Another conservation of momentum problem

In summary, the conversation discusses the problem of determining the final speed of a boat and hunter after 75 bullets are fired from a rifle with a muzzle velocity of 600 m/s. The total weight of the boat and hunter is 1200 N and frictional forces are neglected. The equation m1v1 = m2v2 is used to find the final velocity of the boat, with the assumption that the initial and final momenta are equal. After correcting for the direction of the boat and the mass of the bullets, the final velocity is found to be -0.135 m/s. The conversation also suggests explicitly stating any simplifying assumptions made in the solution.
  • #1
BrainMan
279
2

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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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:
  • #6
Ok thanks I see my mistakes. Thanks!
 

FAQ: Another conservation of momentum problem

1. What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time, unless acted upon by an external force.

2. How is the conservation of momentum related to Newton's laws of motion?

The conservation of momentum is an extension of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. This means that when two objects interact, their total momentum before and after the interaction must be equal.

3. Can momentum be lost or gained in a closed system?

No, according to the conservation of momentum, the total momentum of a closed system must remain constant. This means that momentum cannot be lost or gained, but can only be transferred between objects within the system.

4. How does the conservation of momentum apply to collisions?

In a collision between two objects, the total momentum of the system before the collision must be equal to the total momentum after the collision. This principle is used to calculate the velocities of the objects after the collision.

5. Are there any real-life applications of the conservation of momentum?

Yes, the conservation of momentum is used in various fields such as engineering, sports, and transportation. For example, it is used in designing car airbags and calculating the trajectory of a golf ball after being hit by a club.

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