Answer :
The speeds and the direction of each after the collision:
The proton = -1.692 x 10⁷m/s to the left.
The carbon = 0.3076 x 10⁷ m/s to the right.
The law of conservation of momentum
If there are two object each with masses m₁ and m₂ move with speed v₁ and v₂, then the two objects collide, so after colliding the speed of each object becomes v₁’ and v₂'.
Since there are no external forces acting on the system, then the momentum of the system is conserved, meaning the momentum before and after the same collision.
The equation is:
(m₁)(v₁) + (m₂)(v₂) = (m₁v₁)' + (m₂v₂)'
We have,
velocity of the proton = 2.0 x 10⁷ m/s ⇒ v₁
Mass of the proton = m₁
The velocity of a carbon atom = 0 ⇒ v₂
The mass of the carbon atom = 12 m₁
So, the speeds of proton after the collision:
(m₁)(v₁) + (m₂)(v₂) = (m₁v₁)' + (m₂v₂)'
v₁’ = (m₁ - m₂) / (m₁ + m₂) (v₁)
= (m₁ - 12m₁) / (m₁ + 12m₁) (2.0 x 10⁷)
= (-11 m₁) / (13m₁) (2.0 x 10⁷)
= -1.692 x 10⁷m/s ⇒ it's negative, so to the left.
Now, the speed of the carbon atom after collision:
V₂’ = (2m₁) / (m₁ + m₂)v₁
= 2/13 (2 x 10⁷)
= 0.3076 x 10⁷ m/s ⇒ it's positive, so to the right.
Learn more about law of conservation of momentum here: https://brainly.com/question/29765459
#SPJ4
