Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is

Elastic Collision and Energy Loss

Body A of mass \(4m\) moving with speed \(u\) collides with another body B of mass \(2m\), at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is:

(a) \( \frac{5}{9} \)

(b) \( \frac{1}{9} \)

(d) \( \frac{4}{9} \)

Detailed Answer:

                In an elastic collision, both momentum and kinetic energy are conserved. We are given:
                - Mass of body A = \( 4m \),
                - Mass of body B = \( 2m \),
                - Initial velocity of A = \( u \),
                - Initial velocity of B = \( 0 \).

                Let's calculate the velocities of bodies A and B after the collision using the formula for elastic collisions:
                \[
                v_A = \frac{(m_A - m_B)v_A + 2m_Bv_B}{m_A + m_B}
                \]
                \[
                v_B = \frac{(m_B - m_A)v_B + 2m_Av_A}{m_A + m_B}
                \]
                Substituting the given values:

                \[
                v_A = \frac{(4m - 2m)(u) + 2(2m)(0)}{4m + 2m} = \frac{2mu}{6m} = \frac{u}{3}
                \]
                \[
                v_B = \frac{(2m - 4m)(0) + 2(4m)(u)}{4m + 2m} = \frac{8mu}{6m} = \frac{4u}{3}
                \]

                Now, let’s calculate the initial and final kinetic energies:
                - Initial kinetic energy of A: 
                \[
                KE_{\text{initial A}} = \frac{1}{2} (4m) u^2 = 2mu^2
                \]
                - Initial kinetic energy of B:
                \[
                KE_{\text{initial B}} = 0 \, \text{(since B is at rest)}.
                \]
                - Final kinetic energy of A:
                \[
                KE_{\text{final A}} = \frac{1}{2} (4m) \left(\frac{u}{3}\right)^2 = \frac{4m u^2}{18} = \frac{2m u^2}{9}
                \]
                - Final kinetic energy of B:
                \[
                KE_{\text{final B}} = \frac{1}{2} (2m) \left(\frac{4u}{3}\right)^2 = \frac{2m \cdot 16u^2}{18} = \frac{16m u^2}{18} = \frac{8m u^2}{9}
                \]

                The total initial kinetic energy is:
                \[
                KE_{\text{initial total}} = 2mu^2
                \]
                The total final kinetic energy is:
                \[
                KE_{\text{final total}} = \frac{2m u^2}{9} + \frac{8m u^2}{9} = \frac{10m u^2}{9}
                \]

                The energy lost by body A is:
                \[
                \Delta E_A = KE_{\text{initial A}} - KE_{\text{final A}} = 2mu^2 - \frac{2m u^2}{9} = \frac{16m u^2}{9}
                \]

                The fraction of energy lost by body A is:
                \[
                \text{Fraction of energy lost} = \frac{\Delta E_A}{KE_{\text{initial A}}} = \frac{\frac{16m u^2}{9}}{2mu^2} = \frac{8}{9}.
                \]

                Therefore, the fraction of energy lost by body A is:
                \[
                \boxed{\frac{8}{9}} \ (\text{Option (c)}).
                \]
            