To solve this problem, we can break it down into two parts: the collision between the bullet and the pendulum and the subsequent motion of the pendulum.

Step 1: Determine the velocity of the pendulum immediately after the collision

First, we use the principle of conservation of mechanical energy to analyze the motion of the pendulum after the bullet becomes embedded in it.

Energy at the highest point:

At the highest point, the kinetic energy is converted to potential energy. The potential energy at the maximum height $h = 18 \, \text{cm} = 0.18 \, \text{m}$ is:

$PE = (m_{\text{pendulum}} + m_{\text{bullet}}) \cdot g \cdot h$

where:

• $m_{\text{pendulum}} = 4.7 \, \text{kg}$
• $m_{\text{bullet}} = 3 \, \text{g} = 0.003 \, \text{kg}$
• $g = 9.8 \, \text{m/s}^2$
• $h = 0.18 \, \text{m}$

Substituting values:

$PE = (4.7 + 0.003) \cdot 9.8 \cdot 0.18 = 4.703 \cdot 9.8 \cdot 0.18$

$PE \approx 8.283 \, \text{J}$

At the highest point, all the kinetic energy has been converted into potential energy. Hence, the kinetic energy of the combined system just after the bullet is embedded is:

$KE = PE \approx 8.283 \, \text{J}$

Step 2: Use the kinetic energy to find the velocity after the collision

The kinetic energy of the system is given by:

$KE = \frac{1}{2} (m_{\text{pendulum}} + m_{\text{bullet}}) \cdot v_f^2$

where $v_f$ is the velocity of the combined system immediately after the collision. Solving for $v_f$:

$v_f = \sqrt{\frac{2 \cdot KE}{m_{\text{pendulum}} + m_{\text{bullet}}}}$

Substitute the values:

$v_f = \sqrt{\frac{2 \cdot 8.283}{4.703 + 0.003}} = \sqrt{\frac{16.566}{4.706}}$

$v_f \approx \sqrt{3.52} \approx 1.88 \, \text{m/s}$

Thus, the velocity of the combined system immediately after the collision is approximately 1.88 m/s.

Step 3: Apply the conservation of momentum to find the initial velocity of the bullet

The collision between the bullet and the pendulum is perfectly inelastic, meaning the bullet and pendulum move together after the collision. Therefore, we apply the principle of conservation of momentum before and after the collision.

The total momentum before the collision is equal to the total momentum after the collision:

$m_{\text{bullet}} \cdot v_{\text{bullet}} = (m_{\text{pendulum}} + m_{\text{bullet}}) \cdot v_f$

where $v_{\text{bullet}}$ is the initial velocity of the bullet. Solving for $v_{\text{bullet}}$:

$v_{\text{bullet}} = \frac{(m_{\text{pendulum}} + m_{\text{bullet}}) \cdot v_f}{m_{\text{bullet}}}$

Substitute the values:

$v_{\text{bullet}} = \frac{(4.7 + 0.003) \cdot 1.88}{0.003}$

$v_{\text{bullet}} = \frac{4.703 \cdot 1.88}{0.003} \approx \frac{8.832}{0.003}$

$v_{\text{bullet}} \approx 2944 \, \text{m/s}$

Thus, the initial velocity of the bullet is approximately 2944 m/s.

Final Answer:

The bullet's initial speed is approximately 2944 m/s.

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Would you like more details on any part of the solution or have any questions? Here are a few related questions you might find interesting:

1. How does the principle of conservation of momentum apply to inelastic collisions?
2. What would happen to the result if the collision were elastic instead?
3. How can we calculate the maximum height of the pendulum if the initial speed were different?
4. How does the combined mass of the pendulum and bullet affect the system's energy?
5. Can we calculate the energy lost during the collision?

Tip: Remember that when calculating energy conversions (like from kinetic to potential), it's helpful to consider the entire system, as both the bullet and pendulum are involved in the process.