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Momentum of flashlight in intergalactic space

  1. Nov 7, 2006 #1

    I am looking for some help on the following problem:


    I have some questions on part (b)

    Here is my solution for (a):

    Conservation of linear momentum:

    [tex]\Delta p = p_f - p_i = m_f v_f - m_i v_i [/tex]

    Solving for mass:

    [tex]E = mc^2 \Rightarrow m = \frac{E}{{c^2 }}[/tex]

    Substituting mass into conservation of linear momentum:

    \Delta p = \frac{{E_f }}{{c^2 }} \cdot v_f - \frac{{E_i }}{{c^2 }} \cdot v_i

    Replace final velocity and initial velocity with [tex]c[/tex] since both velocities are the speed of the photons which are constant speed of light:

    [tex]\Delta p = \frac{{E_f }}{{c^2 }} \cdot c - \frac{{E_i }}{{c^2 }} \cdot c


    [tex]\Delta p = \frac{{E_f }}{c} - \frac{{E_i }}{c}[/tex]

    [tex]\Delta p = \frac{{\Delta E}}{c}[/tex]

    I think I may have a solution for part (b):

    Using the conservation of mechanical energy:

    [tex]\Delta E_{internal} = \Delta K [/tex]

    [tex]1500 = \frac{1}{2}mv^2 [/tex]

    [tex]1500 = \frac{1}{2}\left( {1.5} \right)v^2 [/tex]

    [tex]v = \sqrt {\frac{{1500}}{{\left( {0.5} \right)\left( {1.5} \right)}}}[/tex]

    [tex]v = 44.7 \mbox{ m/s}[/tex]

    Does that look like a valid solution for part (b)?
    Last edited: Nov 7, 2006
  2. jcsd
  3. Nov 8, 2006 #2
    Part (b) seems incorrect. Does anyone have any suggestions? The velocity seems high because I would imagine photons possess little momentum.

    Here is my new answer for (b):

    Using the conservation of linear momentum and the fact that the momentum of the release of photons is equal to [tex]
    \frac{{\Delta E}}{c}[/tex], we have

    [tex]\frac{{\Delta E}}{c} = mv[/tex]

    [tex]\frac{{1500}}{c} = 1.5v[/tex]

    [tex]v = \frac{{1500}}{{1.5c}}[/tex]

    [tex]v = 3.33 \mbox{\mu m}[/tex]
    Last edited: Nov 8, 2006
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