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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 1 . If d L , obtain an expression (in terms of d, t 1 and d L ) for the time t 2 the ball takes come back to the position from which it was released. Is the motion of the ball simple harmonic ? 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.

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 t1 . Next, the ball is released and it falls through the same height before striking the surface of a liquid of density of d1.                               
 
  1. If d L, obtain an expression (in terms of d, t1 and dL) for the time t2 the ball takes come back to the position  from which it was released.
  2. Is the motion of the ball simple harmonic ?
  3. If d = dL, 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. 

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
6 years ago
Hello Student,
Please find the answer to your question
(a) Let the ball be dropped from a height h. During fall
V = ut + at = 0 + g t1 / 2 ⇒ t1 = 2v/ g
In the second case the ball is made to fall through the same height and then the ball strikes the surface of liquid of density dL. When the ball reaches inside the liquid, it is under the influence of two force (i) Vdg, the weight of ball in downward direction (ii) VdLg, the upthrust in upward direction.
Note :
The viscous forces are absent (given)
Since, dL > d
The upward force is greater and the ball starts retarding.
For motion B to C
u = V , v = 0, t = t, a = - a
v = u + at ⇒ 0 = v + (- a) t
⇒ t = v / a
Now, a = Fnet / m = VdLg – Vdg / Vd = (dL - d)g / d
⇒ t = vd / (dL – d)g …..(iii)
Therefore,
t2 = t1 + 2t = t1 + 2dv / (dL - d) g
= t1 + 2d / (dl – d )g t1g / 2= t1 [1 + d / dL - d]
⇒ t­2­ = dL t1 / dL – d
(b) Since the retardation is not proportional to displacement, the motion of the ball is not simple harmonic
(c) If d = dL then the retardation α = 0. Since the ball strikes the water surface with some velocity, it will continue with the same velocity in downward direction (until it is interrupted by some other force).
Thanks
Kevin Nash
askIITians Faculty

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