Question

A ball whose density is 0.4*10^3kg/m^3 falls into water from a height of 9 cm. to what depth does the ball sink

Answer 1

A ball whose density 0.4 × 10³ kg/m³ falls into water . Before falling into the water, velocity of ball at the surface of water , u = √{2gh}

Now, use formula for finding acceleration of ball,
a = apparent weight/mass of ball
Apparent weight = V(ρ - σ)g
Where , V is the volume of ball , ρ is density of ball and σ is density of water.
Now, apprent weight = V(0.4 - 1) × 10³ × g = -0.6gV × 10³
∴ a = -0.6gV × 10³/ρV
= -0.6g/0.4 = -3g/2 m/s²

Use formula, v² = u² + 2aS
⇒0 = {√{2gh})² + 2(-3g/2)S
⇒2gh = 3gS
⇒2 × 9 = 3S ⇒S = 6cm

Hence, answer is 6cm

Answer 2

Hello dear.

Here is the answer---

Given---
Density of the Ball(ρ) = 0.4 × 10³ kg/m³.
Let the Volume and the Mass of the Ball be V and m respectively.
∵ Density = Mass/ Volume
∴ Mass(m) = 400 × V 

Apparent Weight of the Ball = V(ρ - σ)g
 = V(400 - 1000)g
 = V(-600)g
 = -600 Vg.

Using the Newton Second Law of the Motion,
 Apparent Weight(Force) = Mass(m) × Acceleration(a).
∴ Acceleration(a) = Apparent Weight/Mass of the Ball.
∴ a = (- 600 V g)/400 V
    = (-3/2)g  m/s²


Now, Using the Equation of the Motion,

v² - u² = 2ah
v = Final Velocity. 
 = 0
[Since the ball will be stop after going into the depth]
a = (-3/2)g.
S = Depth to which the ball will sink.
 
u = Initial Velocity
 = √2gh
[It is the velocity of the ball before striking the water

Now,
v² - u² = 2aS
⇒ (0)² - (√2gh)² = 2(-3/2)g × S
⇒ 2g(0.09) = 3g × S
⇒ 0.18 = 3 × S
⇒ S = 0.18/3
⇒ S = 0.06 cm.
⇒ S = 6 cm.

∴ Depth of the water to which the ball will sink is 6 cm.


Hope it helps.