A ball is dropped from height 10 m . Ball is embedded in sand 1 m and stops, then

Options

A. (a) only momentum remains conserved
B. (b) only kinetic energy remains conserved
C. (c) both momenutm and kinetic energy are conserved
D. (d) neither kinetic energy nor momentum is conserved

Answer

To solve the problem of a ball being dropped from a height of 10 m and then embedding in sand for 1 m before stopping, we will follow these steps:

### Step 1: Calculate the velocity of the ball just before it hits the sand.
Using the equation of motion for free fall, we can find the velocity of the ball just before it hits the ground.

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

Where:
- \( u = 0 \) m/s (initial velocity, as the ball is dropped)
- \( g = 9.81 \) m/s² (acceleration due to gravity)
- \( h = 10 \) m (height from which the ball is dropped)

Substituting the values:

\[ v^2 = 0 + 2 \times 9.81 \times 10 \]
\[ v^2 = 196.2 \]
\[ v = \sqrt{196.2} \approx 14.0 \text{ m/s} \]

### Step 2: Calculate the momentum of the ball just before it hits the sand.
Momentum (\( p \)) is given by the formula:

\[ p = mv \]

Where:
- \( m \) is the mass of the ball (let's assume \( m = m_b \) kg)
- \( v \) is the velocity we just calculated.

So,

\[ p = m_b \times 14.0 \text{ m/s} \]

### Step 3: Calculate the deceleration of the ball as it embeds in the sand.
The ball comes to a stop after embedding in the sand for a distance of 1 m. We can use the equation of motion to find the deceleration (\( a \)).

Using:

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

Where:
- \( v = 0 \) m/s (final velocity, as the ball comes to rest)
- \( u = 14.0 \) m/s (initial velocity before embedding in sand)
- \( s = 1 \) m (distance embedded in sand)

Substituting the values:

\[ 0 = (14.0)^2 + 2a(1) \]
\[ 0 = 196 + 2a \]
\[ 2a = -196 \]
\[ a = -98 \text{ m/s}^2 \]

### Step 4: Calculate the force exerted by the sand on the ball.
Using Newton's second law, the force (\( F \)) can be calculated as:

\[ F = ma \]

Where:
- \( a = -98 \text{ m/s}^2 \) (deceleration)
- \( m = m_b \) (mass of the ball)

Thus,

\[ F = m_b \times (-98) \]

### Final Answer:
The force exerted by the sand on the ball is \( F = -98 m_b \) N, where \( m_b \) is the mass of the ball.

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