Conservation of Linear Momentum?

[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.

 

[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.

 

[Dantes]

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.

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

 

[suspenc3]

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

There are no weights given for the two particles

 

[suspenc3]

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.

Im still not too sure what to do.

How do I get the values of Ke

 

[suspenc3]

Would I use:

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

Also I don't have a $$\vec{v}$$

 

[suspenc3]

?can Anyone Help?

 

[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 let's simplify that a bit.
$$\frac{3}{2}(\frac{1}{2}m_b*(v_b)^2)=60$$

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