If light and a heavy body have an equal kinetic energy of translation, then ________.

Concept:

\(\begin{array}{l} {m_1} > {m_2}\\ \frac{1}{2}{m_1}v_1^2 = \frac{1}{2}{m_2}v_2^2 \Rightarrow \frac{{{m_1}}}{{{m_2}}} = \frac{{v_2^2}}{{v_1^2}} > 1\\ {v_2} = {v_1}\sqrt {\frac{{{m_1}}}{{{m_2}}}} \\ As,\;{m_1} > {m_2},\;{v_2} > {v_1} \end{array}\)

∴ the speed of the smaller object is higher than that of bigger.

The momentum of the smaller object:

\(\begin{array}{l} {m_2}{v_2} = {m_2}{v_1}\sqrt {\frac{{{m_1}}}{{{m_2}}}} = {m_2}{v_1}\sqrt {\frac{{{m_1}}}{{{m_2}}}} .\frac{{{m_1}}}{{{m_1}}} = {m_1}{v_1}\;\sqrt {\frac{{{m_1}}}{{{m_2}}}} .\frac{{{m_2}}}{{{m_1}}} = {m_1}{v_1}\sqrt {\frac{{{m_2}}}{{{m_1}}}} \\ \frac{{{m_2}{v_2}}}{{{m_1}{v_1}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} < 1 \end{array}\)

\({m_2}{v_2} < {m_1}{v_1}\)

The lighter body will have smaller momentum.

Alternate solution:

\(KE=E=\frac{1}{2}mv^2 \)

P = mv

\(E=\frac{1}{2}mv^2 =\frac{p^2}{2m}\)

\(P=\sqrt {2mE}\)

For the same kinetic energy: 

P ∝ √m

The lighter body will have smaller momentum and heavier body will have larger momentum.