A particle P2 chases particle P1 with constant speed.

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Homework Help Overview

The problem involves two particles, P1 and P2, where P1 moves with a constant velocity along the x-axis and P2 chases P1 with a constant speed from a different starting point. The discussion focuses on determining the initial acceleration of P2, the ultimate separation between the two particles, and the nature of the trajectory of P2 as observed from the frame of P1.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the motion using relative velocities and questions the nature of the trajectory. Participants discuss the implications of the rate of change of direction of the velocity vector and the relationship between relative positions.

Discussion Status

Participants are exploring the concepts of angular velocity and trajectory shape, with some providing links to external resources. There is an ongoing examination of the assumptions regarding the inertial frames and the possibility of qualitative arguments for the trajectory without deriving equations. Some participants express uncertainty about the treatment of the two-particle system.

Contextual Notes

There are discussions about the definitions of angular velocity and the nature of the trajectory, with references to geometric arguments and conic sections. The original poster has provided an attachment for further details, indicating that some information may be missing from the discussion.

Ishan Sharma
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Homework Statement


[/B]
A particle P1 moves with a constant velocity v along x-axis, starting from origin. Another particle
P2 chases particle P1
with constant speed v, starting from the point (0, d). Both motion begin
simultaneously. Find -

1. Initial acceleration of P2.
2. Ultimate Separation between P1 and P2
3. Whether the trajectory of P2 in frame of P1 will be Hyperbola or Parabola.

The Attempt at a Solution


Let r be the distance between 1 and 2 at any
time t.
$$vcosθ- v= \frac{dr}
{dt} ... (1) $$
$$vcos\theta= \frac{dx}
{dt} ... (2)$$

For the rest of solution, please refer the attachment.

My questions-

1. This ω, according to me is the rate of change of direction of velocity vector. Now, I couldn't think about dθ/dt directly. So, what I did was, I saw took the ω with respect to P1. Is that right? If not, what's the correct answer?
2. I don't know what to do with the trajectory.

Any help would be appreciated.

Thankyou.
mZnac.png
 

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I am impressed by the elegance of your solution to the first part.
For ω, not sure what your difficulty is. The rate of rotation will be the same in all inertial frames.
For the trajectory, how does the relative x position relate to the terms in your dx=vdt+dr?
 
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haruspex said:
For ω, not sure what your difficulty is. The rate of rotation will be the same in all inertial frames.
For the trajectory, how does the relative x position relate to the terms in your dx=vdt+dr?
I am not sure about ω because the two particle system can not be treated as a rigid body. So how can we say that ω with respect to (w.r.t) ground frame will be same as that w.r.t. any inertial frame?
And for trajectory, can we not give some qualitative or intuitive argument without forming the actual equation..? For eg: If we show that the distance of a particle from a fixed point is constant, we can say it moves in a circle. In case of conics, their definition can be used to find the nature of path. Can we say something similar about our question..?
 
Ishan Sharma said:
I am not sure about ω because the two particle system can not be treated as a rigid body. So how can we say that ω with respect to (w.r.t) ground frame will be same as that w.r.t. any inertial frame?
Yes, rate of rotation is the same in any inertial frame. Indeed, an accelerating frame is ok, as long as it is not itself rotating.
Ishan Sharma said:
And for trajectory, can we not give some qualitative or intuitive argument without forming the actual equation
You can give a geometric argument. After answering my question in post #3, look at https://amsi.org.au/ESA_Senior_Years/SeniorTopic2/2a/2a_2content_10.html.
 
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r+Δx=d
⇒r=d-Δx
Thus the distance between focus and directrix is d and the distance between particles is always equal to distance between P2 and directrix. Thus the path is parabolic. Am I right, sir?
 
Ishan Sharma said:
r+Δx=d
⇒r=d-Δx
Thus the distance between focus and directrix is d and the distance between particles is always equal to distance between P2 and directrix. Thus the path is parabolic. Am I right, sir?
Yes.
 

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