Cons. of momentum and cons. of energy

[musicfairy]

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of A is 5.00 times the mass of B, and the energy stored in the spring was 63 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of each particle?

What I did:

0 = m₁v₁ + m₂v₂

m₁ = 5m₂

v₁ = .2v₂

U = K₁ + K₂

I got stuck here. I'm not given v or m. What am I supposed to do?

 

[LowlyPion]

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of A is 5.00 times the mass of B, and the energy stored in the spring was 63 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of each particle?

What I did:

0 = m₁v₁ + m₂v₂

m₁ = 5m₂

v₁ = .2v₂

U = K₁ + K₂

I got stuck here. I'm not given v or m. What am I supposed to do?

You are given PE and you know the relationship between the velocities basically from Newton's Third Law - action = reaction and from that you figured the Velocity of the heavier is 1/5 that of the the lighter one. (Though I note that it's really v1 = -1/5*v2.)

So you know that the KE of the system is 63 J after the spring unloads.

You also know that

0 = m₁v₁ + m₂v₂

m₁v₁ = - m₂v₂

Multiply both sides by the identity that v1 = - 1/5 V2

m₁v₁² = 1/5 * m₂v₂²

Now look at your PE = KE1 + KE2 = 1/2 * m₁v₁² + 1/2 * m₂v₂²

But we can substitute using the previous relationship and get:

PE = 1/2 * 1/5 * m₂v₂² + 1/2 * m₂v₂²

63 J = 6/5 * 1/2 * m₂v₂²

 

[musicfairy]

Thanks for the explanation.

Using the final equation you gave me, I was able to get the correct answer.

Earlier I substituted into U = K1 + K2 and got the right answer, except I switched the kinetic energies of A and B. Problem solved.

Once again, thanks a lot for the explanation. I understand the concepts better now.

 

[p21bass]

Just wanted to thank you both. I was having a heck of a time with a problem that was very similar to this. I had tried to use the principles of energy, but was getting stuck. Pion's explanation was very helpful.

 

[LowlyPion]

Just wanted to thank you both. I was having a heck of a time with a problem that was very similar to this. I had tried to use the principles of energy, but was getting stuck. Pion's explanation was very helpful.

Welcome to PF.

Glad you found it of value.

Cheers.