WBJEE · Physics · Center of Mass Momentum and Collision
A tennis ball hits the floor with a speed \(v\) at an angle \(\theta\) with the normal to the floor. If the collision is inelastic and the co-efficient of restitution is \(\varepsilon\), what will be the angle of reflection?
- A \(\tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)\)
- B \(\sin ^{-1}\left(\frac{\sin \theta}{\varepsilon}\right)\)
- C \(\theta \varepsilon\)
- D \(\theta \frac{2 \varepsilon}{\varepsilon+1}\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)\)
Step-by-step Solution
Detailed explanation
Hint: \(\tan \theta^{\prime}=\frac{\operatorname{usin} \theta}{\operatorname{eucos} \theta} \quad \tan \theta^{\prime}=\frac{1}{\mathrm{e}} \tan \theta\) Since the floor is smooth. Hence tangential component of velocity remains unchanged.
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