General to specific classical mechanics

  • #1
2
0
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 [itex]m[/itex] is thrown from the origin at t=0 with initial three momentum [itex]p_0[/itex] in the y direction. If it is subject to a constant force [itex]F_0[/itex] in the x direction, find its velocity [itex]\mathbf{v}[/itex] as a function of t, and by integrating [itex]\mathbf{v}[/itex] 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 [itex] \gamma_0 [/itex] at [itex]t=0[/itex], and evolves to [itex]\gamma[/itex]. I see this [itex]\gamma_0[/itex] altering general case.
 

Answers and Replies

  • #2
8
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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.
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  • #3
2
0
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.
 
  • #4
8
2
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.
 
  • #5
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
 
  • #6
8
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when you combine two 1d analysis (x and y separately), you get 2d
 
  • #7
kuruman
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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.
 

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