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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, $v$?

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1 Answer 1

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This is what the collision looks like in the centre of mass frame:

Collision

Because the particles have the same mass and the total momentum in the COM frame is zero both particles are coming in at a speed of $v/2$, and because we know the first particle goes out at 90 degrees the second particle must leave at 90 degrees in the opposite direction. Conservation of energy means all velocities are $v/2$. So just transform back to the lab frame to get the final velocity of $m_2$ in the lab frame.

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You're using conservation of energy. Are you assuming an elastic collision? It's not mentioned in the problem that the collision is elastic. –  Joebevo Aug 27 '12 at 10:24
    
If the collision is inelastic you can't answer the question because you haven't been given the co-efficient of restitution. –  John Rennie Aug 27 '12 at 13:22
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