Question

A ball of mass m hits a floor with a speed $v_0$ making an angle of incidence $\alpha$ with the normal. The coefficient of restitution is e. Find the speed of the reflected ball.

• A. $\sqrt{(v_{0}sin\alpha {)}^2+(e{v}_{0}cos\alpha {)}^2}$
• B. $\sqrt{(v_{0}sin\alpha {)}^2-(e{v}_{0}cos\alpha {)}^2}$
• C. $\sqrt{(v_{0}cos\alpha {)}^2+(e{v}_{0}sin\alpha {)}^2}$
• D. $\sqrt{\left(v_{0}sin\alpha \right)+\left(e{v}_{0}sin\alpha \right)}$

Answer 1

The correct option is A $\sqrt{(v_{0}sin\alpha {)}^2+(e{v}_{0}cos\alpha {)}^2}$

Answer is A.
The component of velocity $v_0$ along common tangential direction $v_0$ sin $\alpha$ will remain unchanged. Let v be the component along common normal direction after collision. Applying, Relative speed of separation = e (Relative speed of approach) along common normal direction, we get
Thus, after collision components of velocity v' are $v_0$ sin $\alpha$ and $e{v}_0$ cos $\alpha$
Therefore, $v^{\prime }=\sqrt{(v_{0}sin\alpha {)}^2+(e{v}_{0}cos\alpha {)}^2}$
Hence, the speed of the reflected ball is $v^{\prime }=\sqrt{(v_{0}sin\alpha {)}^2+(e{v}_{0}cos\alpha {)}^2}$