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Four particles $P_1, P_2,P_3,P_4$ are moving in a plane. At $t=0$, they are at the four corners of a square $ABCD$ of edge length $l$. Each of the particles has a constant speed $v$. The velocity of $P_1$ is always directed to $P_2$, that of $P_2$ is always directed to $P_3$, that of $P_3$ is always directed to $P_4$, that of $P_4$ is always directed to $P_1$. At what time $t$ will all the four particles collide and where will they collide? Also how to calculate the angular acceleration of the line joining the particle and the final point of collision at any instant?

My attempt: First I assumed for any small time $dt$, any of the one particle moves a distance $v$ $dt$. Then for an adjacent particle the change in the angle (very small) $\Rightarrow$ $tan(d\theta)=\frac{vdt}{l}$ and it gives $\omega=\frac{v}{l}$. So all the particles are rotating with this angular frequency. But now how can we use this value of angular frequency? Firstly, I am finding it difficult to visualize the situation due to which I am not able to understand exactly what concept should we apply. Secondly, can we use linear equations of motion to approach this? Also can we view the motion with respect to a particle? enter image description here

Found this on Wikipedia (Mice Problem/Pursuit Curve)

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  • $\begingroup$ Does this answer your question? Why should the particles meet at a common point? $\endgroup$ Commented Jul 4, 2021 at 8:48
  • $\begingroup$ @VincentThacker, I have somewhat tried to figure out that the curve will be spiral, but the main query is that what exactly should we apply here? As that question was closed, so probably we can go in some more depth here. $\endgroup$
    – UNAN
    Commented Jul 4, 2021 at 8:50
  • $\begingroup$ @Community You can find your answer here. The same applies to 4 particles in a square. $\endgroup$ Commented Jul 4, 2021 at 9:03
  • $\begingroup$ @VincentThacker, I am not still able to solve after reading the mentioned post. Sorry $\endgroup$
    – UNAN
    Commented Jul 4, 2021 at 10:04
  • $\begingroup$ See also Wikipedia, Wolfram, and Math.SE. $\endgroup$
    – Qmechanic
    Commented Jul 4, 2021 at 13:48

1 Answer 1

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Hint: Try putting 4 dots on the paths after a short time.

Join the 4 dots and see what happens.

There is a nice simple answer to this...

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  • $\begingroup$ squares of smaller sides? $\endgroup$
    – UNAN
    Commented Jul 4, 2021 at 9:14
  • $\begingroup$ @Community yes, then you've got the same problem again, but this time with a smaller square, then think about how much the side has reduced and how long it took $\endgroup$ Commented Jul 4, 2021 at 9:16
  • $\begingroup$ the new velocities will be directed along the sides of the new square... $\endgroup$ Commented Jul 4, 2021 at 9:24
  • $\begingroup$ Even after using your hint, I am not able to solve @JohnHuner, Really sorry. Also can you please go through the link mentioned by VincentThacker in the above comments? $\endgroup$
    – UNAN
    Commented Jul 4, 2021 at 11:19
  • $\begingroup$ Can you please help further? @JohnHunter $\endgroup$
    – UNAN
    Commented Jul 4, 2021 at 13:45

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