Question

A proton is traveling to the right at 2.0 * 107 m/s. It has a headon perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?

Answer 1

Speed of proton = 1.69 x 10⁷ m/s to the left

Speed of carbon = 3.08 x 10⁶ m/s to the right

Step-by-step explanation:

Elastic collision means momentum and kinetic energy are conserved.

Mass of carbon = 12 x Mass of proton.

Initial velocity of proton = 2 x 10⁷ m/s

Initial velocity of Carbon = 0 m/s

Final velocity of proton = u

Final velocity of carbon = v

Initial momentum = Final momentum.

m x 2 x 10⁷ + 12m x 0 = m x u + 12m x v

u + 12v = 2 x 10⁷

u = 2 x 10⁷ - 12v

Initial K.E = Final K.E

0.5 x m x (2 x 10⁷)² + 0.5 x 12m x 0² = 0.5 x m x u² + 0.5 x 12m x (v)²

u² + 12v² = 4 x 10¹⁴

(2 x 10⁷ - 12v)² + 12v² = 4 x 10¹⁴

4 x 10¹⁴ - 48 x 10⁷v + 144v² + 12v² = 4 x 10¹⁴

156v² = 48 x 10⁷v

v = 3.08 x 10⁶ m/s

u + 12 x 3.08 x 10⁶ = 2 x 10⁷

u = -1.69 x 10⁷ m/s

Speed of proton = 1.69 x 10⁷ m/s to the left

Speed of carbon = 3.08 x 10⁶ m/s to the right