Question

A proton moving at v1​=5.5Mm/s collides elastically and head-on with a second proton moving in the opposite direction at v2​=9.8 Mm/s. Choose positive velocities in the direction of v1​.

Answer 1

The final velocities of both protons after the head-on elastic collision will be 7.65 Mm/s in opposite directions.

Step-by-step explanation:

In an elastic collision, the total momentum and kinetic energy of the particles remain conserved. By applying the conservation of momentum and kinetic energy, we can determine the velocities of the protons after the collision.

Let's assume the mass of the protons to be m.

Given that the initial velocity of the first proton (v1) is 5.5 Mm/s and the initial velocity of the second proton (v2) is 9.8 Mm/s in the opposite direction, the final velocities (v1' and v2') can be calculated using the equations:

v1' = (m*v1 - m*v2) / (m + m)

v2' = (2 * m * v1 + m * v2) / (m + m)

Plugging in the values, we get:

v1' = (m*5.5 Mm/s + m*9.8 Mm/s) / (2m)

v2' = (2 * m * 5.5 Mm/s - m * 9.8 Mm/s) / (2m)

which simplifies to:

v1' = 7.65 Mm/s

v2' = 7.65 Mm/s

Therefore, the final velocities of both protons after the collision will be 7.65 Mm/s in opposite directions.