# Linear Momentum

1. Dec 7, 2008

### teckid1991

1. The problem statement, all variables and given/known data
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?

2. Relevant 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

3. The attempt at a solution

a.) for the final velocities...
- (Vo1 - V02) = Vf2 - Vf1
- (0 - .2) = Vf2 - Vf1
.2 + Vf1 = Vf2

for Vf1...
M1V01 + M2V02 = M1Vf1 + M2Vf2
(.005)(.2) + (.01)(0) = (.005)Vf1 + (.01)(.2 + Vf1)
.001 = .015(Vf1) + .002
-.001 = .015(Vf1)
-.067 [rounded number] = Vf1

plug into first problem
.2 + Vf1 = Vf2
.133 = Vf2

b.) $$\sum$$ K0 = $$\sum$$ Kf [not sure if this is the correct method for solving this equation...]
K01 + K02 = Kf1 + Kf2
.5(.005)(.2)2 + .5(.01)(0)2 = .5(.005)(-.067)2 + .5(.01)(.133)2

Im not sure if its K01/Kf2 or K01/ (Kf1 + Kf2)

2. Dec 7, 2008

### 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.

3. Dec 7, 2008

### teckid1991

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

4. Dec 8, 2008

### 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.

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