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Question: An inelastic ball falls from a height of \(100\) meters. It loses \(20\% \) of its total energy due ...

An inelastic ball falls from a height of 100100 meters. It loses 20%20\% of its total energy due to impact. The ball will now rise to a height of:
A) 80m80\,m
B) 120m120\,m
C) 60m60\,m
D) 9.8m9.8\,m

Explanation

Solution

You can easily solve the question by approaching what happens when the ball makes the impact and what its total energy at the impact will be, even after the impact, if it loses 20%20\% of its total energy, the new height should be easy to find.

Complete step by step answer:
We will be solving the question exactly like we explained in the hint section of the solution to the question.

Firstly, we will check what is the total energy of the ball at the initial point, then we will find its Kinetic energy just before the impact, then we consider in the loss of energy due to inelastic collision of the ball with the floor, and then we will compare how high the ball can rise again by equating the energy finally.

Let us now solve the question:
Let us assume that the ball has a fixed mass mm and since 100m100m is not a large height, we can assume the gravity to be constant throughout our problem, let it be gg
The height at which the ball is initially is given in the question: hi=100m{h_i}\, = \,100\,m
At this height, the potential energy will be: Ui=mgh=100mg{U_i}\, = \,mgh\, = \,100mg
At this point, the ball is initially at rest, hence, Ki=0{K_i} = 0
Just before the impact, all of the energy of the ball will be purely Kinetic because the height of the ball at that moment will be h=0U=0h = 0\,\,\, \Rightarrow \,\,\,U = 0
So, total energy just before the impact remains E=100mgE = 100mg
Now, after the impact, the ball loses 20%20\% of its energy, which means that the ball retains 80%80\% of its total energy. This means that the new total energy can be written as:
En=100mg×80100 En=80mg  {E_n} = 100mg \times \dfrac{{80}}{{100}} \\\ {E_n} = 80mg \\\
We can see that just after the impact, the total energy becomes 80mg80mg
We can also observe that at this moment, its energy is still purely Kinetic and its height, thus, its potential energy is still 00
When the ball rises to the maximum height, all of the total energy would be only Potential and no Kinetic energy.
Hence, we can write:
Uf=En{U_f} = {E_n}
mghf=80mgmg{h_f} = 80mg
Cancelling out on both sides, we get:
hf=80{h_f} = 80 meters

Hence, the right option is the option (A).

Note: The method that we employed is a longer method but clears all the concepts that are required to solve such a question. A faster approach would have been to directly subtract 20%20\% of the energy and equate that to the final potential energy as we already know that the maximum height will be achieved when there is no kinetic energy.