00:01
Question one tells us that a rocket cruises past a laboratory with this velocity here given by vr in the positive x direction, just as a proton is launched with the velocity shown here on both the i and j hat directions.
00:14
We want to find for this question in part a, what are the proton's speed and its angle from the y axis, or the y prime, the moving axis in a, the lab frame and b the rocket frame.
00:27
So in general, we don't find this net velocity.
00:29
So in the laboratory frame, we just can use the components given to us in the velocity of the proton's expression.
00:36
So i can say vp net then in terms of magnitude, it is simply just the components square and then square rooted.
00:45
So that's 1 .41 times 7 .6.
00:48
This is the i component, right? because if remember, i can do this just because i can think of them as a right triangle.
00:54
This is my i component.
00:56
This is my j component.
00:57
So the net velocity here is just my hypotenuse.
01:02
So if i take my velocity in my eye direction or my x direction, 1 .41 times 10 to the 6 meters per second squared, plus the same term again because it's the same magnitude, just in the different direction and squared with the whole thing, i would find that my net velocity in this scenario is 2 .0 times 10 to the 6 meters per second.
01:37
And the angle it asks for in the y direction, start from the y axis would be this angle data here.
01:44
But we can tell just because the magnitude of the i component and j component are equal, that must mean that the angle from the y axis has to be 45 degrees.
01:54
My diagram doesn't quite show it, but a little more realistic, if your components are the exact same on your right triangle, and this has to be on both sides 45 degrees...