00:01
Here, for part a, we're going to use the conservation of energy from equation 817.
00:08
We can say that k sub 1 plus u sub 1 is going to be equal to the kinetic energy at sub 2 plus the potential energy sub 2.
00:21
We can say that here we're going to, we know that this is going to be 0, and we know that essentially we can say that mgl multiplied by 1 minus cosine theta sub 1.
00:46
This would be equal to 1 half mv sub 2 squared plus mgl multiplied by 1 minus cosine of theta sub 2.
00:58
We know that here masses are going to of course cancel out and we know that l is equaling 1 .4 meters, theta sub 1 is equaling 30 degrees, and theta sub 2 is equaling 20 degrees.
01:14
Therefore, we can solve for v .2.
01:18
This is going to be equal to the square root of 2gl, multiplied by cosine of theta sub 2 minus cosine of theta sub 1.
01:31
And after plugging our, plugging our variables, we find that the velocity sub 2 is equaling 1 .4 meters per second.
01:40
This is our answer for part a...