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    Spring mechanics with 2 different masses

    DISREGARD THE LAST POST ITS WRONG. this is a solved problem thanks for the help everyone.
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    Spring mechanics with 2 different masses

    Problem 8 is for the Mass 3m find the time that is required to reach the floor (from the top of the ramp to the floor). I use three formulas v0sin theta =V0y, y = v0yt =1/2 at^2, and the quadratic formula and you solve for t. Problem 9 is Calculate the velocity of the mass 3m at the point of...
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    Spring mechanics with 2 different masses

    thanks tiny tim problem 1 is the same as before for problem two and three we use the formula PE = (mV1^2)/2 + (3mV2^2)/2 to calculate conservation of energy and mv1 =3mv2 to calculate conservation of momentum. this will give you 8.39 m/s for problem 2 and 2.8 m/s for problem three. problem 4...
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    Spring mechanics with 2 different masses

    I'm running out of time can't you just tell me what I need to do? I can do the rest and I have shown an attempt at the problem. Also I know that 1,4,5,and 7 are correct. and their were actually 11 problems and I can solve those If I can find the answer to 2 and 3.
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    Spring mechanics with 2 different masses

    I'm sorry I can't figure this out.
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    Spring mechanics with 2 different masses

    Can you give me a formula or a procedure that I can use, cause I know I'm not doing this right, but I want to know the right way.
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    Spring mechanics with 2 different masses

    OK I'm not quite sure how to do that, But this is my attempt. Because of conservation of momentum I know PE = (.5V1^2)/2 + (1.5V2^2)/2 but that leaves me with two variables, so what do I do? M =.5 and 3M = 1.5, which means I am using the formula KE = (mv^2)/2
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    Spring mechanics with 2 different masses

    why doesn't my answer take into account the conservation of energy and momentum/how do It the right way?
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    Spring mechanics with 2 different masses

    final answers: for problem 1: PE =KE, where PE = potential energy and KE = kinetic energy formula two equals .5K delta X^2. this means PE = (750*.25^2)/2 = 23.44 J for problem 2 I believe you use V = square root (2KE)/m = square root 2(23.44)/.5 = 9.68m/s I believe problem 3 is solved the same...
  10. F

    Spring mechanics with 2 different masses

    so T should = .43 because y=.5at^2, so t = square root y/(.5a)
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    Spring mechanics with 2 different masses

    The way I see it I found the x direction velocity in problem 3 so I need to find V that accommodates two dimensions (the x as well as the y). so i am using the formula v0cos 20 = vx so vx/cos20 = v0 which gives me 5.55m/s. So I'm wondering if my method is correct?
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    Spring mechanics with 2 different masses

    I've already solved that, I'm on step 6 but the answer is 22.44, all the info you need is in the first, second, and fourth post.
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    Spring mechanics with 2 different masses

    I'm confused how does that help me find the speed of mass 3m when it leaves the ramp? Your answer seems to be the perspective I need to look at this problem, is their something wrong with how I have been solving the problems? If so where did I mess up and if not using work/energy formulas how do...
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    Spring mechanics with 2 different masses

    I believe I made a mistake on # 5 so what I think I was suppose to do was take the work of the spring - the work of friction = (mv^2)/2 now multiply the new work by 2 divide by m and square root both sides and you get velocity. so the answer should be 9.568 round to 9.57.
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    Spring mechanics with 2 different masses

    this picture should help
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    Spring mechanics with 2 different masses

    Homework Statement Two masses m and 3m are joined together by a compressed massless spring of constant k = 750 N/m. When spring is released it pushes masses apart and they slide on a horizontal tabletop as shown below. Mass 3m slides to the right and its track is frictionless. Mass m slides to...
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    Static electricity and electrons

    my final answers part a: 5.93 * 10^6 m/s part b: 1.686 * 10^7 s part c: 8.78 * 10^11 m/s^2 part d .0125 m part e: 6.078 * 10^6 m/s part f: 45.71 degrees final questions can I uses 5.93 * 10^6 m/s from part a as V0 in part e? can I uses 6.078 * 10^6 m/s from part e as velocity y in part f
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    Static electricity and electrons

    ok let me change that is the thought process correct for the rest?
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    Static electricity and electrons

    for part d I was suppose to find vertical displacement where gravity is negligible so I used the formula x = (at^2)/2 so my answer was .02496 for part e. i need to find the magnitude of velocity so I use the formula V = V0 + at I used part a's velocity for V0 and I used vertical...
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    Static electricity and electrons

    for part d i used (at^2)/2 . I already solved a and t. so my answer was .02496
  21. F

    Static electricity and electrons

    I mistyped part D the actual queston is find the vertical displacement of the electron the moment it leaves the electric field
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    Static electricity and electrons

    so because it is an electron q = 1.6 *10^-19 C and M = 9.11* 10 ^-31. and the Volts in the parallel plates is 1 and the change in vertical distance is .1 M so it should be (q(1/.1))/M. if this is correct then the answer should be a = 1.756 * 10^12. did i do this right if not please tell me where...
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    Static electricity and electrons

    I was already their my question was if how I found F was correct i applied f = qE, E = k(q/r^2), and V= k(q/r) and derived F = V^2/K. to get the answer you use a= F/m but my question was how to get F.
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    Need help finding distance of skier going down a hill! Help!

    can anyone that just helped mocambo help me my problem is at https://www.physicsforums.com/showthread.php?p=1627736#post1627736 I really need help, anything will be greatly appreciated.
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    Static electricity and electrons

    Ok i now am confused? I don't know how to find a I just try two different ways but I think they are both wrong. first way: ended up using the Pythagorean theorem to find distance traveled and I plugged it into x = vt +.5at^2 where I use my horizontal velocity as the initial velocity. I think...
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    Static electricity and electrons

    ok then t = 1.686 * 10^-7. now for part c do I have to put the kinematic in 2 dimensions?
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    Static electricity and electrons

    I do not think so because the parallel plates voltage is +1 which would change the force which would also means it changes the velocity. unless the x component of velocity is constant which I'm not certain of. also check out that pic I think my reasoning will be clearer if you see it.
  28. F

    Static electricity and electrons

    I can't get the right answer Dick can you help? I seem to be unable to do part B.
  29. F

    Static electricity and electrons

    I'm having trouble finding time. Should I use a kinematics formula?
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