Another conservation of momentum problem

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Homework Help Overview

The problem involves the conservation of momentum in the context of a boat and a hunter firing bullets. The initial scenario describes a boat at rest on a lake, with a specific weight and a series of bullets being fired, prompting a discussion on the resulting speed of the boat.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the momentum conservation formula and question the initial calculations regarding the direction of momentum. There are inquiries about how weight relates to momentum and the need for clear assumptions in the problem setup.

Discussion Status

Participants are actively engaging with the original poster's calculations, pointing out potential mistakes and encouraging the posting of full calculations for clarity. There is a recognition of errors in sign and calculation, but no consensus has been reached on the correct approach or final answer.

Contextual Notes

There is an emphasis on the need to consider the relationship between weight and momentum, as well as the importance of clearly stating assumptions made during the problem-solving process.

BrainMan
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Homework Statement


The combined weight of a small boat and a hunter is 1200 N. The boat is initially at rest on a lake as the hunter fires 75 bullets, each of mass 0.006 kg. If the muzzle velocity of the rifle is 600 m/s and frictional forces between boat and water are neglected, what speed will be acquired by the boat?


Homework Equations


m1v1 = m2v2

The Attempt at a Solution


The first thing I realized is that the total momentum before the bullets are fired is zero and so I realized the momentum after the bullets are fired is zero. So m1v1 = m2v2. Then I found the total momentum of the bullets by multiplying the momentum of one bullet by 75 the total number of bullets. Then I divided that number by 1200 to find the final velocity of the boat. I got .135 m/s and the answer is -2.21 m/s
 
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1st you made a mistake with the sign since the boat and the bullet will go in opposite directions. The formula m1v1=m2v2 does not take that into consideration

2nd you made some calculation mistake somewhere. Hard to tell since you didn't post your calculation. ALWAYS POST YOUR CALCULATION IN FULL. DON"T JUST DESCRIBE YOUR CALCULATION. POST IT.
 
dauto said:
1st you made a mistake with the sign since the boat and the bullet will go in opposite directions. The formula m1v1=m2v2 does not take that into consideration

2nd you made some calculation mistake somewhere. Hard to tell since you didn't post your calculation. ALWAYS POST YOUR CALCULATION IN FULL. DON"T JUST DESCRIBE YOUR CALCULATION. POST IT.

Here is my calculations I did
(75)(.006)(600) = -(1200)(v)
270 = -1200v
v = -.135
 
You are given the boat's and man's weight. How does weight relate to momentum?

It would also be better to explicitly state a simplifying assumption you have made.
 
BrainMan said:
Here is my calculations I did
(75)(.006)(600) = -(1200)(v)
270 = -1200v
v = -.135

What is 270/1200? it is not 0.135...And you have to work with masses. 1200 is the weight of the boat.



ehild
 
Last edited:
Ok thanks I see my mistakes. Thanks!
 

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