- #1

annamal

- 381

- 33

mgh = 0

If instead the ball were moving with an initial velocity v, would the equation be

##mgh + \frac{1}{2}mv^2 = 0##?

If there were an external force that caused the ball to drop with initial velocity v, would conservation of energy still be applicable since there was an external force?

How come with conservation of momentum laws, you can't have one mass m colliding with the second mass ##m_2## and then both masses stopping together after the collision as in:

##mv_1 = 0##. How come momentum is not conserved here?