General to specific classical mechanics

[yang32366]

Homework Statement

Taylor Classical Mechanics 15.84

Relevant Equations

Below

Source = John R. Taylor, Classical Mechanics, page 651 + page 677

Trying to solve,

A mass m is thrown from the origin at t=0 with initial three momentum p_0 in the y direction. If it is subject to a constant force F_0 in the x direction, find its velocity \mathbf{v} as a function of t, and by integrating \mathbf{v} find its trajectory.

Taylor solves this and I slowly worked this problem if mass released from rest.

$$\gamma = \sqrt{1+\bigg(\frac{Ft}{mc}\bigg)^2} $$

$$\mathbf{v}(t)=\frac{\mathbf{p}}{m\gamma}=\frac{\mathbf{F}t}{m\sqrt{1+(Ft/mc)^2}}$$

$$\mathbf{x}(t)=\frac{\mathbf{F}}{m}\left(\frac{mc}{F}\right)^2\left(\sqrt{1+\left(\frac{Ft}{mc}\right)^2}-1\right)$$

I am not sure how I could get this specific.

Thoughts=

There exists \gamma_0 at t=0, and evolves to \gamma. I see this \gamma_0 altering general case.

 

[MrsZee]

You are making it too complicated for no reason; what is speed of light doing in your attempt?
Observe vertical and horizontal motion separately and u will see what needs to be done.
#OnlyPerfectPracticeMakesItPerfect
#MathAndPhysicsHelpOnline

 

[yang32366]

You are making it too complicated for no reason; what is speed of light doing in your attempt?
Observe vertical and horizontal motion separately and u will see what needs to be done.
#OnlyPerfectPracticeMakesItPerfect
#MathAndPhysicsHelpOnline

I am unsure how this help.
Velocity of light is never changing.
Velocity of mass and so velocity of mass’s frame (velocity of S’ frame in special relativity language) is changing.

 

[MrsZee]

velocity of light has nothing to do with the mass m thrown from the origin at t=0 with initial three momentum p0 in the y direction, with additional constant force in x direction.

as usual, the hardest part of the question is to understand what they are talking about.

the question is actually about projectile motion:
- x component comes from the horizontal constant force which will create displacement = Vx times the time in the air
- y component comes from p0 = m Vy, and is the subject to gravitational force; this will determine your time in the air.

hope this helps; i am not allowed here to just solve it for you.

 

[timetraveller123]

i think the OP is required to solve the problem relativistically

@yang32366
the equation you have are for 1d motion
the problem is asking you to solve 2 d version of this problem
it is like throwing a ball horizontally of a cliff. there is constant force acting on the y-axis and you give it a initial $P_o$in the x direction except you have to do it relativistically

one tip would be to solve for $\gamma$ first

 

[MrsZee]

when you combine two 1d analysis (x and y separately), you get 2d

 

[kuruman]

i think the OP is required to solve the problem relativistically

Yes, this is problem 15.84 in Taylor's book, which I happen to own, and appears at the end of the chapter on Special Relativity. Also, OP omitted the last sentence in the problem that reads, "Check that in the non-relativistic limit the trajectory is the expected parabola." Contrary to OP's claim, Taylor does not solve this or at least the solution does not appear in the textbook.