question_answer

A solid ball of density half that of water falls freely under the gravity from a height of 19.6 m and then enters water. Upto what depth will the ball go? How much time will it take to come again to the water surface? Neglect air resistance and Viscosity of water, \[\left( \text{g }=\text{ 9}\text{.8 m}/{{\text{s}}^{\text{2}}} \right)\].

Answer:

                Let m be the mass and p be the density of the ball. Therefore, density of water = \[2\rho \]. Energy of the ball on striking the water surface = mgh = mg x 19.6 J                             ...(i) Net force opposing the motion of the ball in water \[=uptrust\,-weight\,of\,ball\,=\left( \frac{m}{\rho }\times 2\rho \times g \right)-mg=mg\] If d is depth upto which the ball goes in water, then Work done = energy of the ball on striking water surface \[mg\times d=mg\times 19.6\];                        \[\therefore \]   \[d=19.6\,m\] Velocity on striking the water surface \[u=\sqrt{2gh}=\sqrt{2\times 9.8\times 19.6}=19.6m/s\] From  \[\upsilon =u+at\];          \[0=19.6-\left( 9.8 \right)t\]     \[\therefore \]t = 2s. This is the time taken by ball to travel a depth (d) in water. Total time taken by the ball to reach the water surface = 2 + 2 = 4 s