- #1

sfzpilot

- 3

- 0

I am not sure how to do this. I know to change the cm/s to m/s but not sure what equation to use. Any help is appreciated. Thanks!

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- Thread starter sfzpilot
- Start date

- #1

sfzpilot

- 3

- 0

I am not sure how to do this. I know to change the cm/s to m/s but not sure what equation to use. Any help is appreciated. Thanks!

- #2

Tedjn

- 737

- 0

You seem to know that it's about momentum. What happens to momentum during the collision?

- #3

sfzpilot

- 3

- 0

- #4

Tedjn

- 737

- 0

Have you learned about conservation of linear momentum?

- #5

sfzpilot

- 3

- 0

I have an equation x=1/2(vi+vf)t. But that seems too easy.

- #6

Tedjn

- 737

- 0

The equation that you wrote down,

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

is conservation of linear momentum when there is no external force on the two masses (i.e. the system). This holds true, when there is no external force, for all collisions whether they are elastic or not, so try to use this to solve your problem. Remember that velocity is sign sensitive.

For future reference, the difference between elastic and inelastic collisions is that elastic collisions preserve not only linear momentum but total kinetic energy in the colliding objects. Inelastic collisions lose some of that kinetic energy during the collision, usually as heat or sound. All real collisions are at least a bit inelastic since some energy is always lost. The type of inelastic collision you may be thinking of is when two objects stick together after they collide, like a car crash, in which both cars were moving before the accident (had kinetic energy), but are clearly not moving after the crash (lost kinetic energy).

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