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Linear momentum/kinetic energy

  1. Mar 22, 2006 #1
    A certain radioactive (parent) nucleus transforms to a different (daughter) nucleus by emitting an electron and a neutrino. The parent was at rest at the origin of an xy coordinate system. The electron moves away from the origin with linear momentum (-3.3 x 10-22 kg m/s) ; the neutrino moves away from the origin with linear momentum (-2.2 x 10-23 kg m/s) . What are (a) the magnitude and (b) angle (from the +x axis) of the linear momentum of the daughter nucleus? (c) If the daughter nucleus has a mass of 1.8 x 10-26 kg, what is its kinetic energy?

    i know part a and b i did are correct:
    part a. i got 3.3x10^-22 kg m/s
    part b the angle is 3.81 degrees

    but i cant figure out part c? what formula am i supposed to use.
     
  2. jcsd
  3. Mar 22, 2006 #2
    Do you know how to write down the kinectic energy of a particle in terms of its momentum?
     
  4. Mar 22, 2006 #3
    p = mv?

    the KE = (m -mi)V^2

    ?
     
  5. Mar 22, 2006 #4
    K.E = 1/2 mv^2 . Multiply and divide RHS by m. K.E = (mv)^2/2m
     
  6. Mar 22, 2006 #5
    what is RHS?
    K.E = (mv)^2/2m
    i get 9.8x10-70....but that doesnt look right.
     
  7. Mar 22, 2006 #6
    Right hand side.

    You start with [tex] K.E. = \frac{1}{2}mv^2 [/tex]

    Then multiply both sides by m, but to keep them the same you must divide by m, i.e.:
    [tex]K.E. = \frac{1}{2}mv^2 . \frac{m}{m} [/tex]
    [tex]K.E. = \frac{m^2 v^2}{2m} [/tex]
    [tex]K.E. = \frac{(mv)^2}{2m} [/tex]
    Then use the fact that [tex] p = mv [/tex] to get:
    [tex]K.E. = \frac{p^2}{2m} [/tex]

    Can you use that formula to get the answer?
     
  8. Mar 22, 2006 #7
    thanks. i got it now. :blushing:
     
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