A 0.660-kg ball is dropped from rest at a point 2.10 m above the floor. The ball rebounds straight upward to a height of 1.20 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?
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One Response
7.1.2009
Determine speed before collision:
v = sqrt(2ad)
v = sqrt[(2)(9.8)(2.1)]
v = -6.42 m/s
Determine speed after collision:
v = sqrt(2ad)
v = sqrt[(2)(9.8)(1.2)]
v = 4.85 m/s
Impulse = change in momentum
= m(change in speed)
= 0.660(4.85-(-6.42))
= 11.3 N s