Find the average force required to hold the gun in position

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Discussion Overview

The discussion revolves around calculating the average force required to hold a gun in position while it fires bullets at a specified rate and speed. The focus includes the application of impulse and momentum concepts in physics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the impulse equation, suggesting that the average force can be calculated using the relationship between force, time, mass, and change in velocity.
  • Another participant reiterates the impulse equation and emphasizes the need for consistent units when calculating force, recommending the conversion of mass to kilograms for force in Newtons.
  • It is noted that the momentum of a single bullet is the product of its mass and velocity, and the total momentum for six bullets is six times that of one bullet.
  • A participant mentions that calculating the impulse for one bullet over a shorter time interval yields the same result as calculating for six bullets over one second.
  • There is a light-hearted exchange regarding the ease of typing "thanks" versus "tnx" on mobile devices.

Areas of Agreement / Disagreement

Participants generally agree on the application of impulse and momentum concepts, but there is no consensus on the specific calculations or methods to be used, as some details remain unresolved.

Contextual Notes

There are limitations regarding unit conversions and the assumptions made about time intervals for impulse calculations that have not been fully addressed.

Joe_1234
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A gun fires 6 bullets per second into a target. The mass of each bullet is 3g & the speed of 500 m/s. Find the average force required to hold the gun in position.
 
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Impulse equation ...

$F \cdot \Delta t = m \cdot \Delta v$
 
skeeter said:
Impulse equation ...

$F \cdot \Delta t = m \cdot \Delta v$
Tnx
 
You titled this "momentum" so it looks like you already knew the basic idea. The momentum of a single bullet is "mass times velocity" and you are given both of those. Of course, the momentum of 6 bullets is 6 times the momentum of a single bullet. Since this all happens in one second, divide by one second to get the force.

Be careful to use consistent units. You are given the mass in grams and the speed in m per second. Either convert mass to kg to get the force in Newtons or convert the speed to cm per second to get the force in dynes. I recommend the former.
 
Last edited by a moderator:
HallsofIvy said:
You titled this "momentum" so it looks like you already knew the basic idea. The momentum of a single bullet is "mass times velocity" and you are given both of those. Of course, the momentum of 6 bullets is 6 times the momentum of a single bullet. Since this all happens in one second, divide by one second to get the force.

calculating the impulse for one bullet over $\Delta t = \frac{1}{6} \text{ sec}$ yields the same result as six bullets in $\Delta t = 1 \text{ sec}$
 
Joe_1234 said:
Tnx
Serious question - is it harder to type “thanks” or “tnx “ on your phone? On mine the latter will auto correct to something else so it’s actually harder.
 

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