91Ó°ÊÓ

Given data:

The kinetic energies of both bullets are the same.

Assumptions:

Let the mass of the first bullet be \({m_1}\) and the mass of the second bullet be \({m_2}\).

Let the speed of the first bullet be \({v_1}\) and the speed of the second bullet be \({v_2}\).

The lower mass has a greater speed if the kinetic energies of the two particles of different masses are the same because speed varies inversely with mass.

\(\begin{array}{l}K = \frac{1}{2}m{v^2}\\{v^2} \propto \frac{1}{m}{\rm{ }}\left( {k = {\rm{constant}}} \right)\end{array}\)

Let the mass of the first bullet be twice the mass of the second, i.e.,\({m_1} = 2{m_2}\).

According to the question, the kinetic energies of both bullets are the same. Therefore,

\(\begin{array}{c}\frac{1}{2}{m_1}v_1^2 = \frac{1}{2}{m_2}v_2^2\\{\left( {\frac{{{v_1}}}{{{v_2}}}} \right)^2} = \frac{{{m_2}}}{{{m_1}}}\\{\left( {\frac{{{v_1}}}{{{v_2}}}} \right)^2} = \frac{{{m_2}}}{{2{m_2}}}\\{v_2} = \sqrt 2 {v_1}\end{array}\)

Hence, the lighter bullet has a higher speed, and the speed of the lighter bullet is \(\sqrt 2 \) times the speed of the heavier bullet if the heavier mass is double the lighter mass.

You know that work done is equal to the change in kinetic energy.

Since both bullets have the same kinetic energy, they will do an equal amount of work.