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A 10 g bullet moving directly upward at ...

A 10 g bullet moving directly upward at `1000 ms^(-1)` strikes and passes through the center of mass of a 10 kg block initially at rest. The bullet emerges from the block moving directly upwards at `400 ms^(-1)` . What will be the velocity of the block just after the bullet comes out of it ?

A

`1 ms^(-1)`

B

`0.4ms^(-1)`

C

`1.4 ms^(-1)`

D

`0.6ms^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the principle of conservation of momentum. The momentum before the bullet strikes the block must equal the momentum after the bullet passes through the block. ### Step-by-Step Solution: 1. **Identify the given values:** - Mass of the bullet, \( m_b = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} \) - Initial velocity of the bullet, \( u_b = 1000 \, \text{m/s} \) - Final velocity of the bullet after passing through the block, \( v_b = 400 \, \text{m/s} \) - Mass of the block, \( m_{block} = 10 \, \text{kg} \) - Initial velocity of the block, \( u_{block} = 0 \, \text{m/s} \) (since it is at rest) 2. **Write the conservation of momentum equation:** The total momentum before the collision equals the total momentum after the collision. \[ m_b \cdot u_b + m_{block} \cdot u_{block} = m_b \cdot v_b + m_{block} \cdot v_{block} \] 3. **Substitute the known values into the equation:** \[ (10 \times 10^{-3} \, \text{kg}) \cdot (1000 \, \text{m/s}) + (10 \, \text{kg}) \cdot (0 \, \text{m/s}) = (10 \times 10^{-3} \, \text{kg}) \cdot (400 \, \text{m/s}) + (10 \, \text{kg}) \cdot v_{block} \] 4. **Calculate the left side of the equation:** \[ 10 \times 10^{-3} \cdot 1000 = 10 \, \text{kg m/s} \] 5. **Calculate the right side of the equation:** \[ 10 \times 10^{-3} \cdot 400 = 4 \, \text{kg m/s} \] So the equation becomes: \[ 10 = 4 + 10 \cdot v_{block} \] 6. **Rearranging the equation to solve for \( v_{block} \):** \[ 10 - 4 = 10 \cdot v_{block} \] \[ 6 = 10 \cdot v_{block} \] \[ v_{block} = \frac{6}{10} = 0.6 \, \text{m/s} \] 7. **Conclusion:** The velocity of the block just after the bullet comes out of it is \( 0.6 \, \text{m/s} \).
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