Find the average force required to hold the gun in position

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SUMMARY

The average force required to hold a gun in position while firing 6 bullets per second, each with a mass of 3 grams and a speed of 500 m/s, can be calculated using the impulse equation: F · Δt = m · Δv. The momentum of a single bullet is determined by multiplying its mass (0.003 kg) by its velocity (500 m/s), resulting in a momentum of 1.5 kg·m/s. The total momentum for 6 bullets is 9 kg·m/s, and since this occurs over one second, the average force is 9 Newtons.

PREREQUISITES
  • Understanding of the impulse-momentum theorem
  • Basic knowledge of unit conversion (grams to kilograms)
  • Familiarity with Newton's laws of motion
  • Ability to perform calculations involving mass and velocity
NEXT STEPS
  • Study the impulse-momentum theorem in detail
  • Learn about unit conversions in physics, specifically mass and force
  • Explore Newton's second law of motion and its applications
  • Investigate real-world applications of momentum in ballistics
USEFUL FOR

Physics students, engineers, and anyone interested in the mechanics of firearms and projectile motion will benefit from this discussion.

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.
 
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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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