Conservation of Momentum difficulty

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Homework Help Overview

The discussion revolves around a conservation of momentum problem involving a man on a frozen pond who throws a textbook to propel himself. The scenario includes calculating the time it takes for him to reach the opposite shore after throwing the book.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the application of conservation of momentum, questioning the initial and final speeds of the book and the man. There is discussion about the definitions of initial and final speeds in the context of the problem.

Discussion Status

The conversation includes attempts to clarify the definitions of initial and final speeds, with some participants correcting earlier statements. There is an ongoing exploration of the assumptions made in the problem setup, particularly regarding the state of the system before and after the throw.

Contextual Notes

Participants note that the initial state of the system is assumed to be at rest, which raises questions about the interpretation of speed before and after the throw. There is a lack of consensus on the implications of these assumptions for the calculations involved.

dontcare
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I'm not sure if i did this problem correctly. Could someone check it over.

A 730-N man stands in the middle of a frozen pond of a radius of 5.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2 kg physics textbook horizontally toward the north shore at a speed of 5.0 m/s. How long does it take him to reach the south shore?

p_i = p_f
0 = m_1v_1f + m_2v_2f

v_1f = - \frac{m_2}{m_1} * v_2f = - \frac{1.20 kg}{74.5 kg} * 5.0 m/s = -.081 m/s

t = \frac{x}{v} = \frac{5.0 m}{2.46 m/s} = 2.03 s
 
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t = \frac{x}{v} = \frac{5.0 m}{2.46 m/s} = 2.03 s[/QUOTE]

Corrected:
t = \frac{x}{v} = \frac{5.0 m}{.081 m/s} = 62 s[/QUOTE]
 
dontcare said:
Corrected:
t = \frac{x}{v} = \frac{5.0 m}{.081 m/s} = 62 s

That is correct :smile:
 
Thank you for checking my problem.
 
Why is 5.0m/s the final speed of the book but not its initial speed? Wasn't it thrown at a speed of 5.0m/s?
 
Last edited:
we assume that initially all the system is at rest
 
so is it wrong to assume that the initial speed of the book is 5m/s?
 
Yeah, it is. In this case "initial" and "final" mean "before the throw" and "after the throw" respectively.

Of course, there's technically no reason you couldn't pick "initial" and "final" to both be after the throw, but it wouldn't tell you anything useful.
 

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