# General to specific classical mechanics

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.

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

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.

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.

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

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

kuruman