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Another conservation of momentum problem

  1. May 29, 2014 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    m1v1 = m2v2

    3. 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
  2. jcsd
  3. May 29, 2014 #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.
  4. May 29, 2014 #3
    Here is my calculations I did
    (75)(.006)(600) = -(1200)(v)
    270 = -1200v
    v = -.135
  5. May 29, 2014 #4


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    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.
  6. May 29, 2014 #5


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    Homework Helper

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

    Last edited: May 29, 2014
  7. May 30, 2014 #6
    Ok thanks I see my mistakes. Thanks!
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