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Linear momentum

  • Thread starter missrikku
  • Start date

missrikku

Hello again!

I am having trouble with the following linear momentum problem:

A has a mass mA = 9.11x10^31 kg
B has a mass mB = 1.67x10^-27 kg

They are attracted to each other by some electrical force.

Say they are released from rest with a distance between them of Di = 3.0x10^-6 m

When their distance is decreased to D = 1.0x10^-6 m, what is the ratio of

a) their linear momentums

b) speeds

c) kinetic energies

-----

i know:

p = mv

so for a)

I am looking for: Pe/Pp
mAvA/mBvB

I have mA and mB, but not vA or vB

I know that they are released from rest, but how do you find vA and vB?

Can I use V^2 = Vi^2 + 2a(D-Di) ? If so, how do I find the a?

Thanks. I just need to be pointed in the right direction.
 

HallsofIvy

Science Advisor
Homework Helper
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Actually, the phrasing of the questions is a major hint. Notice that the question does not ask for the momenta or velocities of the separate particles but only for the ratios. Also the problem asks first for the ratio of the momenta- if you find that first, the others are easy.
Since the force attracting the two particles to one another is "internal", momentum is conserved- that is, MAvA+ MBvB is a constant.. Since the two particles were initially motionless, what was the initial momentum of each and what was the total momentum (the "constant")?
 

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