Energy Transfer In Elastic Collisions

For a one-dimensional elastic collision in which a mass $m_1$ strikes a stationary mass $m_2$, the fraction of kinetic energy transferred to the target is $\frac{4m_1{m}_2}{(m_1+m_2{)}^2}$, a result derived by combining momentum and kinetic-energy conservation. Substituting $m_1=1\text{ kg}$ and $m_2=3\text{ kg}$ gives $\frac{4\times 1\times 3}{(1+3{)}^2}=\frac{12}{16}=0.75$, or 75%. The option 25% reports the fraction the lighter ball retains rather than transfers. The option 50% would only hold for equal masses, which is not the case here. The option 100% wrongly assumes the incoming ball stops, true only for equal masses. The transfer fraction reaches its maximum value of unity only when the two masses are exactly equal, and it falls off whether the target is made much heavier or much lighter than the projectile. This very principle explains why neutrons are slowed most efficiently by collisions with light nuclei of comparable mass, such as hydrogen, in the moderators of nuclear reactors. This applies the NCERT energy-transfer formula for unequal-mass elastic collisions. As a check, the retained fraction must be ${\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2={\left(\frac{-2}{4}\right)}^2=0.25$, and $0.75+0.25=1$ confirms total energy conservation.