Conservation of Momentum difficulty

Click For Summary
SUMMARY

The discussion centers on a physics problem involving the conservation of momentum, where a 730-N man on a frictionless pond throws a 1.2 kg textbook at 5.0 m/s to reach the opposite shore. The correct calculation for the time taken to reach the south shore is determined to be 62 seconds, using the final velocity of the man after throwing the book. The initial speed of the book is clarified as being 5.0 m/s at the moment of the throw, while the final speed of the man is calculated to be -0.081 m/s. The distinction between initial and final states is emphasized for clarity in momentum calculations.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Basic knowledge of kinematics equations
  • Familiarity with mass and velocity calculations
  • Ability to interpret physics problems involving forces and motion
NEXT STEPS
  • Study the principles of conservation of momentum in closed systems
  • Learn how to apply kinematics equations in real-world scenarios
  • Explore the concept of relative velocity in physics
  • Review examples of momentum problems involving multiple objects
USEFUL FOR

Students studying physics, educators teaching momentum concepts, and anyone interested in solving real-world physics problems involving motion and forces.

dontcare
Messages
12
Reaction score
0
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
 
Physics news on Phys.org
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.
 

Similar threads

Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 27 ·
Replies
27
Views
1K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
998
  • · Replies 3 ·
Replies
3
Views
5K