00:01
Hello, here we have to solve the following assignment and first in question a we have to decide what is the position of pac 2 with respect to the pac 1 when pac 1 crosses the dash line.
00:17
So let's illustrate it.
00:19
So the dash line is here and pac 1 is initially moving at a speed of v1 to the right and the force f is applied to the right.
00:37
And p -2 was moving upwards but the force was applied to the same direction to the right.
00:54
So here let's show the trajectories of motion.
00:59
So let's show the trajectory of p -1 and the trajectory of 2 would be looking like this.
01:15
So that's the trajectory of 2.
01:19
So x, let's introduce x's x's x's, x and y.
01:24
So v1 x as a function of time equals to v1 initial plus f over m times time and v2 x is v1 is um f over m times c and coordinate wise x1 is v1 initial plus plus x1 plus c and coordinate wise x1 is v1 initial plus times t plus f over m times t squared over 2 and x2 as a function of time is b2 is sorry f over m times t squared over 2 as we can see x1 is only flyer than x2 so therefore when 1 crosses so therefore line 2 will be on the left and now we have to show the trajectories again we've showed that trajectory for x1 so that it's simply x1 as a function of time and the trajectory of 2 has two components so the first one is this component and the second part of the trajectory of 2 is times time so therefore here we can yeah let's actually actually let's derive this trajectory more strictly for the second one so for the second one so let's subtract t from the second equation so that is equals to y over v to i that's why x of two equals to f over m times actually f over 2m times t squared which is y2 squared divided by y 2 2 e2 initial squared so that is f over 2m e2 initial squared times y squared over t so that's why we can show it as following so here that will be yeah that actually will be a parabolic trajectory...