# Momentum of flashlight in intergalactic space

• opticaltempest
In summary, the conversation is about a problem involving conservation of linear momentum and energy, specifically for a system involving photons. The solution provided for part (a) involves solving for mass using the equation E=mc^2 and then using conservation of linear momentum to find the change in momentum. For part (b), the initial solution is revised to use conservation of mechanical energy, but is deemed incorrect. A new solution is then proposed using conservation of linear momentum and finding the momentum of the release of photons, resulting in a velocity of 3.33 micrometers.
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:

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

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:
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 $$\frac{{\Delta E}}{c}$$, 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:

Hello,

Thank you for sharing your solution for part (a). It looks like you have correctly applied the conservation of linear momentum equation to solve for the mass of the particles.

For part (b), your solution also looks valid. By using the conservation of mechanical energy equation, you have correctly solved for the final velocity of the particles. However, keep in mind that this solution assumes the particles are traveling in a vacuum with no external forces acting on them. In intergalactic space, there may be other factors at play that could affect the final velocity of the particles. It would be helpful to include a statement acknowledging this assumption in your solution.

Overall, your approach and solutions seem sound. Good job!

## What is momentum?

Momentum is a property of an object that describes its motion and is calculated by multiplying its mass by its velocity.

## Why is the momentum of a flashlight in intergalactic space important?

The momentum of a flashlight in intergalactic space is important because it determines the amount of force that is required to change its motion or stop it completely.

## Does the momentum of a flashlight in intergalactic space change?

Yes, the momentum of a flashlight in intergalactic space can change if there is an external force acting on it, such as gravity or collisions with other objects.

## How does the momentum of a flashlight in intergalactic space compare to its mass?

The momentum of a flashlight in intergalactic space is directly proportional to its mass. This means that the greater the mass of the flashlight, the greater its momentum will be.

## Can the momentum of a flashlight in intergalactic space be measured?

Yes, the momentum of a flashlight in intergalactic space can be measured using the equation p = mv, where p is momentum, m is mass, and v is velocity.

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