Explanation:
The initial kinetic energy [tex]KE_0[/tex] for both blocks is zero. Let [tex]m_1= m[/tex] and [tex]m_2 =3m[/tex]. So using the conservation law of linear momentum, we can write
[tex]0 = m_1v_1 - m_2v_2[/tex]
or
[tex]v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1[/tex]
The final kinetic energies for the two masses are
[tex]KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2[/tex]
[tex]KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)[/tex]
Therefore, the ratio of their kinetic energies is
[tex]\dfrac{\Delta KE_2}{\Delta KE_1} = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}[/tex]
