Two identical balls are dropped from the same height onto a hard surface, the second ball being released exactly when the first ball bollides with the surface. If the first ball has made two more collisions by the time the second one collides. Then the coefficient of restitution between the ball and the surface satisfies :

Options

A. `e gt 0.5`
B. `e = 0.5`
C. `e = (sqrt(3) - 1)/(2)`
D. `e lt (sqrt(3) - 1)/(2)`

Answer

D

Explanation

Velocity of ball before first collision = `sqrt(2gh)`. Time taken between first and second collision `= (2esqrt(2gh))/(g)`
So, `(2esqrt(gh))/(g) + (2e^2sqrt(2gh))/(g) < sqrt((2h)/(g)) , " " 2e^(2) + 2e - 1 < 0 implies e < (sqrt(3) - 1)/(2)` .