A ball of density d is dropped on to 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 released and it falls through the same height before striking the surface of a liquid of density of `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 forces. Assume the depth of the liquid to be large.

Answer

A, B, C, D

Explanation

In elastic collision with the surface, direction of velicity is reversed but its magnitude remains the same,
Therefore, time of fall=time of rise,
or, time of fall `=(t_(1))/(2)`
Hence, velocity of the ball just before it collides with liquids is
`v=g(t_(1))/(2)`
Retardation inside the liquid
`a=("upthrust-weight")/("mass")=(Vd_(L)g-Vdg)/(Vd)`
`=((d_(L)-d)/(d))g`
Time taken to come to rest under this retardation will be
`t=v/a=(g t_(1))/(2a)=(g t_(1))/(2((d_(L)-d)/(d))g)`
`=(dt_(1))/(2(d_(L)-d))`
Same will be the time to come to back on the liquid surface.
therefore,
(a) `(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)`
(b) The motion of the ball is periodic but not simple harminic because the acceleration of the ball is g in air and `((d_(L)-d)/(d)) g` inside the liqudi which is not proportional to the displacenent, which is necessary and sufficient condition for SHM.
(c ) When `(d_(L)=d)`, retardation or acceleration inside teh liquid becones zero (upthrust-weight). Therefore, the ball will continue to move with constant velocity `v=(gt_(1))/(2)` inside the liquid.