A 0.730-kg ball is dropped from rest at a point 1.70 m above the floor. The ball rebounds straight upward to a
A 0.730-kg ball is dropped from rest at a point 1.70 m above the floor. The ball rebounds straight upward to a height of 1.30 m. Taking the negative direction to be downward, what is the impulse of the net force applied to the ball during the collision with the floor?
We had this problem in my physics class. Here’s how I went about it: first find PE initial=mgy use -1.7m for y. Then, remembering that PE=KE, use the formula for KE (1/2mv^2) to solve for v initial.
Now you’ll need to find PE final =mgy, use 1.3m for y. again PE=KE, so you can solve for v final.
we know that the impulse is just the change in momentum. P=mv. So you can find the change in velocity (it will be like adding them, because of the direction change). Then just multiply that by the mass, and you’re done!
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