# Elastic Collision and conservation of momentum

1. Feb 26, 2005

### futb0l

In an elastic collision between two objects. Show that the speed of approach is equal to the speed of seperation.

I know you have two equations, the conservation of momentum and the conservation of kinetic energy, but I wouldn't know how to solve the equations.

2. Feb 26, 2005

### misskitty

You have the two equations right?

M1Vi1+M2Vi2=M1fV1f+M2fV2f.......sorry no subscripts
and
(1/2M1Vi1^2)+(1/2M2Vi2^2)=(1/2M1Vf1^2)=(1/2M2Vf2^2)

All you have to do is take the mass of your objects, substitute them in the correct places and then do the same thing for velocities. Just replace the variables with the numbers. Its basically plug and chug the whole thing out with the info you have.

You'll know if you've done the work properly if both sides of the equation equal on another. The only thing you need to ask yourself is: Is this a perfectly elastic collison?
Because then the equation for the conservation of Kinetic Energy comes in. But if its not then you don't have to worry about the extra equation. Kinetic energy isn't conserved in an elastic collision. Only in Perfectly elastic collisions.

MissKitty

3. Feb 26, 2005

### Andrew Mason

The total momentum is conserved:
$$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$$

which can be rewritten:
(1)$$m_1(v_1 - v'_1) = m_2(v'_2 - v_2)$$

Since energy is conserved:
$$m_1v_1^2 + m_2v_2^2 = m_1v'_1^2 + m_2v'_2^2$$

which can be rewritten:
$$m_1v_1^2 - m_1v'_1^2 = m_2v'_2^2 + m_2v_2^2$$ or

(2)$$m_1(v_1 - v'_1)(v_1 + v'_1) = m_2(v'_2 - v_2)(v'_2 + v_2)$$

so divide (2) by (1)

$$v_1 + v'_1 = v'_2 + v_2$$ or

$$v_1 - v_2 = -(v'_1 - v'_2)$$

This is all one dimensional, of course. The math gets a little cumbersome for 3 dimensions, but you get the idea.

AM

Last edited: Feb 26, 2005
4. Feb 27, 2005

### futb0l

Ooo.. Thanks.
I remembered doing this before a while ago.
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