What is the Maximum Angle of Deflection in a Particle Collision?

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SUMMARY

The maximum angle of deflection in a particle collision involving a mass M colliding with a stationary mass m (where M > m) can be analyzed using the center of mass (CoM) reference frame. In this frame, the initial momentum is zero, leading to the equations p_{CdM} = Mu_1 - mu_2 = 0 before and after the collision. The challenge lies in relating the velocities in the CoM frame (u_1, u_2, u_1', u_2') to the laboratory frame velocities (v_1', v_2'). A clear understanding of four-vector representation is essential for visualizing the scattering angles.

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  • Understanding of particle collision dynamics
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  • Knowledge of momentum conservation laws
  • Proficiency in four-vector notation and transformations
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  • Study the principles of momentum conservation in particle collisions
  • Learn about four-vector representations in relativistic physics
  • Explore the mathematical derivation of scattering angles in CoM frame
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Homework Statement


A particle of mass ##M## and speed ##v_1## bumps a particle of mass ##m##, ##(M >m)##, at rest. Find the maximum angle of deflection of ##M## after the bump.

The Attempt at a Solution


I would like to solve this problem in the reference frame of the center of mass, because it seems that the counts are very little :D
By definition, if I call ##u_1,u_2## the velocities of the particles in the CdM reference frame, I have ##p_{CdM}=Mu_1 -mu_2=0##. Then I call ##u_1 ',u_2 '## the velocities after the bump (the quantity of motion of Cdm is 0 also after the bump). In the image there is a sketch of the situation, with ##\phi## the scattering angle. But now I'm struggling in relating the velocities ##u_1,u_2,u_1 ',u_2 '## with the end velocities of the particles ##v_1 ', v_2'## in the frame of the laboratory. My question is: how to sketch this four vectors?
 

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In the CoM frame the masses can go in any direction. The problem asks for the frame where the larger mass is initially at rest.
 

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