Problem: Suppose we have a particle of mass m moving with speed v that collides with another particle of mass M which is initially at rest. Show that the scattering angle of m after collision can be anywhere between 0 and 180 degress if m < M. Determine the maximum scattering angle when m > M.

Well, first of all I have no idea what they mean by the 'scattering angle'. I assume it's the angle between the old and new velocity vectors of m.

Let [tex]\vec{v} = (v, 0)[/tex] be the velocity vector of m before colliding, [tex]\vec{v'} = (v'_x, v'_y)[/tex] be the velocity vector of m after colliding, and [tex]\vec{u} = (u_x, u_y)[/tex] be the velocity vector of M after the collision.

Using conservation of momentum:

[tex](mv, 0) = m(v'_x, v'_y) + M(u_x, u_y)[/tex]

[tex]0 = mv'_y + Mu_y[/tex]

[tex]mv = mv'_x + Mu_x[/tex]

Using conservation of kinetic energy (since this is an elastic collion):[tex]mv^2/2 = mv'^2/2 + Mu^2/2 \rightarrow Mu^2/m = v^2 - v'^2[/tex]

Now, there are three equations and four unknowns (not counting m < M or M > m). I can derive an equation for the scattering angle in terms of of the velocities, but it doesn't tell me much. What can I do?

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