Elastic Collision and conservation of momentum

[futb0l]

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

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.

 

[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

 

[Andrew Mason]

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

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.

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

 

[futb0l]

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