Question about relative speeds.

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SUMMARY

The discussion centers on the analysis of a collision between two objects with different masses, specifically using the conservation of momentum. When a moving object with speed v_1 and mass m_1 collides with a stationary object of mass m_2, they stick together, leading to a final speed v_2 determined by the equation m_1 v_1 = (m_1 + m_2)v_2. Analyzing the collision from the rest frame of m_1 introduces a different perspective, where m_2 appears to move towards m_1 at speed -v_1. The key conclusion is that while the final speeds appear different in each frame, transforming back to the original frame reconciles the results, confirming the consistency of momentum conservation across reference frames.

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port31
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Lets say I have a moving object that has speed [itex]v_1[/itex]
and mass [itex]m_1[/itex] and it collides with a more massive object of
mass [itex]m_2[/itex] And this mass is at rest and when they collide they stick together.
If I use momentum conservation I would get
[itex]m_1 v_1=(m_1+m_2)(v_2)[/itex] and [itex]v_2[/itex] is the speed after the collision
but what if I wanted to analyze this from the rest frame of [itex]m_1[/itex]
It would look as if the more massive object was moving at me at a speed [itex]v_1[/itex]
So now I would have [itex]m_2(-v_1)=(m_1+m_2)(v_2)[/itex] the final speeds would be different in those 2 cases so what's wrong with my reasoning.
 
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The final speeds would indeed be different because you are using a different reference frame. To check to make sure there is no conflict, compare the before and after speeds of each object. The difference should be the same regardless of which frame you choose.
 
port31 said:
So now I would have [itex]m_2(-v_1)=(m_1+m_2)(v_2)[/itex] the final speeds would be different in those 2 cases so what's wrong with my reasoning.
If you now transform your answer to the original frame (by adding [itex]v_1[/itex]) you'll find that the speeds match.
 

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