Question

A smooth steel ball strikes a fixed smooth steel plate at an angle $\theta $ with the vertical. If the coefficient of restitution is $e$ , the angle at which the re-bounce will take place is

• A. $\theta $
• B. $tan^{- 1} \left[\frac{tan ⁡ \theta }{e}\right]$
• C. $e \, tan \, \theta $
• D. $tan^{- 1} \left[\frac{e}{tan ⁡ \theta }\right]$

Answer 1

Answer: (B)

Since, no force is present along the surface of plate. So, momentum conservation principle for ball is applicable along the surface of plate.

$m\upsilon \, sin \, \theta _{1}=m\upsilon \, sin \, ⁡ \theta _{2}$
Or $\upsilon \, sin \, \theta _{1}=\upsilon_{1} \, sin \, ⁡ \theta _{2}$ …(i)
$e=\frac{\upsilon_{1} \, cos \, \theta _{2}}{\upsilon \, cos \, ⁡ \theta _{1}}=\frac{\upsilon_{1} \, cos \, ⁡ \theta _{2}}{\upsilon \, cos \, ⁡ \theta }$ ( $\theta _{1}=\theta $ )
$∴ \, \, \, \upsilon_{1} \, cos \, \theta _{2}=e\upsilon \, cos \, ⁡ \theta $ …(ii)
$\therefore \frac{\upsilon_{1} \, sin \, \theta _{2}}{\upsilon_{1} \, cos \, ⁡ \theta _{2}}=\frac{\upsilon \, sin \, ⁡ \theta }{e \upsilon \, cos \, ⁡ \theta }=\frac{tan ⁡ \theta }{e}$
$\therefore tan \theta _{2}=\frac{tan ⁡ \theta }{e}$
$\therefore \left(\theta \right)_{2}=\left(tan\right)^{- 1} \left(\frac{tan ⁡ \theta }{e}\right)$