Calculating Stopping Time and Distance Using Impulse Momentum Theorem

AI Thread Summary
The discussion focuses on applying the impulse-momentum theorem to prevent an egg from breaking when dropped from a height. It emphasizes that gradually slowing down the egg using a soft surface, like pillows, reduces the force exerted on it, allowing it to remain intact. The impulse-momentum theorem states that the impulse applied to an object equals its change in momentum, linking force and time. A longer stopping time results in a lower force required to stop the egg. The conversation also transitions to a related problem involving a baseball, seeking assistance in calculating the time and distance it takes to stop under a constant force.
JMarino
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I have a question I am looking for an asnwer: Using the impulse momentum theorum, describe how it is possible to drop an egg from a 2 story high building and not have the egg break.
 
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Sounds like a homework problem to me. Show your thinking so far.
 
Well I know we would need to slow down the object gradually to keep intact, such as using a stack of feathers or pillows. How does the theorum apply?
 
What is the theorem? What is impulse?
 
impulse is a factor of both force and time. So the egg hitting the pillows (let's say) is the impulse and for a period of time to slow the egg down?
 
Impulse is defined as Force X Time; the impulse-momentum theorem tells you that the impulse exerted on an object (by the net force on it) equals its change in momentum.

For the same impulse, if you arrange for the force to be exerted over a greater time (making Time bigger) by using a pillow (for example), what does that imply about the force being exerted on the egg?
 
you need less force to stop the object
 
Right! If you are willing to take a longer time to stop the egg, then you can use less force. Conversely, by using less force--via a softer surface like a pillow--you need more time to stop the egg.
 
Thank you for the help!
 
  • #10
So I have one more problem to solve. Can you help?
A 0.15 kg baseball moving at +26 m/s is slowed to a stop by a catcher who exerts a constant force of 390 N. How long does it take this force to stop the ball? How far does the ball travel before stopping?
- P=mV = 0.15Kg x 26M/s = 3.9Kgm/s. Then what equation?
 
  • #11
JMarino said:
So I have one more problem to solve. Can you help?
A 0.15 kg baseball moving at +26 m/s is slowed to a stop by a catcher who exerts a constant force of 390 N. How long does it take this force to stop the ball? How far does the ball travel before stopping?
- P=mV = 0.15Kg x 26M/s = 3.9Kgm/s. Then what equation?

Look at the first part of post #6 again.
 

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