Relativistic momentum in terms of another relativisic momentum

[Sconlan]

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.

 

[PeroK]

I feel like this should be pretty straightforward knowing all the equations involved but my brain seems be stalling for some reason.

Time to get your brain into gear! You need to make your best effort before we can help.

 

[PeroK]

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.

 

[Sconlan]

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

 

[PeroK]

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)?

 

[PeroK]

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.

 

[Sconlan]

Thank you, I’d overcomplicated it a ridiculous amount because I was rushing 😖 All sorted now 💆🏻‍♂️