Calculate Mass & Velocity of Unknown Element in Proton Collision

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The discussion focuses on calculating the mass and velocity of an unknown element's nucleus during elastic collisions with protons in a particle accelerator. The initial speed of the protons is 4.20×107 m/s, and they rebound at 3.90×107 m/s after colliding with the target nucleus. Using the conservation of momentum and the properties of elastic collisions, the mass of the unknown nucleus can be expressed in terms of the proton mass, and the speed of the nucleus post-collision can be determined through the center of mass calculations.

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You are at the controls of a particle accelerator, sending a beam of 4.20×10^7 m/s protons
(mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 3.90×10^7 m/s . Assume that the initial speed of the target nucleus is negligible and the collision is elastic.

1. Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m.

2. What is the speed of the unknown nucleus immediately after such a collision?


How do I do this? I first thought I would have to use conservation of momentum but I'm not sure. M1U1 + M2U1 = M1V1 + M2V2?
 
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oldspice1212 said:
You are at the controls of a particle accelerator, sending a beam of 4.20×10^7 m/s protons
(mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 3.90×10^7 m/s . Assume that the initial speed of the target nucleus is negligible and the collision is elastic.

1. Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m.

2. What is the speed of the unknown nucleus immediately after such a collision?How do I do this? I first thought I would have to use conservation of momentum but I'm not sure. M1U1 + M2U1 = M1V1 + M2V2?

During all collisions, the velocity of the Centre of Mass remains constant.

During elastic, head-on collisions, if you observe from the frame of reference of the Centre of Mass, each of the masses appear to bounce off at the same speed as they approach the Centre of Mass. [try this calculation with a few of your standard elastic collision problems to confirm it is correct]

That means the Velocity of the Centre of mass is the average of the before and after velocities of each/either body.

Briefly ignoring the 10^7 factor, this proton had an initial velocity of 4.20 in one direction, then after collision had a velocity of 3.90 in the other.
Taking the initial velocity as positive, that means +4.20 and -3.90
The average of those is 0.15

So the Centre of Mass is moving at 1.5 x 106 ms-1
From that value, you can calculate the mass of the nucleus [in terms of the proton mass].
 
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