# Momentum of flashlight in intergalactic space

1. Nov 7, 2006

### opticaltempest

Hello,

I am looking for some help on the following problem:

http://img387.imageshack.us/img387/5481/problemuk7.jpg [Broken]

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:

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

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

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

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)?

Last edited by a moderator: May 2, 2017
2. Nov 8, 2006

### opticaltempest

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 $$\frac{{\Delta E}}{c}$$, we have

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

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

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

$$v = 3.33 \mbox{\mu m}$$

Last edited: Nov 8, 2006