Question

A 35.0-g bullet moving at 475 m/s strikes a 2.5-kg wooden block. The bullet passes through the block, leaving at 275 m/s.

Answer 1

To analyze the scenario where a bullet strikes a wooden block and passes through it, we can use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision, assuming no external forces are acting on the system.

**Given data:**
- Mass of bullet, \( m_b = 35.0 \, \text{g} = 0.035 \, \text{kg} \)
- Initial velocity of bullet, \( v_{b_i} = 475 \, \text{m/s} \)
- Final velocity of bullet, \( v_{b_f} = 275 \, \text{m/s} \)
- Mass of wooden block, \( m_{block} = 2.5 \, \text{kg} \)
- Initial velocity of block, \( v_{block_i} = 0 \, \text{m/s} \) (it is at rest)

**Calculate initial momentum:**
\[
\text{Initial momentum} = m_b v_{b_i} + m_{block} v_{block_i} = (0.035 \, \text{kg})(475 \, \text{m/s}) + (2.5 \, \text{kg})(0 \, \text{m/s}) = 16.625 \, \text{kg m/s}
\]

**Calculate final momentum:**
The final momentum is composed of the momentum of the bullet after it exits the block and the momentum of the block (which starts moving).

Let \( v_{block_f} \) be the final velocity of the wooden block.
\[
\text{Final momentum} = m_b v_{b_f} + m_{block} v_{block_f}
\]
Hence,
\[
\text{Final momentum} = (0.035 \, \text{kg})(275 \, \text{m/s}) + (2.5 \, \text{kg}) v_{block_f}
\]

**Setting initial momentum equal to final momentum:**
\[
16.625 \, \text{kg m/s} = (0.035 \, \text{kg})(275 \, \text{m/s}) + (2.5 \, \text{kg}) v_{block_f}
\]
Calculate the bullet’s momentum after passing through the block:
\[
(0.035)(275) = 9.625 \, \text{kg m/s}
\]
Now substitute this value back into the momentum equation:
\[
16.625 = 9.625 + (2.5)v_{block_f}
\]

**Solving for \( v_{block_f} \):**
\[
16.625 - 9.625 = (2.5)v_{block_f}
\]
\[
7 = (2.5)v_{block_f}
\]
\[
v_{block_f} = \frac{7}{2.5} = 2.8 \, \text{m/s}
\]

Thus, the final velocity of the wooden block after the bullet passes through it is **2.8 m/s**.