Proton bounces back at 3.90×10^7 m/s from unknown nucleus

[oldspice1212]

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?

 

[PeterO]

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].