Velocity of .30 Cal Rifle After Firing a Bullet

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Homework Help Overview

The discussion revolves around a physics problem involving the conservation of momentum in a scenario where a .30 caliber rifle fires a bullet at a block of wood, both mounted on separate carts on a frictionless air track. Participants explore the implications of the setup on the velocities of the rifle and the bullet.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum, questioning the assumptions regarding the mass of the cart and the treatment of the rifle and cart as a single system. There are considerations about the relevance of the block of wood in the problem.

Discussion Status

The discussion is active with various interpretations being explored. Some participants suggest that assuming the cart is massless may be necessary for the problem to be solvable, while others clarify the momentum relationships involved in the scenario.

Contextual Notes

There is uncertainty regarding the mass of the cart and its impact on the momentum calculations. The relevance of the block of wood to the overall problem is also questioned, indicating potential missing information or assumptions that need to be clarified.

StotleD
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Imagine I were to shoots a .30 caliber rifle at a 0.56 kg block of wood. The rifle and wood are mounted on separate carts that sit atop an air track (like a linear air hockey table--ie. frictionless). The 6.4 kg rifle fires a 33 gram bullet at 219 m/s in the positive direction. What would be the velocity of the rifle after the bullet is fired?
 
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It seems like you need to apply conservation of momentum. In order to do that you would need to know the mass of the cart to which the rifle is attatched. Can it be assumed to be massless?
 
Since the air track is frictionless, and every force has an equal and opposite force, the momentum of the rifle should equal the momentum of the bullet, yes?

PL
 
But if the rifle is attached to the cart, much of the rifle's momentum will be transferred to it. They are being treated as one rigid body, I think.
 
Yes, conservation of momentum would be the easiest, and from the information i think you would have to assume the cart is massless, otherwise the question is incomplete. Also the fact that it was shot at a block of wood seems irrelevant, unless there is a part B...
 
The first part:
momentum before is equal to momentum after.
Mass of rifle x velocity of rifle = mass of bullet x velocity of bullet

I know the mass of the rifle and bullet and the velocity of the bullet,
so no I have to find the mass of the rifle.

The second part: (this is an inelastic collision-hit or 'stick collision'.)
Mass of bullet x velocity of bullet = total mass of block and bullet x
velocity of bullet and wood together.

I would have to solve for the velocity of the bullet and wood together.
 
Okay, so do you understand how to do both parts now? You have all the information you need. Solve for the velocity of the rifle in the first one (as it seems the cart is massless), and solve for the velocity of the bullet and wood in the second one.
 
Yes I got it. Thanks all!
 

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