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Forces of motion, newton 2nd law

  1. Oct 8, 2016 #1
    1. The problem statement, all variables and given/known data
    An 8.5 gram slug is fired East from a 0.357 magnum hand gun held 1.5 meters above the ground. The muzzle velocity (as the projectile leaves the gun) is 1400 ft/sec. The bullet is accelerated at a constant rate as it travels through the 10 cm long barrel. After leaving the barrel, the gun powder force drops to 0.

    2 a) [2] What force does the exploding powder exhaust apply to accelerate the slug?

    2 b) [2] What horizontal force is put on the shooters arm by the gun? Sketch a graph of this force as a function of time. (assume the gun is held fixed)

    2. Relevant equations

    u = initial velocity

    v = u + at
    v2 = u2 + 2ad
    F = ma

    3. The attempt at a solution

    a)

    Converting to SI units:
    1400 ft/sec x 12in/ft = 16800 in/s
    16800 in/s x 0.0254m/in = 426.72 m/s
    8.5 grams = 0.0085 kg
    10 cm = 0.1 m

    assuming u = 0 m/s

    426.722 = 0 + 2(0.1)(a)
    a = 910449.7920 m/s2 (what the hell?)

    F = ma
    F = 0.085(910449.7920)
    = 7739 N

    b) Due to Newton's third law, the force exerted by the gun (7739 N) should be the same as on the man's arm.

    My values are very odd, so I feel like what I've done is incorrect however I can't think of how this is wrong. Thanks for your help.
     
  2. jcsd
  3. Oct 8, 2016 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Your answers are correct.

    Uniform acceleration has to be assumed, so your statement: ##\Delta v = a\Delta t## would be correct.

    You should explain where get your equation v2 = u2 + 2ad. This can be derived from the change in kinetic energy or from figuring out the time of travel from the average velocity, and working out the acceleration from ##\Delta v = a\Delta t##

    It takes a very large force to send that bullet out of the barrel at 426 m/s while accelerating for less than .0005 sec. It is easier to see if you take an energy approach: the bullet muzzle energy is .5mv2 = 774 J = force x distance. The distance is .1 m, so the force must be 774/.1 = 7740 N. F = ma so a = F/m = 7740/.0085 = 910449 m.sec-2

    AM
     
  4. Oct 8, 2016 #3
    Ah, I see. That makes a lot of sense when I think about it in terms of an energy perspective. Furthermore, the third and fourth parts of this question are

    c) How far away does the bullet travel (its range) before it hits the ground.
    d) At what angle would such a projectile need to be going initially in order to double the distance it travels before impact, the range.

    Not sure if I have to start a new thread for this since it's apart of the same question but for part c), I wrote:

    u = 426.72 m/s
    v = 0 m/s
    a = -9.8 m/s

    v2 = u2 + 2ad
    426.72^2 = 0 + 2(9.8)(d)

    d = 9290.3 m

    The issue I have with what I did here is that I inputted acceleration in the y direction and velocities in the x direction into the equation to solve for the displacement. I don't think I can do that, can I? If I can't, how would I approach this? Thanks a lot.
     
  5. Oct 8, 2016 #4

    Andrew Mason

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    Homework Helper

    It would take the bullet travelling at 426 m/sec over 20 seconds to travel 9200 m. Does an object take that long to fall 1.5 m?

    The problem with using a formula without going through the important exercise of explaining why you are using it becomes evident. This is the formula relating muzzle speed to acceleration during its transit down the barrel and the length of the barrel.

    You have to work out the time that it would take for the bullet to fall 1.5 m using h = ½at2. Then use d = vt to determine the range. Of course, you have to assume that the bullet does not slow down at all, which is not true at all, so the actual range is quite a bit less than that.
    For part d), let the range be 2R (where R is the range you found in the first part). If you think about it, what you want is the bullet to prescribe a parabolic arc, going up some distance and then back to 1.5 m. above the ground for the first part of the journey and then drop the remaining 1.5 m over the second part. Where would you aim the rifle in order to do that? You have to keep in mind that for the second part it is already dropping at some speed. So it takes less time to fall that last 1.5 m. than it did in part c).You can also find this angle by breaking the problem down this way and using the range formula for projectiles.

    AM
     
    Last edited: Oct 9, 2016
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