TS EAMCET · Physics · Center of Mass Momentum and Collision
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
- A \(e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}\)
- B \(e=\left[\frac{h_0}{h_n}\right]^{1 / 2 n}\)
- C \(e=\frac{1}{n} \sqrt{\frac{h_n}{h_0}}\)
- D \(e=\frac{1}{n} \sqrt{\frac{h_0}{h_n}}\)
Answer & Solution
Correct Answer
(A) \(e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}\)
Step-by-step Solution
Detailed explanation
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…
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