Conservation of Energy Problem

[Mushrooomboy]

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I understand that momentum is conserved in this problem, but can anyone explain to me how kinetic energy is conserved as well? The answer says that it's an elastic collision

 

[Cawolf]

Use the conservation of momentum formula to solve for the momentum of the second ball - which turns out to be 4/3(mv) in the same direction.

Then you can solve for the initial and final kinetic energies of the balls to compare if they are the same.

The initial K = mv^2

The final K = (1/9)mv^2 + (8/9)mv^2

These are equal so kinetic energy is conserved and we can state that the collision is fully elastic.

 

[justadream]

@Cawolf

I also got the 4/3 mv part.

But how did you get the (1/9) and the (8/9) in your final K calculation?

I presume the 1/9 comes from the velocity of the bigger ball being 1/3 of the original (1/3)^2 = 1/9. But how did you get 8/9?

 

[Cawolf]

You know the initial/final masses and the inital/final velocities.

Just plug those into K = 1/2 mv^2.

K1 + K2 = K1' + K2'

(0.5)(2m)(v)^2 + o = (0.5)(2m)(1/3v)^2 + (0.5)(m)(4/3v)^2

mv^2 = 1/9 mv^2 + 8/9 mv^2

mv^2 = mv^2

So the kinetic energy is conserved. If these values were different we would know how much energy was lost to some other process.

 

[The Brown Knight]

Apply momentum and energy conservation equations.
If you get an identity equation (i.e. v = v, 4/3 = 4/3), it's elastic, which it is in this case.
If you get an inequality, it will be inelastic.

 

[Czarcasm]

Curious, anyone mind explaining how you guys arrived at 4/3v? I feel like I'm missing something fairly obvious here.

 

[Lighthill-FFT]

Calculate total momenta of the system. Define the mass of Ball 1 to be 2m, mass of Ball 2 to be m. Total momenta is 2m*v=2mv. Ball 1 is now moving at 1/3 *v. Its momenta is 2m*v/3=2mv/3. The momenta of Ball 2 is therefore mv (2-2/3)=mv(4/3).