# Homework Help: Conservation of Linear Momentum?

1. Nov 16, 2005

### suspenc3

Particle A & B are held together with a compressed spring between them. When they are released the spring pushes them apart and they fly off in opposite directions free of the spring. The mass of A is twice that of B, and the energy stored in the spring was 60J. Once the transfer is complete, what are the kinetic energies of particle A & B

I have no clue where to start on this one.

Can Anyone Point Me In The Right Direction?

Thanks.

2. Nov 16, 2005

### Kazza_765

The title of your post gives it away.
There are two conditions that must be met:
That the momentum of particle A is equal and opposite to the momentum of particle B. (conservation of momentum)
That the combined kinetic energy of particle A & B is equal to 60J (conservation of energy)

Write out this two conditions in terms of mass and velocity and you should be able to solve.

3. Nov 16, 2005

### Dantes

Just wondering you know if the answer is in symbolic form or do you have numbers such as the weight of b?

4. Nov 16, 2005

### suspenc3

There are no weights given for the two particles

5. Nov 16, 2005

### suspenc3

Im still not too sure what to do.

How do I get the values of Ke

6. Nov 16, 2005

### suspenc3

Would I use:

$$W=\Delta {E}_m_e_c+\Delta {E}_t_h_e$$

Also I dont have a $$\vec{v}$$

Last edited: Nov 16, 2005
7. Nov 16, 2005

### suspenc3

?can Anyone Help?

8. Nov 17, 2005

### Kazza_765

Sorry, been away. Ok, advanced reply doesn't seem to be working so hopefully my latex comes out ok.
We'll go through the two conditions (conservation of momentum/energy) seperately then combine them.

Conservation of momentum:
The momentum of the system doesn't change. Assuming the system is stationary to begin with.

$$P_(total)=0$$

$$m_a*v_a + -m_b*v_b=0$$ <- the negative is because they are in opposite directions

$$m_a*v_a = m_b*v_b$$

$$(2m_b)*v_a = m_b*v_b$$

$$v_a=\frac{1}{2}v_b$$

Now for conservation of energy. The systems total inital energy is 60J. After the exchange is complete all of that energy is kinetic.
$$E_(total)=60$$

$$\frac{1}{2}m_a*(v_a)^2 + \frac{1}{2}m_b*(v_b)^2 = 60$$

$$\frac{1}{2}(2m_b)*(\frac{1}{2}(v_b))^2 + \frac{1}{2}m_b*(v_b)^2 = 60$$

Now lets simplify that a bit.
$$\frac{3}{2}(\frac{1}{2}m_b*(v_b)^2)=60$$

$$\frac{3}{2}KE_b=60$$ where $$KE_b=\frac{1}{2}m_b*(v_b)^2$$

Last edited: Nov 17, 2005