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Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force applied by the floor on the ball?

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Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - https://brainly.com/question/12946012#readmore

Eduard22sly

The magnitude of the average force applied by the floor on the ball is 60 N

From the question given above, the following data were obtained:

Mass (m) = 0.15 Kg

Initial velocity (u) = 6.5 m/s

Final velocity (v) = 3.5 m/s

Time (t) = 0.025 s

Force (F) =?

The magnitude of the average force applied by the floor on the ball can be obtained as follow:

[tex]F = \frac{m(v + u) }{t} \\\\F = \frac{0.15(3.5 + 6.5) }{0.025} \\\\F = \frac{0.15(10)}{0.025}\\\\[/tex]

F = 60 N

Thus, the magnitude of the average force applied by the floor on the ball is 60 N

Learn more: https://brainly.com/question/231466

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