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PhoenixFire
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How do you calculate the momentum of two objects immediately after a perfectly elastic collision?
Do you need to find the velocities of the two objects after collision and then use the formula of p=mv. where p is momentum vector, m is mass of the object, and v is the velocity vector?
Or is there another way to find the linear momentum?
 one year ago
 one year ago
PhoenixFire Group Title
How do you calculate the momentum of two objects immediately after a perfectly elastic collision? Do you need to find the velocities of the two objects after collision and then use the formula of p=mv. where p is momentum vector, m is mass of the object, and v is the velocity vector? Or is there another way to find the linear momentum?
 one year ago
 one year ago

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Shane_B Group TitleBest ResponseYou've already chosen the best response.0
It would be easier to explain with an actual problem but for the most part, you just have to remember that momentum and KE are conserved in a perfectly elastic collision. This should help: http://hyperphysics.phyastr.gsu.edu/hbase/colsta.html and http://hyperphysics.phyastr.gsu.edu/hbase/elacol.html#c4
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
I have a problem. object 1: m=10kg v=(6, 1) p=mv=(60, 10) object 2: m=100kg v=(6, 2) p=mv=(600, 200) They collide, what's the momentum of each object immediately after collision.
 one year ago

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
So what I got from those two links was basically that I do need to calculate the final velocities. \[v_{1f}={(m_1  m_2)v_{1i} + 2m_2 v_{2i} \over m_1+m_2}\]\[v_{2f}={(m_2m_1)v_{2i}+2m_1v_{1i} \over m_1+m_2}\] Then apply the momentum formula\[\mathbf {\vec p}=m {\mathbf{\vec v}}\] Kinda strange since the next question asks me to find the velocities.. You'd think they would put it in the right order to help you learn the method.
 one year ago

Shane_B Group TitleBest ResponseYou've already chosen the best response.0
The problem you posted it's a 2dimensional collision and momentum is still conserved but you have to account for the momentum in 2 dimensions. That's a bit tougher and I admit that I haven't done one of those in a while :)
 one year ago
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