Elastic Collision of Two Boxes: Kinetic Energy Transfer

[teckid1991]

Homework Statement

A 5 g box moving to the right at 20 cm/s makes an elastic head on collision with a 10 g box initially at rest.

Mass 1 = .005 kg
initial Velocity 1 = .2 m/s
Mass 2 = .01 kg
initial Velocity 2 = 0 m/s

a.) what velocity does each box have after the collision?
b.) what fraction of the initial kinetic engergy is transferred to the 10 g box?

Homework Equations

- (initial velocity 1 - initial velocity 2) = velocity final 2 - velocity final 1
Mass 1 (initial velocity 1) + Mass 2 (initial velocity 2) = Mass 1 (velocity final 1) + Mass 2 (velocity final 2)
Kinetic energy = .5(mass)(velocity)²

The Attempt at a Solution

a.) for the final velocities...
- (V_o1 - V₀₂) = V_f2 - V_f1
- (0 - .2) = V_f2 - V_f1
.2 + V_f1 = V_f2

for V_f1...
M₁V₀₁ + M₂V₀₂ = M₁V_f1 + M₂V_f2
(.005)(.2) + (.01)(0) = (.005)V_f1 + (.01)(.2 + V_f1)
.001 = .015(V_f1) + .002
-.001 = .015(V_f1)
-.067 [rounded number] = V_f1

plug into first problem
.2 + V_f1 = V_f2
.133 = V_f2

b.) [tex]\sum[/tex] K₀ = [tex]\sum[/tex] K_f [not sure if this is the correct method for solving this equation...]
K₀₁ + K₀₂ = K_f1 + K_f2
.5(.005)(.2)² + .5(.01)(0)² = .5(.005)(-.067)² + .5(.01)(.133)²

Im not sure if its K₀₁/K_f2 or K₀₁/ (K_f1 + K_f2)

 

[rl.bhat]

a.) for the final velocities...
- (Vo1 - V02) = Vf2 - Vf1
- (0 - .2) = Vf2 - Vf1
.2 + Vf1 = Vf2
The above is true when the masses are equal.

 

[teckid1991]

Oh I see. I'm unsure now how this would work since there are 2 variables that I need to find.

 

[rl.bhat]

Use (.005)(.2) + (.01)(0) = (.005)Vf1 + (.01)(Vf2) ... (1)
and .5(.005)(.2)2 + .5(.01)(0)2 = .5(.005)(Vf1)2 + .5(.01)(Vf2)2 ...(2)
Solve the above equations to get the final velocities.