shorter height that it will reach on the bounce back ! So that can be mgh2 = 0.6mgh, where h2 is the new(decreased) height. As you can see mg cancels out on each side. This will give you the following : h2 = 0.6h, which is 6 meters. This also means that the height achieved on the bounce back is only dependent on the initial height from where it drops, and the energy-loss factor in a given environment, and not the mass or the acceleration due to gravity !! In other words if the ball were to be on the moon where the value of gravitational acceleration is 1/6th of earth’s, similar heights will be achieved for the same mass or any mass, with even a different g. The heights will be actually some more due to no atmospheric friction, as the moon has little or no atmosphere, and due to little or no sound energy produced during the collisions, again due to no air like on earth, and far less heat generation !! Things will move far more slowly though taking about 6 times the time taken as on earth, like in slow motion compared to earth, and several times more up and down cycles will result before coming to a standstill due to much less energy lost per cycle ! You only need to know the initial potential energy (mgh) and the energy loss factor to find the successive heights here ! So going back to the calculations another much faster way of doing it is to do it mentally in this case with easy numbers : Since Gravitational PE = mass times acceleration times height(mgh), and it is a linear relationship, if you know what amount or percentage of energy is left on the bounce back, or what amount is expended, then the height achieved will be the percentage as defined by the percentage of the remaining energy, or decreased by the same percentage as the energy lost which here will give you the same answer mentally which is 6 meters ! And you can predict what the subsequent heights will be as the ball continues to lose energy with every collision against the ground, and bounces back up to shorter and shorter heights till it comes to a standstill… It will lose about 40% of its energy with each collision with the ground in the above setting, and 60% of the potential energy will remain of the height it drops from each successive time. Calculating that will give you (0.6)(6m) which is 0.36m on the second rebound or bounce back, then (0.6)(0.36) which is 2.16m on the third bounce back, then (0.6)(2.16m) which is 1.29m on the 4th, then O.774m on the 5th, then 0.464m on the 6th, then 0.279m on the 7th, then 0.167m on the 8th, then 0.100m on the 9th, then 0.060m on the 10th, then 0.036m on the 11th, then 0.022m on the 12th and so on till you can’t see the minute heights achieved anymore, and the ball has lost all its energy and come to a standstill ! In more real time and space the ball will experience some resistance due to air both on its way down and and on its way up on the rebound, and more energy will be lost due to frictional resistance due to air, and less energy will remain on each bounce back and slightly less height will be attained each time and the ball will come to a standstill somewhat faster…
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