A ball strikes a horizontal floor at an angle ${45}^0$ The value of 'e' between the ball and the floor is 1/2. Find the fraction of K.E. loss during the collision

The correct option is C $\frac{3}{8}$
$x$ component of velocity will remain same & for $y$ component
$\left(V_y\right)final-0=-e.V_y$ (velocity in y direction)

so, $K.E.$ lost $=\frac{\frac{1}{2}m{\left(\frac{v}{\sqrt{2}}\right)}^2-\frac{1}{2}m{\left(\frac{ev}{\sqrt{2}}\right)}^2}{2\times \frac{1}{2}m{\left(\frac{v}{\sqrt{2}}\right)}^2}$

Here, the horizontal component of velocity willn't undergo any change.

$=\frac{1-e^2}{2}$

$=\frac{1-0.25}{2}=\frac{0.75}{2}=\frac{3}{8}$