Physics question?

Question

A proton is traveling to the right at 2.0×10^7m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. A.) What are the speeds of each after the collision? B.) What are the directions of each?

Answer 1

(A)

Velocity of the proton is V₁ 2 * 10^7 m/s .

Let the mass of proton is M₁ = x .

The velocity of carbon atom is 0 .                

[since the second mass is at rest in Head on collision so velocity is zero] 

The mass of the carbon atom is M₂ = 12x . 

So the velocity of the proton after collision is 

V₁^’ = [( M₁ - M₂ ) / ( M₁ + M₂ )] V₁

V₁^’ = [( x - 12x ) / ( x + 12x )] 2 * 10^7 m/s

V₁^’ = [( -11x ) / ( 13x )] 2 * 10^7 m/s

V₁^’ = [-11/13] 2 * 10^7 m/s

V₁^’ = [-0.846] 2 * 10^7 m/s

V₁^’ = -1.692 * 10^7 m/s

Here negative sign indicates that proton repeals back .

So the velocity of proton after collision is  1.692 * 10^7 m/s .

The velocity of the carbon atom after collision is 

V₂^’ = [( 2M₁ ) / ( M₁ + M₂ )] V₁

V₂^’  = [( 2x ) / ( x + 12x )] 2 * 10^7 m/s

V₂^’  = [( 2x ) / ( 13x )] 2 * 10^7 m/s

V₂^’  = [2/13] 2 * 10^7 m/s

 

V₂^’  = [0.153] 2 * 10^7 m/s

V₂^’ = 0.3076 * 10^7 m/s

So the velocity of the carbon atom after collision is  0.3076 * 10^7 m/s .

Answer 2

(B)

To estimate the directions , first we have to find their velocities after collision .

Velocity of the proton is V₁ 2 * 10^7 m/s .

Let the mass of proton is M₁ = x .

The velocity of carbon atom is 0 .                

[since the second mass is at rest in Head on collision so velocity is zero] 

The mass of the carbon atom is M₂ = 12x . 

So the velocity of the proton after collision is 

V₁^’ = [( M₁ - M₂ ) / ( M₁ + M₂ )] V₁

V₁^’ = [( x - 12x ) / ( x + 12x )] 2 * 10^7 m/s

V₁^’ = [( -11x ) / ( 13x )] 2 * 10^7 m/s

V₁^’ = [-11/13] 2 * 10^7 m/s

V₁^’ = [-0.846] 2 * 10^7 m/s

V₁^’ = -1.692 * 10^7 m/s

Here the negative sign indicates that the proton repels back after collision .

So the proton atom moving in backward direction with reference to its previous state .

The velocity of the carbon atom after collision is 

V₂^’ = [( 2M₁ ) / ( M₁ + M₂ )] V₁

V₂^’  = [( 2x ) / ( x + 12x )] 2 * 10^7 m/s

V₂^’  = [( 2x ) / ( 13x )] 2 * 10^7 m/s

V₂^’  = [2/13] 2 * 10^7 m/s

V₂^’  = [0.153] 2 * 10^7 m/s

V₂^’ = 0.3076 * 10^7 m/s

So the carbon atom moving in forward direction with reference to its previous state .