How Does Holding an Air Rifle Affect Recoil Speed?

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Homework Help Overview

The discussion revolves around a physics problem involving the recoil speed of an air rifle when fired, considering the mass of the rifle, the bullet, and the person holding it. The subject area includes concepts of momentum and recoil in mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the conservation of momentum and question how to set up the equations for the recoil speed. There are discussions about the correct mass to use for the system and the implications of initial velocities.

Discussion Status

Participants are actively engaging with the problem, raising questions about the setup of the equations and the assumptions regarding initial velocities. Some guidance has been offered regarding the conservation of momentum, but there is still uncertainty about specific terms and values to include in the calculations.

Contextual Notes

There is a mention of confusion regarding the initial velocity of the system and how to account for the combined mass of the girl and the rifle when calculating recoil. The discussion reflects a lack of clarity on certain aspects of the problem setup.

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Kim holds a 2.0 kg air rifle loosely and fires a bullet of mass 1.0 g. The muzzle velocity of the bullet is 150 m/s. What is the recoil spped of the rifle? The next shot the girl holds the rifle tightly against her body. Calculate the new recoil speed assuming the girl has a mass of 48kg.

I'm really not sure how to do this problem, any help would be appreciated.
Thanks
 
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Well you know that momentum will play a part in this question, so will momentum be conserved as the bullet leaves the gun?
 
Yes so is this the right way to set up the equation for the second part:
(50)(v)=(.001)(150)
v=-.003 m/s - v is negative because moving backwards

I'm not sure if 50 kg is the right mass though. I took it as the girl and the gun becoming one system so their masses are added.
 
dlilpyro said:
Yes so is this the right way to set up the equation for the second part:
(50)(v)=(.001)(150)
v=-.003 m/s - v is negative because moving backwards

I'm not sure if 50 kg is the right mass though. I took it as the girl and the gun becoming one system so their masses are added.


Her initial mass would be 48+2+0.001 since she is holding the gun with the bullet in it. However, what is her initial velocity if she is not moving?

After the bullet is fired, you are right with the 0.001(150) but remember she will move backwards with some velocity. So you need to have an additional term on the right side.
 
I'm not sure what the additional term is because we didn't have time to get the equation during class today. But if the initial velocity is zero how can you solve for v final on the left side?
 
dlilpyro said:
I'm not sure what the additional term is because we didn't have time to get the equation during class today. But if the initial velocity is zero how can you solve for v final on the left side?

The v on the left side will be zero since the girl is not moving.

There will be a different V on the right side since the bullet will move forward and she will move backward due to the recoil.

on the right side you will need the moment of bullet (which you have already) + momentum of girl (which will have the V to solve).
 
Alright I think I have it now. Thanks for all the help.
 

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