Finding the coefficient of restitution

1. The problem statement, all variables and given/known data
A body is fired from point P and strikes at Q inside a smooth circular wall as shown in figure. It rebounds to point S (diametrically opposite to P). The coefficient of restitution will be:

(Ans: ##\tan^2\alpha##)

2. Relevant equations

3. The attempt at a solution
Let ##v## (along PQ) be the velocity before collision and ##v’## (along QS) be the velocity after collision.

The coefficient of restitution (e) is defined as:
$$e=\frac{\text{Relative speed after collision}}{\text{Relative speed before collision}}$$

The numerator is ##v’\sin\alpha## and denominator is ##v\cos\alpha##. (The relative speed is measured along normal at the point of collision)

Hence,
$$e=\frac{v’\sin\alpha}{v\cos\alpha}=\frac{v’}{v}\tan\alpha$$

From conservation of linear momentum along PS:
$$mv\cos\alpha=mv’\sin\alpha \Rightarrow \frac{v’}{v}=\cot\alpha$$
Hence,
$$e=1$$
:confused:

Any help is appreciated. Thanks!

Attached Images
File Type: png cor problem.png (9.5 KB)

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