Search any question & find its solution
Question:
Answered & Verified by Expert
A ball is let fall from a height $h_0$. It makes $n$ collisions with the earth. After $n$ collisions it rebounds with a velocity $v_n$ and the ball rises to a height $h_n$, then coefficient of restitution is given by
Options:
Solution:
2845 Upvotes
Verified Answer
The correct answer is:
$e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}$
The velocity of ball after $n$th rebound will be
$v_n=e^n v_0 \quad\left(\text { where } v_0=\sqrt{2 g h_0}\right)$
Therefore, the height after $n$th rebound will be
$\begin{aligned}
h_n & =\frac{v_n^2}{2 g}=\frac{e^{2 n} v_0^2}{2 g}=\frac{e^{2 n} \times 2 g h_0}{2 g} \\
\Rightarrow \quad e^{2 n} & =\frac{h_n}{h_0} \text { or } e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}
\end{aligned}$
$v_n=e^n v_0 \quad\left(\text { where } v_0=\sqrt{2 g h_0}\right)$
Therefore, the height after $n$th rebound will be
$\begin{aligned}
h_n & =\frac{v_n^2}{2 g}=\frac{e^{2 n} v_0^2}{2 g}=\frac{e^{2 n} \times 2 g h_0}{2 g} \\
\Rightarrow \quad e^{2 n} & =\frac{h_n}{h_0} \text { or } e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.