Elastic Collision and conservation of momentum

In summary, in an elastic collision between two objects, the conservation of momentum and kinetic energy are used to show that the speed of approach is equal to the speed of separation. This can be done by substituting the mass and velocity values into the equations and checking if the two sides equal each other. If the collision is perfectly elastic, the conservation of kinetic energy equation applies, but if not, it is not necessary.
  • #1
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.
 
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  • #2
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
futb0l said:
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.

The total momentum is conserved:
[tex]m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2[/tex]

which can be rewritten:
(1)[tex]m_1(v_1 - v'_1) = m_2(v'_2 - v_2)[/tex]

Since energy is conserved:
[tex]m_1v_1^2 + m_2v_2^2 = m_1v'_1^2 + m_2v'_2^2[/tex]

which can be rewritten:
[tex]m_1v_1^2 - m_1v'_1^2 = m_2v'_2^2 + m_2v_2^2[/tex] or

(2)[tex]m_1(v_1 - v'_1)(v_1 + v'_1) = m_2(v'_2 - v_2)(v'_2 + v_2)[/tex]

so divide (2) by (1)

[tex]v_1 + v'_1 = v'_2 + v_2[/tex] or

[tex]v_1 - v_2 = -(v'_1 - v'_2)[/tex]

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

AM
 
Last edited:
  • #4
Ooo.. Thanks.
I remembered doing this before a while ago.
 

1. What is an elastic collision?

An elastic collision is a type of collision where the total kinetic energy of the system is conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

2. What is conservation of momentum?

Conservation of momentum is a fundamental law in physics that states that the total momentum of a closed system remains constant. In other words, the total momentum before a collision is equal to the total momentum after the collision.

3. How is momentum conserved in an elastic collision?

In an elastic collision, both the total kinetic energy and the total momentum of the system are conserved. This means that the sum of the momenta of the objects before the collision is equal to the sum of their momenta after the collision.

4. What are the equations used to calculate momentum and kinetic energy in an elastic collision?

The equations used to calculate momentum and kinetic energy in an elastic collision are:

  • Momentum: p = mv, where p is momentum, m is mass, and v is velocity
  • Kinetic energy: KE = 1/2 * m * v^2, where KE is kinetic energy, m is mass, and v is velocity

5. Can momentum and kinetic energy be conserved in an inelastic collision?

No, in an inelastic collision, the total kinetic energy is not conserved because some of the energy is lost in the form of heat, sound, or deformation. However, momentum is still conserved in an inelastic collision.

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