1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding velocity with changing distance and acceleration

  1. Apr 14, 2008 #1
    I have a physics problem I have been trying to figure out for a week and I'm now to the point of way past frustrated. The question is:

    An object with the mass 1.44*10^-27 kg is propelled at an initial speed of 3.84*10^5 m/s toward a fixed nucleus 4.80 m away. The object is repelled by the nucleus with the force F= a/x^2 where x is the seperation between the object and the nucleus and a = 2.59*10^-26 Nm^2. What is the speed of the object when it is 7.46*10^-10 from the nucleus?
    I have tried many ways to solve this to the point of I dont know what I'm doing anymore and just trying to plug in numbers into equations and still not getting the right answer of 3.15*10^5.
    final speed^2=initial speed^2+2*acceleration*distance
  2. jcsd
  3. Apr 14, 2008 #2
    Use conservation of energy.
  4. Apr 14, 2008 #3
    as in work= .5 mass* velocity^2-.5 mass*velocity^2?
  5. Apr 14, 2008 #4
    as in...
    energy = kinetic energy + potential energy = constant
  6. Apr 15, 2008 #5
    m = 1.44*10^-27 kg, initial speed v1 = 3.84*10^5 m/s, x1 = 4.80 m,
    force F= a/x^2 where a = 2.59*10^-26 Nm^2. x2 = 7.46*10^-10 m, find v2.
    note force is not a constant, but just like gravitational force f = GmM/x^2 except f is attractive but F is repulsive. Remember the corresponding potential energy for f is -GmM/x, then the corresponding potential energy for F is a/x,
    initial total energy = (1/2)mv1^2 + a/x1
    final total energy = (1/2)mv2^2 + a/x2
    energy conservation: (1/2)mv1^2 + a/x1 = (1/2)mv2^2 + a/x2
    then get v2 = 3.15*10^5 m/s
    Last edited: Apr 15, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook