Horizontal Recoil Speed of gun

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SUMMARY

The discussion focuses on calculating the horizontal recoil speed of a 1200kg gun firing an 8kg projectile at a muzzle velocity of 600m/s at a 30-degree angle. The key principle involved is the conservation of momentum, which states that the total momentum before and after the firing event remains constant in a closed system. Participants emphasize the importance of considering the correct system, including the gun, bullet, and Earth, to accurately apply momentum conservation laws. Misunderstandings about momentum and impulse are common among learners, highlighting the need for clear explanations of these concepts.

PREREQUISITES
  • Understanding of conservation of momentum
  • Basic knowledge of projectile motion
  • Familiarity with kinetic energy equations
  • Ability to analyze systems involving external forces
NEXT STEPS
  • Study the conservation of momentum in closed systems
  • Learn about impulse and its relationship to momentum
  • Explore projectile motion calculations at various angles
  • Review kinetic energy equations and their applications
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in understanding the principles of momentum and recoil in projectile systems.

Chica1975
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Homework Statement


a 1200kg gun mounted on wheels shoots an 8kg projectile with a muzzle velocity of 600m/s at an angle of 30degrees above the horizontal. find the horizontal recoil speed of the gun


Homework Equations



KE= .5mvfE2 - .5MViE2

The Attempt at a Solution


to be honest I have read this several times and really do not understand what to do please help me.
 
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What is happening is that the projectile is being shot out of the gun, so the impact of the projectile leaving is causing the recoil.

What principle do you know involves collisions and impacts?
 
elasticity?
 
or perhaps momentum?
 
Chica1975 said:
or perhaps momentum?

Right and in any sort of collision momentum is conserved. However, you have the gun at an angle, so you need to apply the conservation law in both horizontal and vertical directions.
 
can you please explain the conservation law to me - I have read the book several times but our language is not attuned.

In a simple way please tell me what it means. thanks

I have tried everything I can think of with momentum but I don't understand about the law of conservation of momentum and impulse - I think this is my problem.
 
Chica1975 said:
can you please explain the conservation law to me - I have read the book several times but our language is not attuned.

In a simple way please tell me what it means. thanks

I have tried everything I can think of with momentum but I don't understand about the law of conservation of momentum and impulse - I think this is my problem.

Essentially it is that the total momentum in a closed system is zero.

or in easier terms

the momentum before the event = momentum after the event
(in a given direction).
 
Remember that momentum is only conserved when no external force acts on the system.
momentum along X is anyway conserved
But for Y, if you take gun and bullet as system then its not conserved
its conserved only when you system is GUN+BULLET+EARTH

well you'll say that it doesn't matter what i consider as system as momentum is conserved anyway but in more complex problems especially when friction is involved, not considering right system makes problems difficult and maybe also lead to wrong answer!
 

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