Answer: The correct option is D.
Explanation: We are given two gas molecules with different masses. The collision between them is elastic, hence the Total Kinetic energy of the system is conserved.
Initial Kinetic Energy:
[tex]K.E_i=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2[/tex]
Final Kinetic Energy:
[tex]K.E_f=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
where,
[tex]m_1[/tex] = mass of first molecule
[tex]m_2[/tex] = mass of second molecule
[tex]u_1[/tex] = initial velocity of first molecule
[tex]u_2[/tex] = initial velocity of second molecule
[tex]v_1[/tex] = final velocity of first molecule
[tex]v_2[/tex] = final velocity of second molecule
Elastic Collision:
[tex]K.E_i=K.E_f[/tex]
That is,
[tex]\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
[tex]m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2[/tex] .....(1)
We are given in the question that the final velocity of first molecule is decreased by 1/4, which means:
[tex]v_1=\frac{u_1}{4}[/tex]
Putting this value in equation 1 , we get
[tex]m_1u_1^2+m_2u_2^2=\frac{m_1u_1^2}{16}+m_2v_2^2[/tex]
[tex]m_1u_1^2-\frac{m_1u_1^2}{16}+m_2u_2^2=m_2v_2^2[/tex]
[tex]m_1(\frac{15u_1^2}{16})+m_2u_2^2=m_2v_2^2[/tex]
taking [tex]m_2[/tex] on other side, we get
[tex]\frac{m_1}{m_2}(\frac{15u_1^2}{16})+u_2^2=v_2^2[/tex]
From the above relation, it is visible that the velocity change of the second molecule depends on both mass and velocity of the first molecule.
