Momentum of flashlight in intergalactic space

[opticaltempest]

Hello,

I am looking for some help on the following problem:

http://img387.imageshack.us/img387/5481/problemuk7.jpg

I have some questions on part (b)

Here is my solution for (a):

Conservation of linear momentum:

\Delta p = p_f - p_i = m_f v_f - m_i v_i

Solving for mass:

E = mc^2 \Rightarrow m = \frac{E}{{c^2 }}

Substituting mass into conservation of linear momentum:

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

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

\Delta p = \frac{{E_f }}{{c^2 }} \cdot c - \frac{{E_i }}{{c^2 }} \cdot c<br />

Simplify:

\Delta p = \frac{{E_f }}{c} - \frac{{E_i }}{c}

\Delta p = \frac{{\Delta E}}{c}
I think I may have a solution for part (b):

Using the conservation of mechanical energy:

\Delta E_{internal} = \Delta K

1500 = \frac{1}{2}mv^2

1500 = \frac{1}{2}\left( {1.5} \right)v^2

v = \sqrt {\frac{{1500}}{{\left( {0.5} \right)\left( {1.5} \right)}}}

v = 44.7 \mbox{ m/s}

Does that look like a valid solution for part (b)?

 

[opticaltempest]

Part (b) seems incorrect. Does anyone have any suggestions? The velocity seems high because I would imagine photons possesses 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 <br /> \frac{{\Delta E}}{c}, we have

\frac{{\Delta E}}{c} = mv

\frac{{1500}}{c} = 1.5v

v = \frac{{1500}}{{1.5c}}

v = 3.33 \mbox{\mu m}