00:01
Good day, the topic is about conservation of momentum.
00:04
In the absence of external forces, momentum before and after certain process must be conserved.
00:13
So in this case, we're given with a cannon that launches a cannon ball at the speed of 5, 450 meters per second at an angle of 60 degrees above the horizon.
00:26
So let's try to draw that first.
00:28
Say this is your canon bowl and then launch as a canon bowl which by the way mass is m and then speed of v and then a certain angle of 6theta above the horizontal so as a result your canon with this mass of m would recoil and the recoil and the recoil um speed of the cannon is what we're looking for in this case so fine v recoil now suppose the cannon ball starts from rest so but that's uh let's say v capital v just have that as capital v for the for the canon for the canon and small v for the cannon box starts from rest so that's zero meters per second so let's start filling up this value so for the canon shell let's 98 kilograms for the canon itself is 5 times 10 to the 4 kilograms launching speed and launching angle so v recoil for the cannon so by conservation of momentum we say that change in momentum of your cannonball shell or cannon ball plus the change in momentum of your canon is equal to zero.
02:16
All the way you're putting it is that change of the cannon ball is equal to the negative change and momentum of your canon.
02:25
So let's define first the change in momentum of your canon ball.
02:30
So that's going to be mass times change in the velocity, where the change in the velocity is equal to the velocity final, minus velocity initial.
02:52
Canon ball starts from rest, so this is going to be zero, and then the final velocity is given to be in vector form is vf cosine theta ihat plus vf sine theta j hat...