A proton colliding with a stationary carbon nucleus

[HandcuffedKitty]

A proton of mass m undergoes a head-on elastic collision with a stationary carbon nucleus of mass 12m. The velocity of the proton is 395 m/s in the positive x direction. The velocity of the center of mass of the system is 30.38 m/s.

What is the velocity of the proton after the collision?

 

[turin]

"Head-on" tells you the angle. "Elastic" tells you that KE is conserved.

 

[M98Ranger]

To determine the velocity of the proton after the collision, we can use the conservation of momentum and kinetic energy laws. Since the collision is elastic, both laws will apply.

First, let's calculate the initial momentum of the system. The momentum of the proton before the collision can be calculated as:

p1 = m*v1 = m*395 m/s = 395m kg/s

The momentum of the carbon nucleus is zero, since it is stationary.

The initial momentum of the system is therefore:

p1 = 395m kg/s

Next, let's calculate the final momentum of the system. The final momentum of the proton can be calculated as:

p2 = m*v2

Since the collision is elastic, the proton will bounce back in the opposite direction with the same speed. Therefore, v2 = -395 m/s.

The final momentum of the carbon nucleus can be calculated as:

p3 = 12m*v3

Since the carbon nucleus was initially stationary, its final momentum will be in the positive x direction, with the same magnitude as the initial momentum of the proton. Therefore, v3 = 395/12 = 32.92 m/s.

The final momentum of the system is therefore:

p2 + p3 = m*(-395 m/s) + 12m*(32.92 m/s) = -395m kg/s + 395m kg/s = 0

Since the initial and final momentum of the system are equal, the conservation of momentum law is satisfied.

Next, let's calculate the initial and final kinetic energies of the system. The initial kinetic energy of the system can be calculated as:

KE1 = (1/2)*m*v1^2 = (1/2)*m*(395 m/s)^2 = 77,787.5 m^2/s^2

The final kinetic energy of the system can be calculated as:

KE2 = (1/2)*m*v2^2 + (1/2)*12m*v3^2 = (1/2)*m*(-395 m/s)^2 + (1/2)*12m*(32.92 m/s)^2 = 77,787.5 m^2/s^2

Since the initial and final kinetic energies of the system are equal, the conservation of kinetic energy law is also satisfied.

Therefore, the velocity of the proton after the collision can be calculated as:

v2 = -395 m/s