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Question involving the conservation of momentum

1. Homework Statement
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2. Homework Equations
The conservation of momentum states that if there are no external forces acting on a system, then momentum is conserved (i.e., the momentum before an event is equal to the momentum after an event). (Note: We assume that the internal forces follow Newton's third law (i.e., forces are third law pairs) and that the sum of the internal forces is equal to zero.)

3. The Attempt at a Solution
If we are trying to solve for the speed of the shell relative to the ground, then we should consider only the shell moving while the gun is stationary. Therefore, the speed of the shell relative to the ground, and thus the momentum of the shell relative to the ground, will include a factor of the momentum of the gun. This is the approach that I think makes most sense. However, I am stuck. Can someone provide insight?
 

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Orodruin

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then we should consider only the shell moving while the gun is stationary.
No, this is incorrect. It would violate conservation of momentum. The momentum in the gun-bullet system after the firing must be the same as before since there are no forces from the ground. You need to dress these words in equations along with the requirement that the velocity of the bullet relative to the gun is v.
 

Andrew Mason

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1. Homework Statement

3. The Attempt at a Solution
If we are trying to solve for the speed of the shell relative to the ground, then we should consider only the shell moving while the gun is stationary. Therefore, the speed of the shell relative to the ground, and thus the momentum of the shell relative to the ground, will include a factor of the momentum of the gun. This is the approach that I think makes most sense. However, I am stuck. Can someone provide insight?
What is missing here is the fact that the gases expelled from the muzzle at high speed carry momentum. It looks like you are supposed to ignore that and assume that the momentum of the shell + momentum of the gun after the shell leaves the muzzle will sum to the momentum of the gun before firing, all relative to the ground (i.e. 0).

This must be a large artillery gun as it is firing shells (which explode). Normally artillery guns are fired up into the air so you are also supposed to ignore the force of the gun recoil on the earth.

(1) Set up the equation that relates the momentum of the shell to the momentum of the gun as the shell leaves the muzzle, assuming the momentum of the muzzle gases is negligible. (ie. it would be the same as if the shell was launched by a spring inside the gun).

(2) Then express the relative velocity of the shell to gun (vsg) in terms of the velocities of the gun relative to the earth (vge) and of the shell to the earth (vse)

(3) Substitute the value of (vge) in (2) into your equation in (1) and solve for vse

AM
 
Last edited:

CWatters

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What Andrew said..

The trick with this problem is that before firing the velocity of the gun and bullet are specified relative to the ground. Whereas after firing the velocity of the bullet has been specified relative to the gun not the ground. You just need to write your conservation of momentum equations relative to the ground and then adjust them for this difference.
 
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Alternatively you could "reverse time" the 'firing event' (explosion) which would yield a perfectly inelastic collision. Collision impulse is then μΔv where Δv is the relative velocity and μ is the reduced mass ##\frac{m_1m_2}{m_1+m_2}## of the colliding objects - which both come to rest relative to the ground. Now you know the change in momentum, just divide by mass of bullet.
 

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