what is the impulse of the net force applied to the ball during the collision with the floor?
filed in Talk about dog on Jul.01, 2009
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?
July 1st, 2009 on 5:12 pm
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