Answer:
4v/3
Explanation:
Assume elastic collision by the law of momentum conservation:
[tex]m_1v = m_1v_1 + m_2v_2[/tex]
where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively
Substitute [tex]m_2 = m_1/2 \& v_1 = v/3[/tex]
[tex]m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}[/tex]
Divide both side by [tex]m_1[/tex], then multiply by 6 we have
[tex]6v = 2v + 3v_2[/tex]
[tex]3v_2 = 4v[/tex]
[tex]v_2 = \frac{4v}{3}[/tex]
So the final speed of the second car is 4/3 of the first car original speed
