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Momentum is a crucial concept in physics, especially when analyzing collisions. In an elastic collision, like the one we're considering with the croquet ball, momentum is always conserved. This means that the total momentum before the collision is equal to the total momentum after the collision.

When analyzing momentum, we denote the mass of objects as \( m \) and their velocities as \( v \). For our problem, the first croquet ball, with mass \( m_1 = 0.280 \ \text{kg} \), initially moves with velocity \( v_1 \) while the second ball is at rest (\( v_2 = 0 \)). After the collision, the first ball slows down to \( v_1' \) and the second ball starts moving with \( v_2 = \frac{1}{2}v_1 \).

The conservation of momentum equation for this scenario is:

• \( m_1 v_1 = m_1 v_1' + m_2 v_2 \)

Substituting the known quantities and solving it helps us find the mass of the second ball \( m_2 \). This illustrates how momentum conservation provides a powerful tool to solve collision problems.

Elastic collisions are fascinating because they preserve another essential quantity: kinetic energy. Kinetic energy, which is the energy of motion, is also conserved in the elastic collision we're studying. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \).

In our scenario, the initial kinetic energy of the moving croquet ball \( KE_i \) is calculated using:

• \( KE_i = \frac{1}{2}m_1 v_1^2 \)

After the collision, both balls have kinetic energy:

• \( KE_{1f} = \frac{1}{2}m_1 (v_1')^2 \)
• \( KE_{2f} = \frac{1}{2}m_2 v_2^2 \)

Using the relationships discovered, such as \( v_1' = \frac{1}{2}v_1 \) and understanding that \( m_2 = m_1 \), it becomes evident that the total initial kinetic energy equals the total final kinetic energy, thus proving the conservation of kinetic energy in this elastic collision.

In the given problem, understanding how kinetic energy transfers during an elastic collision is essential. Here, we want to determine how much of the original kinetic energy of our moving croquet ball is transferred to the second ball.

The fraction of kinetic energy transferred to the second ball can be found by comparing the kinetic energy of the second ball after the collision to the initial kinetic energy of the first ball:

• Transferred Kinetic Energy: \( KE_{2f} = \frac{1}{2}m_2 (v_2)^2 \)
• Initial Kinetic Energy: \( KE_i = \frac{1}{2}m_1 v_1^2 \)

From the exercise, \( m_2 = m_1 \) and \( v_2 = \frac{1}{2}v_1 \), leading to:

• \( \frac{\Delta KE}{KE_i} = \frac{1}{4} \)

Thus, only one-fourth of the original kinetic energy is transferred to the second ball. This insight reveals how energy redistributes in an elastic collision, where not all energy transfers to the initially stationary object.