NICE CONCEPTUAL PROBLEM OF RELATIVE VELOCITY AND MIN DISTANCE BETWEEN PARTICLES || JEE Mains & ADV

Added: 2026-05-10

[00:00:01]Okay, welcome to Jstar IIT GP YouTube channel. So here I discuss a very nice problem, a very nice conceptual problem about relative velocity and displacement. Okay. So what is the question? So here there is a particle there is a particle this particle is going in this direction and this particle is going in this direction with velocity v_sub_1 this v1 and this is with v_sub_2 and initially they are at distance l2 from this origin we have defined the axis here okay and this angle is theta given and this is l1 okay so and now so first of all uh I defined these position vectors for these two. So for this one the position vector is L2 ICAP. For particle one the position vector is we are breaking it into component L1. So this is going to be L1 sin theta. Okay L1 sin theta and this is going to be uh like this one is going to be L1 cos theta. Okay.

[00:01:14]And also we have written the velocity vectors al for in that case too I have broken this component this v1 into its components perpendicular components along this axis and these are the the uh initial velocity vectors. Okay.

[00:01:31]So the relative velocity and relative uh displacement between these particles like the relative distance between these particles I have written as v1 - v1 x1 - x2. So these are the ones. Okay.

[00:01:44]Now let us calculate the first. So understand the thing. So after a time t what will happen? If we are seeing this from here. Okay. Sorry.

[00:01:56]Okay.

[00:01:59]So if we are seeing from this point.

[00:02:02]Okay. If we sit at this particle two. So I am seeing relative to this relative to the particle two. So this particle two will be at rest. But this particle one will move along the direction of relative velocity. So after a time t a a point of time will come when I will have the minimum distance between this and uh these two particles. Okay. And in that case when uh I'll get the minimum distance when this uh displacement vector between this like or or you can say the uh distance vector or the displacement vector between these particles the relative distance between these particles when this vector and the relative velocity vector after time t will be perpendicular to each other.

[00:02:44]Okay. So we have written the condition here v12 at time t and x12 at time t and the dot product is zero because these two are perpendicular at the minimum condition and this one after a time t will be this one is constant. Okay and x12 after a time t will be this is the x12 initial plus this is the v1 to t. So this one plus this one by triangle vector addition. So we'll get this x minimum vector. So we can write it as this one. Okay. And we are getting the dot product. So this is and from here we can write the t minimum equals to this one. Okay. And if you put the all the values of these vectors so you will get this one you can calculate yourself.

[00:03:32]Okay. So for x minimum like uh for finding the minimum distance between these particles. So for in that case you can uh do this also uh putting the value of this t minimum but this will be very lengthy. Okay. So what is a very good uh good method to do this is for x minimum.

[00:03:50]Okay we know at about this point about this point if this point is at rest and this is moving at a constant velocity this is the velocity isn't changing right. So we can say that the angular momentum will be conserved because the net torque over this point is zero.

[00:04:07]external torque about this point is zero. So we can concept the angular momentum. So that is what I have done here. X12 cross V12 is this the initial angular momentum and this is the final angular momentum when it will be at minimum distance. Okay. And as this this and this are perpendicular. So I can I have written like this. Okay. Fine. And also we need not like write like this.

[00:04:32]We can do the direct cross productduct of these two vectors. That will be easier. Okay. And from here we will get the x minimum uh the magnitude of x12 initial cross v12 initial by v12 initial uh and its magnitude. Okay. So if you put these values and calculate this cross product you'll get a direct answer of this x minimum. So this is a uh I think a better method of doing this.

[00:04:57]Okay. In case you are given this v1 v2 these all values you'll be able to calculate these values easier and faster. as I have taken the variable so it is looking a bit lengthy but when you uh when the datas will be given in numbers and like this you can calculated very fast okay so that's it for today uh subscribe to my channel for more interesting topics and more interesting problems