Question

relative density of a ball is 0.8. it is at a height of 2 metre from the water surface. what will the depth to which the ball will go in water?

Answer 1

WHAT WE SHOULD DO (Steps) :
(1)find the velocity at the surface of water from the mechanical energy conservation .
(2) calculate the retardation offered by the water .
(3) apply [v²-u² = 2as ] and calculate 's'. that will be our answer.
now let us take these one by one :
[1] CALCULATION OF INITIAL VELOCITY (say v):
Let the volume of ball V m³;
Density of ball = 0.8×1000 Kg /m³.
{Relative density = (density of ball )/ (density of water at 4°C.)}
then mass of ball = M= 800×V.
From mechanical energy conservation:
PE at 2m height from the water surface = KE at the surface of water.
2×M×g = ½ ×M× v².
=> v= √(4g).
[2] CALCULATION OF RETARDATION OFFERED BY WATER(say a ) :
inside water there are two forces act on the ball :
*weight of ball (= Mg = 800Vg) downwards;
*buoyant force by water (1000Vg) upwards;
but these are in opposite directions hence net force on ball : 1000Vg-800Vg
= 200Vg (Upwards).
=> (Retardation)a= (200Vg)/M = (200Vg)/(800V) = -g/4 m/s².
[3] CALCULATION FOR THE MAXIMUM DEPTH (say s) FROM SURFACE OF WATER OUR BALL CAN REACH :
at the maximum depth final velocity of ball =0.
applying v²-u² = 2as we get ;
4g = 2×(g/4 )×s.
=> s= 8 m.

Answer 2

above answer is correct..