Recent content by potatogirl
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Calculate electric force using Coulomb's law (vector components are the struggle)
Something like this?- potatogirl
- Post #3
- Forum: Introductory Physics Homework Help
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Calculate electric force using Coulomb's law (vector components are the struggle)
So for this problem I think I am doing something weird with the trig and/or vector components. I calculated the problem like this: First drew a picture, q1 and q2 on the x axis. q3 located equidistant between them but negative .300m in the y direction. First finding magnitude of Electric...- potatogirl
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- Coulomb Electric Force
- Replies: 3
- Forum: Introductory Physics Homework Help
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Potential Energy - 2D inelastic collision - ballistic pendulum
I assume I use trig for the y-comp to get the height. Is that right? Like 0.3m * cos25 . But I am running into a fundamental confusion about what rcm represents I think. Does it represent the rest height? AKA the lowest point?- potatogirl
- Post #8
- Forum: Introductory Physics Homework Help
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Potential Energy - 2D inelastic collision - ballistic pendulum
Momentum = mass * velocity. So without velocity how do I know momentum? Similarly, conservation of momentum says the final momentum (mf1vf1 + mf2vf2) is equal to initial momentum (mi1vi1 + mi2vi2) but again without velocity, how would I go about that? The mass of the bullet-pendulum lump should...- potatogirl
- Post #7
- Forum: Introductory Physics Homework Help
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Potential Energy - 2D inelastic collision - ballistic pendulum
Ah, okay good to know. Thanks. So from here, I am having a hard time determining how to calculate anything without velocity. Also, I am not sure I understand the given variable rcm I am used to seeing xcm for the location of center of mass, or vcm for velocity of center of mass. Is it just...- potatogirl
- Post #4
- Forum: Introductory Physics Homework Help
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Potential Energy - 2D inelastic collision - ballistic pendulum
I really can't find anything in my textbook about how PE relates to 2D collisions, but here is what I do know... ΔPE = mgΔh And I know that since it is a completely inelastic equation kinetic energy is not conserved but momentum is conserved. The change in kinetic energy should be equal to the...- potatogirl
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- 2d collision Impulse momentum Inelastic collision
- Replies: 10
- Forum: Introductory Physics Homework Help
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When is the acceleration due to gravity negative and when is it positive?
I know to break it down into its x and y components and then use Pythagorean: Acceleration in the x direction is Fx/m ---> (7.50 x 10^6*cos55) / (4.50 x 10^5 kg) = 9.56 m/s^2 Acceleration in the y direction is: (Fy - mg)/m ---> ((7.50 x 10^6*sin55) - (4.5 x 10^5* 9.8 m/s^2)) / (4.5 x 10^5 kg)...- potatogirl
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- Acceleration Gravity Positive
- Replies: 13
- Forum: Introductory Physics Homework Help