Kinetic energy..

• The mass and velocity of Munem is $m_1 = 40$ kg and $v_1$ .
• The mass and velocity of Shahriar is $m_2 = 30$ kg and $v_2$ .

According to the condition, $m_1 v_1 = m_2 v_2$ .

So,

The kinetic energy of Munem, $E_{k'} = \dfrac 12 m_1 (v_1)^2 = \dfrac{(m_1 v_1)^2}{2m_1}$ .

The kinetic energy of Shahriar, $E_{k''} = \dfrac12 m_2 (v_2)^2 = \dfrac{(m_2 v_2)^2}{2m_2}$ .

Now,

$\large \dfrac{E_{k'}}{E_{k''}} = \dfrac{\frac{(m_1v_1)^2}{2m_1}}{\frac{(m_2 v_2)^2}{2m_2}} = \dfrac{(m_1 v_1)^2}{2m_1} \times \dfrac{2m_2}{(m_2 v_2)^2} = \dfrac{m_2}{m_1} = \dfrac{30}{40} = \dfrac 34$

Since $E_ {k''} > E_{k'},$ the kinetic energy of Shahariar is greater.