Two objects, one initially at rest, undergo a one-dimensional elastic collision. If half the kinetic energy of the initially moving object is transferred to the other object, what is the ratio of their masses?

Let's denote the initial object's mass as \(m_1\) and its initial velocity as \(v_1\). The stationary object's mass is \(m_2\) and its initial velocity is 0 since it's at rest. After the collision, they will have velocities \(v'_1\) and \(v'_2\) respectively. From the conservation of momentum, we have: \(m_1 \cdot v_1 = m_1 \cdot v'_1 + m_2 \cdot v'_2\). And from the problem statement, we have the conservation of kinetic energy expressed as: \(\frac{1}{2} m_1 \cdot v_1^2 = \frac{1}{2} m_1 \cdot (v_1')^2 + \frac{1}{2} m_2 \cdot (v_2')^2\). Here also it is mentioned that half the kinetic energy of the initially moving object is transferred to the other object, means \( \frac{1}{2} m_1 \cdot {v_1}^2 = \frac{1}{2} m_2 \cdot (v_2')^2\)

From Step 1, we have equation \( \frac{1}{2} m_1 \cdot {v_1}^2 = \frac{1}{2} m_2 \cdot (v_2')^2\). This equation can be simplified to \(v_2' = v_1 \cdot \sqrt{\frac{m_1}{m_2}}\)

From Step 1, we have the momentum equation: \(m_1 \cdot v_1 = m_1 \cdot v'_1 + m_2 \cdot v'_2\). Substitute \(v_2' = v_1 \cdot \sqrt{\frac{m_1}{m_2}}\) (from step 2) into momentum equation, we get \(m_1 \cdot v_1 = m_1 \cdot v'_1 + m_2 \cdot v_1 \cdot \sqrt{\frac{m_1}{m_2}}\)

Solving the equation from Step 3, we get \(m_1/m_2 = 3\)