Answer:
[tex]v_B=\frac{v}{3}[/tex]
Explanation:
Given that:
mass of object A, [tex]m_A=M[/tex]
mass of object B, [tex]m_B=2M[/tex]
speed of object A, [tex]v_A=v[/tex]
So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.
[tex]m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B[/tex]
[tex]M\times v+0 = M\times \frac{v}{3}+2M\times v_B[/tex]
[tex]2M\times v_B= \frac{2M\times v}{3}[/tex]
[tex]v_B=\frac{v}{3}[/tex]
