Question

A ball of relative density 0.8 falls into water from a height of 2m. The depth to which the ball will sink is (neglect viscous forces):

Answer 1

(Note: In this solution subscript B means for ball and subscript W means for water)

Given that, relative density of the ball is 0.8,

$\longrightarrow \sf \dfrac{\rho_B}{\rho_W} = 0.8$

$\longrightarrow \sf \dfrac{\rho_W}{\rho_B} = \dfrac{1}{0.8}$

$\longrightarrow \sf \dfrac{\rho_W}{\rho_B} = 1.25 \quad \quad \dots(1)$

We can divide this problem into two parts - first, before entering water, and second, after entering water.

Before entering water:-

Using third equation of motion,

$\sf v^2 = u^2 + 2as$

Putting $\sf v = v_1, u = 0\:m/s, s=-2\:m$ and $\sf a = -g,$

$\longrightarrow \sf (v_1)^2 = 0^2 + 2(-10)(-2)$

$\longrightarrow \sf (v_1)^2 = 40$

$\longrightarrow \sf v_1 = \sqrt{40} \quad \quad \dots (2)$

After entering water:-

According to the F.B.D.,

$\sf (m_Bg) - ({F}_{(Buoyant)}) = m_Ba$

As Buoyant force is equal to the weight of the fluid displaced,

$\longrightarrow \sf (m_Bg) - (m_W\times g) = m_Ba$

$\longrightarrow \sf (m_Bg) - (\rho_W\times V_W \times g) = m_Ba$

As the same volume of water is displaced as much is the volume of the ball,

$\longrightarrow \sf (m_Bg) - (\rho_W\times V_B \times g) = m_Ba$

As we know that Volume = Mass/Density,

$\longrightarrow \sf (m_Bg) - (\blue{\rho_W} \times \dfrac{m_B}{\blue{\rho_B}} \times g) = m_Ba$

From (1),

$\longrightarrow \sf (m_Bg) - (1.25\times m_B \times g) = m_Ba$

$\longrightarrow \sf m_Bg(1 - 1.25) = m_Ba$

$\longrightarrow \sf a = g \times -0.25$

$\longrightarrow \sf a = -2.5 \:m/s^2\quad\quad\dots (3)$

(Negative sign shows the acceleration is in opposite direction of the motion of the ball)

Now again, after the ball has entered water,

Applying third equation of motion,

$\sf v^2 = u^2 + 2as$

Putting $\sf v = v_2 = 0\:m/s, u = v_1$ and $\sf a = -2.5\:m/s^2$

$\longrightarrow \sf (v_2)^2 = (v_1)^2 + 2as$

From (2) and (3),

$\longrightarrow \sf 0 = (\sqrt{40})^2 + 2(-2.5)(s)$

$\longrightarrow \sf -40 = - 2\times 2.5 \times s$

$\longrightarrow \sf 40 = 2 \times 2.5 \times s$

$\longrightarrow \sf s = \dfrac{40}{2\times 2.5}$

$\longrightarrow \sf \underline{\underline{\sf{\green{ s = 8\:m}}}}$