The adjacent figure shows an infinite frame of two sides in a gravity-free space. What is the final constant kinetic energy expected of the ball of mass m projected as shown with an initial velocity v0 ? The coefficient of restitution for the collision between the ball and the frame is e= text0.5 .

Question

The adjacent figure shows an infinite frame of two sides in a gravity-free space. What is the final constant kinetic energy expected of the ball of mass m projected as shown with an initial velocity v0 ? The coefficient of restitution for the collision between the ball and the frame is e= text0.5 . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-huvopvrsjt2dsoo9.jpg" /> (A)  text0.5mv02 (B)  text0.25mv02 (C)  text0.125mv02 (D)  text0.325mv02

Tardigrade Admin · Accepted Answer

The component v=v0sin45° =(v0/√2) perpendicular to the walls is finally lost after a long time. Hence, final remaining K.E.=(1/2)m((v0/√2))2

$0.5 m v_{0}^{2}$

$0.25 m v_{0}^{2}$

$0.125 m v_{0}^{2}$

$0.325 m v_{0}^{2}$

The component $v = v_{0} s in 4 5^{\circ} = \frac{v _{0}}{2}$ perpendicular to the walls is finally lost after a long time.
Hence, final remaining $K . E . = \frac{1}{2} m (\frac{v _{0}}{2})^{2}$

0.5mv02​

0.25mv02​

0.125mv02​

0.325mv02​

The component v=v0​sin45∘=2​v0​​ perpendicular to the walls is finally lost after a long time. Hence, final remaining K.E.=21​m(2​v0​​)2

$0.5 m v_{0}^{2}$

$0.25 m v_{0}^{2}$

$0.125 m v_{0}^{2}$

$0.325 m v_{0}^{2}$

The component $v = v_{0} s in 4 5^{\circ} = \frac{v _{0}}{2}$ perpendicular to the walls is finally lost after a long time.
Hence, final remaining $K . E . = \frac{1}{2} m (\frac{v _{0}}{2})^{2}$