Question

A ball whose density is $0.4 \times {10^3}kg/{m^3}$ falls into water from a height of 9cm. To what depth does the ball sink?
A.9cm
B.6cm
C.4.5cm
D.2.25cm

Answer 1

In order to solve this problem,we use the principle of conservation of energy and also use the equation of motion so that we can arrive at the result.

Complete Step by Step Answer:
Considering the ball is at a height of 9cm from the surface of water. Then, according to conservation of energy we have $mgh = \dfrac{1}{2}m{u^2}$ where m=mass of ball and u=initial velocity of the ball. Substituting given values in this equation we get
$\Rightarrow gh = \dfrac{{{u^2}}}{2} \\\ \Rightarrow u = \sqrt {2gh} = \sqrt {g \times 2 \times 9} = \sqrt {g18} cm/\sec \\\ \\\$ ……eq1
Once the ball immerses in the water, it experiences an upthrust/buoyant force which in turn retards the velocity of the ball in the water. Now, we know this retarding force will be ${a_{retarding}} = \dfrac{{{w_{apparent}}}}{{mass}}$ where ${w_{apparent}}$ = apparent weight of ball= (true weight of ball)-(weight of liquid displaced by the ball). From this we get:
$\Rightarrow {a_{retarding}} = \dfrac{{(V{\rho _b}g) - (V{\rho _w}g)}}{{V{\rho _b}}} = \left( {\dfrac{{{\rho _b} - {\rho _w}}}{{{\rho _b}}}} \right)g$
Where, ${\rho _w}$ = density of water, ${\rho _b}$ = density of ball. Substituting known values:
$\Rightarrow {a_{retarding}} = \left( {\dfrac{{0.4 - 1}}{{0.4}}} \right) \times \dfrac{{{{10}^{ - 3}}}}{{{{10}^{ - 3}}}} \times g = \dfrac{{ - 0.6}}{{0.4}}g = \dfrac{{ - 3}}{2}g$ ……eq2
Once the ball reaches its maximum depth, it will stop due to the retarding force and the final velocity, v will become 0. So from eq1 and eq2, we finally have $v = 0$ cm/sec, $u = \sqrt {18g}$ cm/sec and $a = \dfrac{{ - 3}}{2}g$. Here, we use the equation of motion ${v^2} - {u^2} = 2as$ and by substituting known values we get:
$\Rightarrow s = \dfrac{{{v^2} - {u^2}}}{{2a}} = \dfrac{{0 - {{\left( {\sqrt {18g} } \right)}^2}}}{{2 \times ( - \dfrac{3}{2}g)}} = \dfrac{{18}}{3} = 6cm$
Hence, when dropped in water, the ball will stop after it travels a distance of 6cm.

Note: The value of the force ${a_{retarding}}$ comes out to be negative, as it is a decelerating force which retards the speed of the ball in water.