You are at the controls of a particle accelerator, sending a beam of 3.00×107 m/sm/s protons (mass m𝑚) 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 2.70×107 m/sm/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.Part AFind the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m𝑚.Express your answer as a multiple of m𝑚.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typemel𝑚el =nothingm𝑚SubmitRequest AnswerPart BWhat is the speed of the unknown nucleus immediately after such a collision?Express your answer in meters per second.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typevnucleus𝑣nucleus =

This problem involves the conservation of momentum and kinetic energy because it's an elastic collision.

Part A:

The initial momentum of the system is the momentum of the proton because the nucleus is initially at rest. The final momentum of the system is the sum of the momentum of the proton and the nucleus after the collision.

The initial kinetic energy of the system is the kinetic energy of the proton because the nucleus is initially at rest. The final kinetic energy of the system is the sum of the kinetic energy of the proton and the nucleus after the collision.

By setting the initial momentum equal to the final momentum and the initial kinetic energy equal to the final kinetic energy, we can solve for the mass of the nucleus in terms of the mass of the proton.

Let's denote the mass of the proton as m and the mass of the nucleus as M. The initial speed of the proton is 3.00×10^7 m/s and the final speed of the proton is 2.70×10^7 m/s. The speed of the nucleus after the collision is unknown.

The initial momentum is m * 3.00×10^7 = m * 3.00×10^7.

The final momentum is m * 2.70×10^7 + M * v, where v is the speed of the nucleus after the collision.

The initial kinetic energy is 0.5 * m * (3.00×10^7)^2 = 0.5 * m * (3.00×10^7)^2.

The final kinetic energy is 0.5 * m * (2.70×10^7)^2 + 0.5 * M * v^2.

Setting the initial momentum equal to the final momentum and the initial kinetic energy equal to the final kinetic energy, we get two equations:

m * 3.00×10^7 = m * 2.70×10^7 + M * v

0.5 * m * (3.00×10^7)^2 = 0.5 * m * (2.70×10^7)^2 + 0.5 * M * v^2

Solving these two equations for M, we get M = 4m.

So, the mass of the nucleus is four times the mass of the proton.

Part B:

Substituting M = 4m into the momentum equation, we get:

m * 3.00×10^7 = m * 2.70×10^7 + 4m * v

Solving for v, we get v = 0.75×10^7 m/s.

So, the speed of the nucleus immediately after the collision is 0.75×10^7 m/s.