Finding force without acceleration

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SUMMARY

The discussion centers on calculating the average force exerted on a door by bullets fired from a machine gun. The bullets, weighing 22 g and traveling at 260 m/s, impact a wooden door at a rate of 40 bullets per second. The average force exerted on the wooden door is calculated to be 229 N using the impulse-momentum theorem, which states that impulse (I) equals the change in momentum (Δmv) over the time interval (Δt). In the case of elastic collision with a steel door, the average force remains the same due to unchanged mass and final velocity of the bullets.

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  • Understanding of impulse-momentum theorem
  • Basic knowledge of Newton's second law (F = ma)
  • Familiarity with concepts of elastic and inelastic collisions
  • Ability to perform calculations involving mass, velocity, and force
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  • Learn about elastic and inelastic collisions in physics
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RedDanger
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Homework Statement


A machine gun fires 22 g bullets horizontally at 260 m/s at a constant rate of
40 bullets/s. (a) If the bullets embed themselves in a thick wooden door, what is the average force
exerted on the door? (b) If the bullets hit a steel door and rebound elastically, what is the average
force exerted on the door?


Homework Equations


F = ma
Vf = Vi + at


The Attempt at a Solution



The door shouldn't experience any force at all, because the bullets are not accelerating horizontally. Is this correct?
 
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RedDanger said:
The door shouldn't experience any force at all, because the bullets are not accelerating horizontally. Is this correct?
Not at all correct. The bullets are going at some velocity, and then they are stopped by the door right? Thus they have some acceleration.

You should think about this in terms of `impulse' (I)
<br /> I \equiv \Delta mv = F \Delta t<br />
The impulse, which is defined as a change in momentum, is the force applied times the time over which it is applied.
 
Okay, so for the first part I solved for F and got 229N. For the second part, the amount of force exerted on the door should be the same because neither the mass of the bullets nor their final velocity changes due to the type of collision. Is this correct?
 

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