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Question One ---. A 0.024 kg ball is dropped to the floor from a height of 1.75 meters. Just after the impact with the floor, the ball has an upward velocity of 5.2 m/s. If the ball was in contact with the floor for exactly 0.023 seconds, find the following:

a.) The impulse delivered to the ball, from the floor.
b.) The average force of impact.

Question TWO --- Two identical objects A and B of mass 4.0 kg move on a horizontal and frictionless surface. Object B initially move to the right with a speed of 0.5 m/s. Object A initially moves to the right with the speed of 1.5 m/s, so that it collides with object B.

a.) Determine the total momentum of teh two cart system.
b.) A student predicts that the collision will be totally inelastic (they stick together after the collision). Assuming this is true, determine the velocity of each object immediatly after the collision.

PLEASE HELP
YOUR HELP IS GREATLY APPRECIATED!!!!

2007-12-10 11:22:38 · 1 answers · asked by mimi j 1 in Science & Mathematics Physics

1 answers

v^2 = 2gh = (2)(9.80665)(1.75
v = √((2)(9.80665)(1.75))
p = 0.024√((2)(9.80665)(1.75))
∆p = 0.024(5.2 + √((2)(9.80665)(1.75)))
a.) ∆p ≈ 0.2654066 N•s ≈ 0.265 N•s
b.) F ≈ 0.2654066 / 0.023 ≈ 11.5394 N ≈ 11.5 N

TWO
a.) ∑p = (4.0 kg)(1.5 m/s + 0.5 m/s) = 8.0 kg•m/s
b.) v = (8.0 kg•m/s) / (8.0 kg) = 1.0 m/s

2007-12-10 12:01:44 · answer #1 · answered by Helmut 7 · 0 0

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