00:01
Since in this question, a collision occurred between 2 kg particles traveling with the velocity va1, that is equals to va1, that is equals to 2 minus 4 eyecap, minus 5 j cap, right? so minus of 4 meter per second, icap and plus minus of 5 meter per second, obviously, in a side unit into j cap.
00:27
And mass of this a particle that is equals to 2kg, right? and the second mass that of equals to 4 kg that is moving with the velocity vb right is equal to 6 and minus 2 so b b 1 that is equals to 6 meter per second iqcab and plus minus of 2 meter per second j gap so this have a mass of 4kgg right so basically we need to find after the collision the move though both the mass connect so that means we need to find their velocity in unit reactor notation right so what we're going to do is over here that what how can we find that so first of all you know that there is no extra force that means momentum should be conserved right so that means momentum along x and y should be conserved so let's say that velocity after collision is x i cap and plus y j gap right so that means uh before the collision what is the momentum that is due to this one 2 into minus 4 so that means 2 into minus 4 and plus of 4 into 6, 4 into 6, and that is in a cap equals, let's say this is mass will be x into mass of the, when these two combine, that will be 6 icap, right? so that means this will be 24 minus of oil, that means 16 icap equals to 6x i cap, right? so that means x will be nothing but that have a component that is equals to 16 upon 6.
02:08
So that will be 8 upon 3, right? and similarly, similarly, when we equate these two, that means minus 5 into 2, so 10 and minus 4 into 8...