Elastic collision between alpha particle and gold Nuclei

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In an elastic collision between a 3.94 MeV alpha particle and a gold nucleus at rest, both conservation of momentum and energy must be applied to determine the kinetic energies of the recoiling nucleus and the rebounding alpha particle. The initial kinetic energy of the alpha particle is converted into the kinetic energy of both particles after the collision, with the total energy remaining constant at 3.94 MeV. The discussion emphasizes the importance of using conservation of momentum to find the velocities of the particles post-collision. The kinetic energy of the recoiling nucleus and the rebounding alpha particle can then be calculated from these velocities. Understanding these principles is crucial for solving the problem accurately.
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Homework Statement


When an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose a 3.94 MeV alpha particle has a head-on elastic collision with a gold nucleus that is initially at rest. What is the kinetic energy of (a) the recoiling nucleus and (b) the rebounding alpha particle? The gold nucleus has a mass of 197 u and the alpha has a mass of 4.0 u.

Homework Equations


K1+K2=3.94MeV

The Attempt at a Solution


Since the alpha particle will momentarily stop when the repulsion between the two particles is enough to stop it, the kinetic energy from the particle will all be converted into potential energy so K1f+K2f=3.94Mev. But I'm not sure where to go from here? I feel like I'm missing something simple
 
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You are missing conservation of momentum.

At the point where the alpha particle stops the gold nucleus is already moving, but details of the collision process are not relevant here.
 
as mfb hints...you need first to consider conservation of MOMENTUM to find the recoil velocities...the calculate the KEs
 

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