A particle of mass $M$ moving in a straight line with speed $v$ collides with a stationary particle of the same mass. In the center of mass coordinate system, the first particle is deflected by 90 degrees. Find the speed of the second particle after collision in the laboratory system.
My effort:
Considering the lab frame, we have, by conservation of linear momentum $$Mv=Mv_{1f}+Mv_{2f}\tag1$$ which implies that $$v=v_{1f}+v_{2f}\tag2$$ We want to know $v_{2f}$. Therefore, we have to find $v_{1f}$.
From my textbook, I get
$$v_{1f}=v+v_{1f}'\tag3$$
where $v_{1f}'$ is the post-scattering velocity of the incident particle in the center of mass system and $v$ is the velocity of the center of mass in the laboratory system.
Also from my text, I get
$$V=\frac{v_{1i}'}{2}=\frac{v}{2}\tag4$$
Therefore, it looks like I have everything I need to solve the problem except $v_{1f}'$. All I am given is that the post-scattering angle of the incident particle in the center of mass frame is 90 degrees. How do I find the velocity of the same particle in terms of the incident velocity in the lab system, $v$?