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

AI Thread Summary
The discussion centers on a collision between a 0.400 kg bead and a 0.600 kg bead on a frictionless surface. The smaller bead, initially moving at 3.50 cm/s, collides elastically with the larger bead, which is at rest, resulting in the smaller bead reversing direction with a velocity of 0.70 cm/s. Participants discuss the conservation of linear momentum, emphasizing that momentum is conserved during the collision despite changes in velocity. The equation for distance traveled, x = 1/2(vi + vf)t, is mentioned, but participants note that after the collision, there is no acceleration due to the absence of external forces. Understanding the difference between elastic and inelastic collisions is also highlighted, with a focus on how kinetic energy is conserved in elastic collisions.
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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,

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

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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