To solve this problem, we need to use the principle of conservation of momentum and the principle of conservation of kinetic energy, both of which apply to elastic collisions.
- Conservation of Momentum:
The total momentum before the collision is equal to the total momentum after the collision. Let \( m_1 = 2\, \text{kg} \) be the mass of the first body, and \( u \) be its initial velocity. The initial velocity of the second body \( m_2 \) is zero since it is at rest. \(m_1 \times u + m_2 \times 0 = m_1 \times v_1 + m_2 \times v_2\)
Given, \( v_1 = \frac{u}{4} \). Thus, \(2 \times u = 2 \times \frac{u}{4} + m_2 \times v_2\)
Simplifying gives: \(2u = \frac{u}{2} + m_2 \times v_2\)
\(4u - u = 2m_2 \times v_2\)
\(3u = 2m_2 \times v_2\) - Conservation of Kinetic Energy:
The total kinetic energy before collision is equal to the total kinetic energy after collision. \(\frac{1}{2} m_1 u^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\)
Substituting the values: \(\frac{1}{2} \times 2 \times u^2 = \frac{1}{2} \times 2 \times \left(\frac{u}{4}\right)^2 + \frac{1}{2} \times m_2 \times v_2^2\)
Simplifying gives: \(u^2 = \frac{u^2}{8} + \frac{m_2 \times v_2^2}{2}\)
Perform algebraic simplifications: \(8u^2 = u^2 + 4m_2 \times v_2^2\)
\(7u^2 = 4m_2 \times v_2^2\) - Solving the Equations:
From the equation of momentum: \(v_2 = \frac{3u}{2m_2}\)
Substitute this expression into the kinetic energy equation: \(7u^2 = 4m_2 \times \left(\frac{3u}{2m_2}\right)^2\)
\(7u^2 = \frac{36u^2}{4m_2}\)
Simplifying gives: \(7 = \frac{36}{4m_2}\)
\(28m_2 = 36\)
Thus, \( m_2 = \frac{36}{28} = 1.2857 \approx 1.2 \, \text{kg} \) (considering the options given).
Therefore, the mass of the second body is 1.2 kg.