Newton's 3rd law question -- A gun firing a bullet

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In the discussion about Newton's 3rd law related to a gun firing a bullet, the force exerted by the gun on the bullet is 250N directed to the left. According to Newton's 3rd law, the bullet exerts an equal and opposite force of 250N back onto the gun, directed to the right. This interaction illustrates the principle that forces between two bodies are equal in magnitude and opposite in direction. Although it may seem surprising that a small bullet can exert a force on the gun, this is a fundamental aspect of Newton's laws. Understanding this interaction is crucial for grasping the dynamics of the system involved.
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Homework Statement
A 10.0 kg gun applies a force of 250 N left on a 0.0200 kg bullet. What is the force on the gun?
Relevant Equations
F=ma
I'm confused by which forces should be equal to each other and what I solve for. How should I approach this problem?
 
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well the data of the problem (we can ignore the given masses for the purpose of answering the question posed by the problem) can be summarized as follows:

The force from the gun to the bullet is equal to 250N and towards the left.

Using Newton's 3rd law you should be able to complete the following sentence
The force from the bullet to the gun is equal to ... and towards the ...
 
You basically have two bodies (the gun and the bullet) which interact. The gun applies a force to the bullet, hence by Newtons 3rd law, the bullet applies an equal and opposite force to the gun.

I know it might seem counter intuitive to you that the tiny bullet applies a force to the gun, but that is what Newton's 3rd law tell us and this is pretty much what is happening in reality if we omit some details (the details are that the gun and the bullet interact through the gas mixture produced by the ignited gunpowder which has a Pressure and a Temperature, but we can omit these details for now).
 
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