Finding force without acceleration

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The discussion revolves around calculating the average force exerted on a door by bullets fired from a machine gun. For the first scenario, where bullets embed in a wooden door, the average force calculated is 229 N, emphasizing the importance of considering impulse and momentum change. In the second scenario, where bullets rebound elastically from a steel door, the average force remains the same as the first scenario since the mass and final velocity of the bullets do not change. The initial misconception that the door experiences no force due to lack of horizontal acceleration is corrected by understanding the role of impulse. The conversation highlights the significance of applying physics principles correctly to solve real-world problems.
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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