A ball of density d is dropped onto a horizontal solid surface. It bounces elastically from the surface and returns to its original position in a time `t_(1)`. Next, the ball is relasesed and it falls through the same height before striking the surface of a liquid of density `d_(L)`.
(a) If `dltd_(L)`, obtain an expression ( in terms of `d, t_(1)` and `d_(L))` for the time `t_(2)` the ball takes to come back to the position from which it was released.
(b) Is the motion of the ball simple harmonic?
(c ) If `d=d_(L)` how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative force. Assume the depth of the liquid to be large.

Options

A. ` 2t_1 `
B. ` (dt_1 )/((d_L- d )) `
C. ` (dt_1)/(2 (d_L- d )) `
D. ` (2dt_1 )/(( d _L - d )) `

Answer

C

Explanation

In elastic collision with the surface, direction of velocity is reversed but its magnitude remains the same. Therefore,
time of fall = time of rice. Or time of fall ` = (t_1 )/(2) `
Hence, velocity of the ball just before it collides with liquid is ` v = g (t _ 1 )/(2) " " ...(i) `
Retardation inside the liquid ` a = ("Upthrust - weight")/("mass") = (Vd_L g - V dg )/(Vd ) = ((d_L - d )/(d)) g " " `...(ii)
Time taken to come to rest under this retardation will be ` t = (V)/(a) = (g t _ 1 )/( 2 a ) = (g t _1)/(2((d_L- d )/(d ))g ) = (d t _ 1)/(2(d _ L - d )) `
same will be the time to come back on the liquid surface ` because ` a is constant. Therefore,
` t_2 ` = time the ball takes to came back to the position from where it was released
` = t _ 1 + 2t = t _ 1 + (dt _ 1 )/(d_L - d ) = t _ 1 [ 1 + (d)/(d_L - d )] or t_2 = (t_1 d _ L)/(d _ L - d ) `