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

Options

A. 9 cm
B. 6 cm
C. 4.5 cm
D. 2.25 cm

Explanation

To solve the problem of how deep the ball sinks in water after being dropped from a height of 9 cm, we can follow these steps:

### Step 1: Calculate the velocity of the ball just before it hits the water
The ball falls from a height \( h = 9 \, \text{cm} \). We can use the equation of motion to find the velocity just before it hits the water:

\[ v = \sqrt{2gh} \]

Where:
- \( g \) is the acceleration due to gravity. In CGS units, \( g \approx 980 \, \text{cm/s}^2 \).
- \( h = 9 \, \text{cm} \).

Substituting the values:

\[ v = \sqrt{2 \times 980 \, \text{cm/s}^2 \times 9 \, \text{cm}} = \sqrt{17640} \approx 132.5 \, \text{cm/s} \]

### Step 2: Determine the buoyant force acting on the ball
The density of the ball is given as \( \rho_b = 0.4 \times 10^3 \, \text{kg/m}^3 = 400 \, \text{kg/m}^3 \) and the density of water is \( \rho_w = 1 \times 10^3 \, \text{kg/m}^3 = 1000 \, \text{kg/m}^3 \).

The buoyant force \( F_b \) acting on the ball is equal to the weight of the water displaced by the ball:

\[ F_b = V \cdot \rho_w \cdot g \]

Where \( V \) is the volume of the ball. The weight of the ball \( W_b \) is:

\[ W_b = V \cdot \rho_b \cdot g \]

### Step 3: Calculate the apparent weight of the ball
The apparent weight \( W_a \) of the ball when submerged is given by:

\[ W_a = W_b - F_b \]

Substituting the expressions for \( W_b \) and \( F_b \):

\[ W_a = V \cdot \rho_b \cdot g - V \cdot \rho_w \cdot g = V \cdot g (\rho_b - \rho_w) \]

### Step 4: Calculate the retardation of the ball
The retardation \( a \) can be calculated using the apparent weight:

\[ a = \frac{W_a}{m} = \frac{V \cdot g (\rho_b - \rho_w)}{V \cdot \rho_b} = g \frac{\rho_b - \rho_w}{\rho_b} \]

Substituting the values:

\[ a = g \frac{0.4 \times 10^3 - 1 \times 10^3}{0.4 \times 10^3} = g \frac{-0.6 \times 10^3}{0.4 \times 10^3} = -\frac{3}{2}g \]

### Step 5: Use the equations of motion to find the distance sunk
Using the equation of motion:

\[ v^2 = u^2 + 2as \]

Where:
- \( v = 0 \) (final velocity when the ball stops sinking),
- \( u = 132.5 \, \text{cm/s} \) (initial velocity),
- \( a = -\frac{3}{2}g \).

Substituting the values:

\[ 0 = (132.5)^2 + 2 \left(-\frac{3}{2}g\right) s \]

Rearranging gives:

\[ s = \frac{(132.5)^2}{3g} \]

Substituting \( g \approx 980 \, \text{cm/s}^2 \):

\[ s = \frac{17556.25}{3 \times 980} \approx \frac{17556.25}{2940} \approx 5.97 \, \text{cm} \]

### Conclusion
The ball sinks approximately \( 6 \, \text{cm} \) into the water.

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