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

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Homework Help Overview

The problem involves a 70-g steel ball that falls vertically onto a steel plate, contacts it for a brief period, and rebounds elastically. Participants are exploring the implications of the given time intervals and the relationship between impulse and momentum.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning how to calculate the velocity of the ball at various points, the significance of the total round trip time, and the relationship between change in momentum and impulse. There is confusion regarding the time intervals and their relevance to the calculations.

Discussion Status

There is an active exploration of the problem with participants raising questions and attempting to clarify their understanding of the concepts involved. Some guidance has been offered regarding the neglect of the brief contact time when calculating the speed at impact.

Contextual Notes

Participants are grappling with the implications of the given time intervals, particularly the total round trip time of 2.00 s and the contact time of 0.5 ms. There is uncertainty about how these factors influence the calculations of velocity and momentum.

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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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