How Does Bead Momentum Affect Post-Collision Movement on a Frictionless Surface?

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Homework Help Overview

The discussion revolves around a problem involving two beads colliding elastically on a frictionless surface. The original poster describes the initial conditions, including the masses and velocities of the beads, and seeks assistance in determining the distance the larger bead moves after the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of momentum and its conservation during the collision. There are questions about how to relate momentum changes to the distance traveled by the larger bead post-collision. Some participants express uncertainty about the relevance of certain equations and whether acceleration is present after the collision.

Discussion Status

The conversation is ongoing, with participants exploring the principles of conservation of momentum and the implications of a frictionless environment. Guidance has been offered regarding the application of momentum conservation equations, but no consensus has been reached on the specific approach to solve the problem.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the differences between elastic and inelastic collisions, as well as the implications of a frictionless surface on the motion of the beads.

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A 0.400 kg bead slides on a straight frictionless wire with a velocity of 3.50 cm/s to the right. The bead collides elastically with a larger 0.600 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. Find the distance the larger bead moves along the wire in the first 5.0 seconds following the collision.

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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You seem to know that it's about momentum. What happens to momentum during the collision?
 
Well since I know that Momentum=mass x velocity, the mass stays the same but velocity goes down. Therefore, the momentum goes down. But I am not sure on how that all fits into the distance.
 
Have you learned about conservation of linear momentum?
 
Yes I have learned about them. I have learned about the elastic collision (m1v1+m2v2=m1v1'+m2v2') and also Interlastic (m1v1+m2v2=(m1+m2)v')
I have an equation x=1/2(vi+vf)t. But that seems too easy.
 
It is true that x = 1/2(vi + vf)t for constant acceleration. In this case, however, is there any acceleration or force on either bead after the collision? Would anything happen to the velocity of either bead after the collision (remember the wire is frictionless)?

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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