Two protons (each with rest mass m = 1.67 times 10^-27 kg) are initially moving

Question

Two protons (each with rest mass m = 1.67 times 10^-27 kg) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that produces an eta^0 particle. The rest mass of the eta^0 is m_eta = 9.75 times 10^-28 kg. Part A If the two protons and the eta^0 are all at rest after the collision, find the initial speed v of the protons. Express your answer as a fraction of the speed of light to three significant figures. Part B What is the kinetic energy E_k of each proton? Express your answer in millions of electron volts to three significant figures. Part C What is the rest energy E_r of the eta^0? Express your answer in millions of electron volts to three significant figures.

Explanation / Answer

A)
energy created = mn*c^2 = (9.75*10^-28)*c^2

Let speed of each proton be v

Let initial energy = 2*(0.5*mp*v^2)
= mp*v^2
= (1.673*10^-27)*v^2

use:
initial energy = final energy of n particle
(1.673*10^-27)*v^2 = (9.75*10^-28)*c^2
v= 0.763c
Answer: 0.763 c

B)
E of each proton = 0.5*m*v^2
= 0.5*(1.673*10^-27)*(0.763c)^2
= 0.5*(1.673*10^-27)*(0.763*3*10^8)^2
=1.76*10^-10 J
= (1.76*10^-10)/(1.6*10^-19)
=1.10*10^10 eV
= 1.10*10^4 MeV
Answer:1.10*10^4 MeV

C)
rest energy = initial energy of protons
= 2*1.10*10^4 MeV
= 2.20*10^4 MeV
Answer: 2.20*10^4 MeV

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.