Today I came across the question below. I thought to write about it explaining how to find the total distance traveled by a bouncing object.
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Question:
A particle falls from h height and bounces. Prove that the total distance traveled is \(\frac{\left ( 1+e^{2} \right )}{\left ( 1-e^{2} \right )}h \). and the total time is \( \sqrt{\frac{2h}{g}} \cdot \frac{(1+e)}{1-e}\)
where e is the elasticity coefficient.
we can find the particle's velocity when it hits the ground for the first time using the law of mechanical energy conservation as shown below.
\(mgh = \frac{1}{2}mv^{2}\)
\(\Rightarrow v = \sqrt{2gh}\)
and we can use Newton's law to find out the velocity (v) of the particle when it bounces for the first time.
\(e = \frac{v}{\sqrt{2gh}}\)
\(\Rightarrow v = e\sqrt{2gh} \)
According to this, we can determine the particle's speeds when it hits the ground are in a geometric series as follows.
\( e\sqrt{2gh}, e^{2}\sqrt{2gh}, e^3\sqrt{2gh},....\)
The n th term can be expressed as \(e^{n}\sqrt{2gh}\) .
Now we have to calculate the total distance traveled in the nth bounce. To calculate this distance we can use the energy conservation law as well as the motion formula \(v^{2} = u^{2} +2as \). As we want to find the total distance (not the displacement) we have to calculate the maximum height (h') the particle reaches in the nth bounce and multiplicate it by two.
\(mgh' = \frac{1}{2}m(e^{n}\sqrt{2gh})^2\)
\(h' = e^{2n}h\)
\(2h' = 2 e^{2n}h\)
Now we have the total distance traveled by the particle in its nth bounce and all we have to do is calculate the sum of all 2h' s (d') and add it to the distance of the first fall (h).
\(d' = \sum_{1}^{r}2 e^{2n}h = 2h \sum_{1}^{r} e^{2n}\)
\(\sum_{1}^{r} e^{2n}\) is a geometric series which can be written as below.
\(e^{2}+e^{4}+e^{6}+e^{8}+......\)
in this geometric series,
- First term (a) = \(e^{2}\)
- r' = \(e^{2}\)
so \(\sum_{1}^{r} e^{2n} = \frac{a(1-r'^{r})}{1-r'}\)
\(\Rightarrow \sum_{1}^{r} e^{2n}\ = \frac{e^{2}(1-(e^{2})^r)}{1-e^{2}}\)
as e < 1 we can calculate the sum of the series when the r goes to infinity as follows.
\(\Rightarrow \sum_{1}^{r} e^{2n}\ = \frac{e^{2}(1-(0))}{1-e^{2}}\)
\(\Rightarrow \sum_{1}^{r} e^{2n}\ = \frac{e^{2}}{1-e^{2}}\)
\(\therefore d' = 2h \frac{e^{2}}{1-e^{2}} \)
\(\therefore d = 2h \frac{e^{2}}{1-e^{2}} + h \)
\(\therefore d =\frac{\left ( 1+e^{2} \right )}{\left ( 1-e^{2} \right )}h \)
The total time can be calculated as the same as the total distance. All you have to do is to calculate the total time for one bounce and find the sum of time for all bounces and add the time for the first fall. If you have learned something important from this share this on your social media platforms.
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