Correct Answer – C
Fraction of `KE` lost in collision
`Delta K % = Delta K % = ((1)/(2) mu^(2) – (1)/(2) mv^(2))/((1)/(2) m u^(2)) = 1 – ((v)/(u))^(2) = (1)/(4)` (given)
`v = y sqrt((3)/(4)`
The ball strikes at `45^(@)` will hot change while component of velocity normal to wall will change.
`v_(x) = u cos 45^(@) = (u)/(sqrt(2)`
`v_(y) = eu cos 45^(@) = (eu)/(sqrt(2)`
`v = sqrt(v_(x)^(2) + v_(y)^(2)) = [((u)/(sqrt2))^(2) + ((eu)/(sqrt2))^(2)])^(1)/(2)`
`implies v = u [(1)/(2) + (e^(2))/(2)]^(1)/(2)`
Solving (i) and (ii), we get
`e = (1)/(sqrt2)`.