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Impulse-momentum gunshot

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    A guy standing on a frictionless surface has a mass of 69.5kg including the gun he is holding. He fires a 4.2g (0.0042 kg) round at 955 m/s.

    A.)Find the recoil velocity of the hunter if he fires horizontally.
    B.)Find the recoil velocity of the hunter if he fires 54 degrees above the horizontal?

    2. Relevant equations
    Impulse momentum theorem

    3. The attempt at a solution
    A.)Find the recoil velocity of the hunter if he fires horizontally.

    The other problems that I have are collision based. In this one, I simply don't understand how to set up the problem.

    I get that before the shot the total momentum of the hunter and bullet are zero, because neither has a velocity, and momentum - mv. After the shot, I get that:
    [tex]69.5(v_{hunter}) + 0.0042(v_{bullet})[/tex]

    So in total I have:
    [tex]0_{before shot} = 69.5(v_{hunter}) + 0.0042(955)[/tex]
    [tex]0 = 69.5(v_{hunter}) + 4.011[/tex]
    [tex]- 4.011 = 69.5(v_{hunter})[/tex]
    [tex]-\frac{4.011}{69.5} = (v_{hunter})[/tex]
    [tex]v_{hunter} = -0.0577[/tex]

    Am I even doing this remotely correct?

    For part B. I assume I take the horizontal component of the velocity of the bullet, but then does the hunter experience a recoil that is down and left, (if he is shooting right and up)?
  2. jcsd
  3. Oct 30, 2011 #2
    Yeah thats done right. Keep in mind that you're getting a negative sign because you assumed the velocities are in the same direction ( keeping both m1v1 and m2v2 positive).

    Drawing a diagram, you'll see that the hunter will go one way and the bullet the other. So one should be negative. Its all direction convention, but you got the magnitude correct.

    B.) I'm sure the question is asking for just the horizontal change in momentum. So yes, just take the horizontal component of v and do the same.
  4. Oct 30, 2011 #3
    Okay, so for part B.

    955cos(45) is the new velocity for the bullet which represents the x component of it's actual velocity. This gives me the equation:
    [tex]0_{before-shot}= 69.5(v) + (0.0042)955cos(45)[/tex]
    [tex]-69.5(v)= (0.0042)955cos(45)[/tex]
    [tex]v= \frac{(0.0042)955cos(45)}{-69.5}[/tex]
    [tex]v = -0.0408[/tex]

    Is that correct?

    Edit: No that is not correct! What am I doing wrong?
  5. Oct 30, 2011 #4
    A minor mistake. Take a look what values you're using, and what was given.

    But what made you think you did it wrong? 0.0408 is less than 0.0577 which is a reasonable result. It is wrong, but in the reasonable direction. If it had been more than your answer in a), that would of been obvious.
  6. Oct 30, 2011 #5
    Oh geez, cos(54), not 45.... (Thank you)

    I find that the velocity is = -0.0340 m/s now!

    Does that seem correct? For the first part I input my negative velocity (because the hunter was recoiling back) and it told me that it should be positive. How am I supposed to know if this should be input as positive or negative?
  7. Oct 30, 2011 #6


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    It doesn't matter. You can choose whichever sign convention (either "right is positive" or "left is positive") that you like, as long as you use it consistently.

    You set up the equation just fine. 0 = m1v1 + m2v2 is just fine -- a proper expression of conservation of momentum. The values of v1 and v2 will have intrinsic signs that tell you their directions.

    In your first post, you implicitly decided that the direction in which the bullet was fired was "positive" by entering in a value of v_bullet = +955 m/s. As a result, you got a negative result as the velocity of the hunter, since he must move in the opposite direction to the bullet.

    You could just as well have decided that the direction in which the bullet was going should be defined to be the negative direction, in which case you would have entered a negative velocity of -955 m/s for the bullet, and you would have obtained a positive velocity for the hunter. In this second case, nothing about the physics or the situation has changed. You have merely adopted the opposite sign convention from the first case.
  8. Oct 30, 2011 #7
    Thank you cepheid, I do understand the direction naming convention that I chose. Masteringphysics has it's own unique way of picking the opposite direction for positive x as I do though! I guess they really want the magnitude? The problem did not specify any direction.
  9. Oct 30, 2011 #8
    If I understand what you're saying, then:
    The negative sign is telling you your assumption is wrong.

    They way you wrote the equation: 0=m1v1+m2v2 is stating that the velocities v1 and v2 are in the same direction. As in, when he shoots the gun, he follows the bullet! This is what the equation says the way you wrote it. When it spits out a negative sign, it is saying errr! Your assumption is wrong, change direction 180 degrees.

    So if you draw your diagram with gun pointing to the right, you will see that the bullet v2 will be positive, but the hunter v1 will be negative. So you then write your equation 0=(-m1v1)+m2v2 which gives you m1v1=m2v2 and a positive v1 meaning your assumption (hunter v1 goes left (- x direction) is right!

    If I misunderstood your post, disregard all that.
  10. Oct 30, 2011 #9
    Thanks frozen,
    That makes sense to me. For example. it's intuitive that the velocity of the hunter will be backwards from the direction he shoots. If I pick a positive x to the right, then the velocity I return for his will be negative, and the opposite is true. Therefor, my solution was a negative velocity, but the online homework program wanted it as a positive, even though it never specified what direction you should call positive x. I think it's an oversight with the homework in this case. A frustrating oversight, since positive and negative velocities would have been correct for the problem.

    I get this problem now though. Thank you both for the help.
  11. Oct 30, 2011 #10


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    If MasteringPhysics wants a positive velocity for the answer, then I guess it chose the opposite sign convention from what you did (either that, or it just wanted the magnitude). I don't know what else to say about that.

    Maybe this is just semantics, but I wouldn't assume that the equation above was saying that a positive velocity is being assumed for both, and then when a negative sign comes out, it "tells you that your assumption was wrong."

    I would say the above equation 0 = m1v1 + m2v2 assumes nothing about the sign of v1 and v2, which are 1D scalar velocities that can be either positive or negative in value, depending on their direction (this is what I meant when I said that v1 and v2 have "intrinsic" signs). The fact that one turns out to be negative and the other positive is exactly what is expected, given that the two things are moving in opposite directions. Nothing has been done "wrong."
  12. Oct 30, 2011 #11
    Yeah I understand.. But it is what helped me bridge the gap, and it sometimes helps other people bridge the gap.
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