Answer:
Explanation:
Given
Object 1 is of mass [tex]\frac{m}{2}[/tex]
Object 2 has mass of [tex]m[/tex]
They undergone inelastic collision and do not move after collision i.e. the collision is perfectly inelastic
Final momentum of both the object is zero
suppose [tex]v_1[/tex] and [tex]v_2[/tex] are the velocities of object 1 and 2 respectively then
[tex]m_1v_1+m_2v_2=0[/tex]
[tex]\frac{v_1}{v_2}=-\frac{m_2}{m_1}[/tex]
[tex]\frac{v_1}{v_2}=-2[/tex]
[tex]v_1=-2v_2[/tex]
i.e. velocity of object 1 is twice the velocity of object 2 but opposite to the direction of object 2
The comparison of the velocity of both objects before the collision is [tex]u_1=-2u_2[/tex]
According to the law of conservation of momentum, the momentum of a body before collision is equal to the momentum after collision. Mathematically;
[tex]m_1v_1 + m_2u_2=m_1v_1+m_2v_2[/tex]
Since both objects do not move after the collision, the equation becomes:
[tex]m_1v_1 + m_2u_2=0[/tex]
Also if Objects 1 has half the mass of object 2, then [tex]m_1=0.5m_2[/tex]
Substitute into the formula above:
[tex]0.5m_2u_1 + m_2u_2=0\\0.5m_2u_1 =-m_2u_2\\0.5u_1=-u_2\\u_1=-2u_2[/tex]
The comparison of the velocity of both objects before the collision is [tex]u_1=-2u_2[/tex]
Learn more here: https://brainly.com/question/24634142
