A buttle weighting 10 g and moving with a velocity `800 ms^(-1)` strikes a 10 kg block resting on a frictionless surface. The speed of the block after the perfectly inelastic collision is approximately

Options

A. `8 cm s^(-1)`
B. `80 cm s^(-1)`
C. `8 ms ^(-1)`
D. `0.8 cm s^(-1)`

Answer

To solve the problem of a bullet colliding with a block in a perfectly inelastic collision, we can follow these steps:

### Step 1: Identify the given values
- Mass of the bullet, \( m_1 = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} \)
- Velocity of the bullet, \( u_1 = 800 \, \text{m/s} \)
- Mass of the block, \( m_2 = 10 \, \text{kg} \)
- Initial velocity of the block, \( u_2 = 0 \, \text{m/s} \)

### Step 2: Write the conservation of momentum equation
In a perfectly inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. The equation is:

\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) V \]

Where \( V \) is the final velocity of the combined mass after the collision.

### Step 3: Substitute the known values into the equation
Substituting the values we have:

\[ (10 \times 10^{-3} \, \text{kg}) \times (800 \, \text{m/s}) + (10 \, \text{kg}) \times (0 \, \text{m/s}) = (10 \times 10^{-3} \, \text{kg} + 10 \, \text{kg}) V \]

This simplifies to:

\[ (10 \times 10^{-3} \times 800) = (10 \times 10^{-3} + 10) V \]

### Step 4: Calculate the left side of the equation
Calculating the left side:

\[ 10 \times 10^{-3} \times 800 = 8 \, \text{kg m/s} \]

### Step 5: Simplify the right side of the equation
The right side becomes:

\[ (10 \times 10^{-3} + 10) = 10.01 \, \text{kg} \quad \text{(approximately 10 kg)} \]

### Step 6: Set up the equation to solve for \( V \)
Now we can set up the equation:

\[ 8 = 10.01 V \]

### Step 7: Solve for \( V \)
To find \( V \):

\[ V = \frac{8}{10.01} \approx 0.8 \, \text{m/s} \]

### Step 8: Convert to centimeters per second
To convert \( V \) from meters per second to centimeters per second:

\[ V = 0.8 \, \text{m/s} \times 100 = 80 \, \text{cm/s} \]

### Conclusion
The speed of the block after the perfectly inelastic collision is approximately \( 80 \, \text{cm/s} \).