Question:

Two subatomic particles are accelerated in opposite directions in a circular particle accelerator with a radius of 500 meters. Particle A has a charge of +1e and a mass of 9.1 x 10^-31 kg, while particle B has a charge of -2e and a mass of 1.6 x 10^-27 kg. Particle A is accelerated to a speed of 0.8c, where c is the speed of light in a vacuum. Particle B is accelerated to a speed of 0.6c.

a) Calculate the magnitude of the linear momentum for each particle. b) Determine the kinetic energy of each particle in electron volts (eV). c) Find the total energy of the system.

Given: Charge of an electron = e = 1.6 x 10^-19 C Speed of light in a vacuum = c = 3.0 x 10^8 m/s

Answer:

a) To calculate the linear momentum of each particle, we can use the formula:

linear momentum (p) = mass (m) x velocity (v)

For particle A: mass (m_A) = 9.1 x 10^-31 kg velocity (v_A) = 0.8c = 0.8 x 3.0 x 10^8 m/s

linear momentum (p_A) = m_A * v_A

linear momentum (p_A) = (9.1 x 10^-31 kg) * (0.8 x 3.0 x 10^8 m/s)

linear momentum (p_A) = 21.84 kg m/s

For particle B: mass (m_B) = 1.6 x 10^-27 kg velocity (v_B) = 0.6c = 0.6 x 3.0 x 10^8 m/s

linear momentum (p_B) = m_B * v_B

linear momentum (p_B) = (1.6 x 10^-27 kg) * (0.6 x 3.0 x 10^8 m/s)

linear momentum (p_B) = 28.8 kg m/s

Therefore, the magnitude of the linear momentum for particle A is 21.84 kg m/s, and the magnitude of the linear momentum for particle B is 28.8 kg m/s.

b) To determine the kinetic energy of each particle in electron volts (eV), we can use the formula:

kinetic energy (KE) = (1/2) * mass (m) * velocity^2 (v^2)

For particle A: mass (m_A) = 9.1 x 10^-31 kg velocity (v_A) = 0.8c = 0.8 x 3.0 x 10^8 m/s

kinetic energy (KE_A) = (1/2) * m_A * v_A^2

kinetic energy (KE_A) = (1/2) * (9.1 x 10^-31 kg) * (0.8 x 3.0 x 10^8 m/s)^2

kinetic energy (KE_A) = 5.832 x 10^-14 J

To convert to electron volts (eV), divide by the charge of an electron (e):

kinetic energy (KE_A) = (5.832 x 10^-14 J) / (1.6 x 10^-19 C)

kinetic energy (KE_A) = 3.645 x 10^5 eV

For particle B: mass (m_B) = 1.6 x 10^-27 kg velocity (v_B) = 0.6c = 0.6 x 3.0 x 10^8 m/s

kinetic energy (KE_B) = (1/2) * m_B * v_B^2

kinetic energy (KE_B) = (1/2) * (1.6 x 10^-27 kg) * (0.6 x 3.0 x 10^8 m/s)^2

kinetic energy (KE_B) = 2.592 x 10^-14 J

To convert to electron volts (eV), divide by the charge of an electron (e):

kinetic energy (KE_B) = (2.592 x 10^-14 J) / (1.6 x 10^-19 C)

kinetic energy (KE_B) = 1.62 x 10^5 eV

Therefore, the kinetic energy of particle A is 3.645 x 10^5 eV, and the kinetic energy of particle B is 1.62 x 10^5 eV.

c) The total energy of the system is the sum of the kinetic energies of the two particles:

total energy of the system = kinetic energy of particle A + kinetic energy of particle B

total energy of the system = (3.645 x 10^5 eV) + (1.62 x 10^5 eV)

total energy of the system = 5.265 x 10^5 eV

Therefore, the total energy of the system is 5.265 x 10^5 eV.