Finding velocity and momentum of a bullet and recoiling gun

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SUMMARY

The discussion focuses on calculating the velocity and momentum of a bullet and the recoiling gun based on a given kinetic energy of 90 J. The bullet's final velocity is determined to be 134.16 m/s, resulting in a momentum of 1.3416 kg m/s. The recoiling gun, with a mass of 4 kg, has a calculated momentum of -1.3416 kg m/s and a kinetic energy of 0.225 J. The barrel length of 50 cm is deemed irrelevant for these calculations.

PREREQUISITES
  • Understanding of kinetic energy equations (KE = 1/2mv^2)
  • Knowledge of momentum calculations (p = mv)
  • Familiarity with conservation of momentum principles
  • Basic physics concepts related to motion and forces
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  • Explore the relationship between kinetic energy and work done in physics
  • Learn about significant figures and their importance in scientific calculations
  • Investigate the effects of barrel length on bullet velocity in firearms
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in the dynamics of projectile motion and recoil effects in firearms.

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Homework Statement


A bullet leaves the barrel of a gun with a kinetic energy of 90 J. The barrel of the gun is 50 cm long. The gun has a mass of 4 kg, the bullet 10 g. (a) Find the bullets final velocity. (b) Find the bullets final momentum. (c) Find the momentum of the recoiling gun. (d) Find the kinetic energy of the recoiling gun.

Homework Equations


KE = 1/2mv^2
p = mv

The Attempt at a Solution


(a) Since I know the kinetic energy of the bullet as it leaves the barrel, I attempted to find the velocity with this information.

90 J = (1/2)(.01 kg)v^2
which gives v = 134.16 m/s

(b) To find momentum of the bullet I plugged this value for v into p= mv, this gives the result p = 1.3416 kg m/s.

for (c) I used conservation of momentum, since the initial momentum of both the bullet and gun are zero.
0 = (1.3416 kg m/s) + (4 kg)v
solving this give v = -.3354 m/s

(d) KE =(1/2)mv^2 so the kinetic energy of the gun is KE = (1/2)(4 kg)(-.3354 m/s)^2 = .225 J

I'm not entirely sure if I solved this correctly. My main problem is that it gives the length of the barrel of the gun and I did not use this in the solution. Is this unnecessary information? or did I do this incorrectly? I thought that the 50 cm could be used to find the work done on the gun, but 50 cm is how far the bullet moved, not how far the gun moved.
 
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Your calculations look good. The barrel length is irrelevant as you have surmised. You might want to round your results to better reflect the number of significant figures provided in the given data.

In future, please retain the formatting template headings when you put together your post. I've re-inserted the headers for you this time.
 
Thank you very much! I planned to round my final results but just kept them that way in case I needed to use them in further calculations. Thanks again for the input!
 

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