How to Solve the Impulse Problem with a 70-g Steel Ball in Contact with a Plate

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A 70-g steel ball falls and strikes a steel plate, remaining in contact for 0.5 ms before rebounding elastically. The total round trip time for the ball is 2.00 seconds, allowing for the neglect of the brief contact time when calculating impact velocity. The velocity upon impact is approximately 9.8 m/s, and the ball rebounds at the same speed due to elastic collision. The change in momentum is crucial for determining the average force exerted during contact, which involves careful attention to vector direction. Understanding impulse and momentum is key to solving the problem accurately.
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Homework Statement



A 70-g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 2.00 s. In this situation, the average force exerted on the ball during contact with the plate is closest to:



The Attempt at a Solution



Im not sure about a lot in this problem. Do I calculate the velocity has it falling for 1 sec or .75 sec or I shouldn't need to do that? I am so confused on this one.
 
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I don't get why do you have those 2.00 s here. What is balls velocity when it hits the plate? When it re-bounces? How is change in momentum related to to impulse?
 
So does it fall down in 1 sec? Is it then moving at 9.8 m/s. Does it then go up at 9.8 m/s?
 
preluderacer said:
Do I calculate the velocity has it falling for 1 sec or .75 sec or I shouldn't need to do that?
You're told that the total round trip time is 2.00 s and that the time in contact with the floor is 0.5 ms, so as far as computing the speed as it hits the floor you can safely neglect that 0.5 ms.

preluderacer said:
So does it fall down in 1 sec? Is it then moving at 9.8 m/s. Does it then go up at 9.8 m/s?
Yes.
 
so what I did was have (0.070kg(9.8m/s)-0.070kg(9.8m/s) / 0.005 s ? This seems very wrong to me
 
preluderacer said:
so what I did was have (0.070kg(9.8m/s)-0.070kg(9.8m/s) / 0.005 s ? This seems very wrong to me
Careful with signs. Momentum is a vector--direction matters.
 
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