Relativistic momentum in terms of another relativisic momentum

In summary, the conversation involves someone seeking help with a physics problem involving equations and algebraic exercises. They eventually figure out the solution by simplifying the equations and calculating for gamma factors.
  • #1
Sconlan
3
0
Homework Statement
When the ship passes at velocity V1, the shuttle pilot measures the magnitude of the momentum of the spaceship to be p1.

On a subsequent mission, the spaceship again passes the shuttle, this time at twice the previous velocity, V2 = 2V1 = 0.840c.

The shuttle pilot again measures the momentum of the spaceship. In terms of p1, what is the magnitude of the measured momentum of the spaceship this time?
Relevant Equations
The relativistic momentum p of a material particle of mass m and velocity v is defined by the following vector equation:

p=mv/√(1-v^2/c^2 )

The total relativistic energy of a particle m travelling at speed v is given by:

E_tot=(mc^2)/√(1-v^2/c^2) = E_trans+E_mass.

Where:

E_trans = ((mc^2)/(1-v^2/c^2))-mc^2

And

E_mass = mc^2

The relationship between the total relativistic energy and the magnitude of the relativistic momentum can be expressed as:

(E_tot)^2 = p^2c^2+m^2c^4
I feel like this should be pretty straightforward knowing all the equations involved but my brain seems be stalling for some reason.
 
Physics news on Phys.org
  • #2
Sconlan said:
I feel like this should be pretty straightforward knowing all the equations involved but my brain seems be stalling for some reason.
:welcome:

Time to get your brain into gear! You need to make your best effort before we can help.
 
  • #3
Hint. It's a nice algebraic exercise to express ##p_2## in terms of ##p_1##, but you can always do this particular problem numerically.
 
  • #4
So far I've solved for p1 (I think) but I can't figure out how to write p2 in terms of p1.

E_tot^2 = (p1^2 c^2) + (m^2 c^4)

p1^2 c^2 = (E_tot^2) - (m^2 c^4)

p1^2 = (E_tot^2 - m^2 c^4) / c^2

p1 = √(E_tot^2 - m^2 c^4 ) / c

Where E_tot = (mc^2) / √(1 - (v^2/c^2) ) = γmc^2

p1 = √((γm^2c^4 ) - (m^2 c^4 )) / c
 
  • #5
Sconlan said:
So far I've solved for p1 (I think) but I can't figure out how to write p2 in terms of p1.

E_tot^2 = (p1^2 c^2) + (m^2 c^4)

p1^2 c^2 = (E_tot^2) - (m^2 c^4)

p1^2 = (E_tot^2 - m^2 c^4) / c^2

p1 = √(E_tot^2 - m^2 c^4 ) / c

Where E_tot = (mc^2) / √(1 - (v^2/c^2) ) = γmc^2

p1 = √((γm^2c^4 ) - (m^2 c^4 )) / c
Isn't ##p_1 = \gamma_1 mv_1##, where ##v_1 = 0.42c##? And ##p_2 = \gamma_2 mv_2##, where ##v_2 = 0.84c##?

Can you just do that numerically (with the ##m## cancelling out)?
 
  • Like
Likes nasu
  • #6
PS if you write ##\dfrac {p_2}{p_1} = \dfrac{\gamma_2 mv_2}{\gamma_1 mv_1} = \dfrac{2\gamma_2}{\gamma_1}## I'm not sure that simplifies much further, so you just have to calculate the two gamma factors.
 
  • #7
Thank you, I’d overcomplicated it a ridiculous amount because I was rushing 😖 All sorted now 💆🏻‍♂️
 

1. What is relativistic momentum?

Relativistic momentum is a concept in physics that takes into account the effects of special relativity on the momentum of an object. It is a measure of an object's motion and is dependent on its mass, velocity, and the speed of light.

2. How is relativistic momentum different from classical momentum?

Relativistic momentum takes into account the effects of special relativity, such as time dilation and length contraction, which are not accounted for in classical momentum. This means that relativistic momentum can be significantly different from classical momentum at high speeds.

3. Can relativistic momentum be expressed in terms of another relativistic momentum?

Yes, relativistic momentum can be expressed in terms of another relativistic momentum. This is because relativistic momentum is a vector quantity and can be broken down into its components, which can then be expressed in terms of other relativistic momentum components.

4. What is the formula for calculating relativistic momentum?

The formula for calculating relativistic momentum is p = mv/√(1-v^2/c^2), where p is the relativistic momentum, m is the mass of the object, v is the velocity, and c is the speed of light.

5. How does relativistic momentum change at high speeds?

At high speeds, relativistic momentum increases significantly compared to classical momentum, as the effects of special relativity become more pronounced. This means that objects with high relativistic momentum can have significantly different behavior and properties compared to objects with low relativistic momentum.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
521
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
803
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
23
Views
3K
  • Introductory Physics Homework Help
Replies
12
Views
595
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
972
  • Introductory Physics Homework Help
2
Replies
45
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
817
Back
Top