Head-on Proton Collision: Finding Speed

AI Thread Summary
In a head-on collision of two protons, one proton has double the speed of the other, and they approach each other to a closest distance of 3.0 x 10^14 m. The discussion highlights the need to apply the conservation of energy principle to solve for their initial speeds. The total energy equation includes kinetic energy and potential energy due to their interaction. Participants express confusion about how to handle the unknowns: mass, velocity, and charge. The conversation emphasizes the importance of breaking down the problem into manageable parts to find a solution.
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Question - Two protons, separated by a large distance, are fired directly at each other. The closest approach in this head-on collision is 3.0 x 10^14 m. One proton starts with two times the speed of the other. Assuming there is no delection from the original path, determine the initial speed of each proton.

Solution - I read through my notes and textbook but I couldn't find any way to connect speed with this question. Is there any specific equations to use?
 
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kevykevy said:
Question - Two protons, separated by a large distance, are fired directly at each other. The closest approach in this head-on collision is 3.0 x 10^14 m. One proton starts with two times the speed of the other. Assuming there is no delection from the original path, determine the initial speed of each proton.

Solution - I read through my notes and textbook but I couldn't find any way to connect speed with this question. Is there any specific equations to use?
Conservation of energy.

the total energy is {1 \over 2} m_1 v_1^2 + {1 \over 2} m_2 v_2^2 - {k q_1 q_2 \over r} and it is conserved.
 
So, since I don't know what the mass is or the speed, would I split the question up into two parts, the first being at collision so I can solve for mass since speed is zero. Then I can use the law of conservation of energy to find velocity at start?
 
Actually, how would I solve that if there is 3 unknowns: mass, velocity, and charge?
 
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