The first problem you did.. the equation you used is for inelastic situations.. but the diagram and the situation was elastic. so there should be two different velocities. not one.
I don't think the way you set up the last part is right. It should be s^2 instead of s^3 because you have to view that 4kgm/s^2 as a constant. If you don't, you get a complicated derivative quotient rule. So you have to think it as the derivative of (4)(t^2).
why the change of momentum is gonna equal to 0??
At 4:00, why should the total change in momentum be equal to 0
every1 makes math videos, we need more physics videos. thanks for this
The first problem you did.. the equation you used is for inelastic situations.. but the diagram and the situation was elastic. so there should be two different velocities. not one.
Thanks a lot..please, do more videos for physics 2
I don't think the way you set up the last part is right. It should be s^2 instead of s^3 because you have to view that 4kgm/s^2 as a constant. If you don't, you get a complicated derivative quotient rule. So you have to think it as the derivative of (4)(t^2).
Very understandable and I love the accent! Makes me feel alive while studying. Cheers mate! :))
THankzzzzzzzzzz
you da bomb! this helped so much
thanks!!!!