Machine gun conservation of momentum problem

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  • #1
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Homework Statement


A machine gun held by a soldier fires bullets at the rate of three per second. Each bullet has a mass of 30 g and a speed of 1200 m/s. Find the average force exerted on the soldier.


Homework Equations


FΔt = mgf-mgi


The Attempt at a Solution


I attempted to do this by plugging it into the above equation 1200(.03)/ 3 = 12 to find the average force. The actual answer is 108. I know that the 3 should actually be multiplied to find the force but I am not sure why. Wouldn't that be finding the average momentum per second?
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


A machine gun held by a soldier fires bullets at the rate of three per second. Each bullet has a mass of 30 g and a speed of 1200 m/s. Find the average force exerted on the soldier.


Homework Equations


FΔt = mgf-mgi


The Attempt at a Solution


I attempted to do this by plugging it into the above equation 1200(.03)/ 3 = 12 to find the average force. The actual answer is 108. I know that the 3 should actually be multiplied to find the force but I am not sure why. Wouldn't that be finding the average momentum per second?

Isn't the impulse equation:

FΔt = mΔv ?

What Δt are you using?

Remember, the cyclic rate of the gun is 3 rounds per second.
 
  • #3
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Isn't the impulse equation:

FΔt = mΔv ?

What Δt are you using?

Remember, the cyclic rate of the gun is 3 rounds per second.

Yes it is the title was a typo. I am using 3 as the change in time.
 
  • #6
Nathanael
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because I thought that was the Δt.

You could make Δt 3 if you want to, but then would the "1200*0.3" be correct? (To me that implies Δt=1/3 since 0.3 is the mass of one bullet and it takes 1/3 of a second for 1 bullet to be shot.)

Edit:
What is the mass of all the bullets shot in Δt=3 seconds?
 
Last edited:
  • #7
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You could make Δt 3 if you want to, but then would the "1200*0.3" be correct? (To me that implies Δt=1/3 since 0.3 is the mass of one bullet and it takes 1/3 of a second for 1 bullet to be shot.)

Edit:
What is the mass of all the bullets shot in Δt=3 seconds?

OK I see the time would be 1/3 because it takes 1/3 of a second to fire a bullet therefore you have to divide by 1/3. Thanks!
 
  • #8
Nathanael
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OK I see the time would be 1/3 because it takes 1/3 of a second to fire a bullet therefore you have to divide by 1/3. Thanks!

Yes, correct.

But you could use any Δt you want. If Δt=3 then instead of 1200*0.03 you'll have 1200*0.27 (because 0.27 is the mass of all of the bullets shot in 3 seconds).

Then divide by 3 and you'll get the same answer.
 

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