Question

A body falls from a height '\(h\)' on a horizontal surface and rebounds. Then it falls again and rebounds and so on. If the coefficient of restitution is \(\dfrac{1}{2}\), then the total distance travelled by the ball before coming to rest is (neglect air resistance)

Answer 1

When a body falls from height \(h\) it hits the ground with the velocity \(v = \sqrt{2gh}\)

\(\Rightarrow h = \dfrac{v^2}{2g}\)

Now the ball undergoes collision, loses some energy and rebounds with a velocity \(ev\) where \(e\) is the coefficient of restitution. It will now rise to a height \(h_1\) and come back by covering the same distance \(h_1\)

\(h_1=\dfrac{(ev)^2}{2g}\)

Now the ball will again strike the ground and undergo collision and will rebound with a velocity \(e^2v\). If it rises to a height \(h_2\)

\(h_2=\dfrac{(e^2v)^2}{2g}\)

Thus, the total distance is given as

\(d = h + 2h_1 + 2h_2 +~.~.~.\)

\(d = \dfrac{v^2}{2g} + \dfrac{2e^2v^2}{2g} + \dfrac{2e^4v^2}{2g}+~.~ .~ .\)

\(\Rightarrow d = \dfrac{v^2}{2g} + \dfrac{2e^2v^2}{2g}(1 +e^2+e^4+~.~.~.)\)

Here \((1 +e^2+e^4+~.~.~.)\) forms an infinite GP with first term \(a=1\) and common rartio \(r = e^2\)

So, sum \(S = \dfrac{a}{1-r}=\dfrac{1}{1-e^2}\)

\(\Rightarrow d = \dfrac{v^2}{2g} + \dfrac{2e^2v^2}{2g}\dfrac{1}{1-e^2}\)

\(\Rightarrow d = \dfrac{v^2}{2g} \left[ 1 + \dfrac{2e^2}{1-e^2}\right]\)

\(\Rightarrow d = \dfrac{v^2}{2g} \left[ \dfrac{1-e^2+2e^2}{1-e^2}\right]\)

\(d=h\left [ \dfrac{1+e^2}{1-e^2} \right ]\)

\(d=h\left [ \dfrac{1+\left ( \dfrac{1}{2} \right )^2}{1-\left ( \dfrac{1}{2} \right )^2} \right ]\) (\(\because e=\dfrac{1}{2}\))

\(d=h\left [ \dfrac{5/4}{3/4} \right ]\)

\(d=\dfrac{5h}{3}\)