# A ball bounces to 80% of its original height. What fraction of potential energy is lost in each bounce?

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Hint: In order to solve this question, we are first going to consider the initial height and the corresponding initial potential energy, then finding the height after the first bounce, and thus, the corresponding potential energy is found. Then, their difference is found and thus fraction of energy lost.

Formula used:
If the mass of the ball is$m$ and its original height is $h$, thus, its initial potential energy will be
${U_i} = mgh$

Complete step-by-step solution:
Let the mass of the ball be$m$ and its original height be$h$, thus, its initial potential energy will be given by the equation
${U_i} = mgh$
Now, after that one first bounce, the height is going to be$80\% \,of\,h$, i.e. the original height of the ball.
Thus, the final height is
${h_f} = \dfrac{{80}}{{100}} \times h = 0.80h$
The final potential energy of the ball after each bounce will be
${U_f} = 0.80mgh$
The potential energy lost in each bounce is equal to
${U_i} - {U_f} = mgh - 0.80mgh = 0.20mgh$
Therefore, fraction of the potential energy lost by the ball in each bounce is equal to: $0.20$

Note:It is important to note that the energy lost in each bounce is inversely proportional to the height to which the ball reaches after one bounce. The more is the height of the ball after bouncing off, the less is the energy lost by the ball in that one bounce and if the height after bouncing remains almost same, the energy lost is also almost same.