Calculating Linear Momentum of a System with a Bullet and a Rod Collision

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SUMMARY

The discussion focuses on calculating the linear momentum of a system consisting of a bullet and a rod after a collision. The initial angular momentum is given by the equation mva, while the moment of inertia of the system is calculated as I = (1/3)*M*L^2 + m*(a^2). The correct approach involves finding the angular velocity (ω) after the collision and using the center of mass (CM) to determine the total linear momentum, expressed as P = (m+M)*V_cm. The key takeaway is that the velocity of the end of the rod does not represent the velocity of the entire system, necessitating the use of the center of mass for accurate calculations.

PREREQUISITES
  • Understanding of linear momentum and angular momentum concepts
  • Familiarity with moment of inertia calculations, specifically for rods
  • Knowledge of center of mass and its significance in momentum calculations
  • Basic proficiency in physics equations related to collisions
NEXT STEPS
  • Study the concept of center of mass in multi-body systems
  • Learn how to calculate angular momentum before and after collisions
  • Explore the relationship between angular velocity (ω) and linear velocity (V)
  • Investigate the application of conservation laws in collision problems
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Students studying physics, particularly those focusing on mechanics, as well as educators and professionals involved in teaching or applying concepts of momentum and collisions in real-world scenarios.

Parallel
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Homework Statement



a rod with mass 'M' and length 'L' is pivoted about a frictionless axle through it's end .a bullet with mass 'm' and speed 'v' is shot and sticks to the rod a distance 'a' from the axle.

I need to find the LINEAR momentum of the system just after the hit.

Homework Equations



moment of inertia of a rod about it's center of mass: I=(1/12)*M*L^2
L = I*(omega)
p = (m+M)*V

The Attempt at a Solution



the intial angular momentum is: mva (about the pivot)
angular momentum after the hit is just: I*omega
where I = (1\3)*M*L^2 + m*(a^2)

equating those two gives omega,by the relation V=omega*L
we can get the speed V,so P = (m+M)*V

and this is not the answer

thanks for the help
 
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The velocity you calculated is not the velocity you need. Your equation for P should give you a hint about the velocity you do need.
 
I really don't see,how the equation for P should give me a hint,can you be more specific?

can you tell me what is the velocity I calculated?
 
Parallel said:
I really don't see,how the equation for P should give me a hint,can you be more specific?

can you tell me what is the velocity I calculated?

You calculated the velocity of the end of the rod. The only thing moving with that velocity is the tiny bit of mass at that end. Everything else is moving slower.

The linear momentum of a system of particles is the sum of the linear momenta of each bit of mass in the system. The concept of the center of mass is important because you can find this total momentum by finding the velocity of the center of mass and treating all the mass as if it had that same velocity [P = (m+M)*V_cm]
 
I tried to work in the center of mass frame,(although it's messy),but how can I find V_cm?..should I just compute angular momentum before the hit and after the hit,in the CM frame,and equate them?

thanks for your help
 
Parallel said:
I tried to work in the center of mass frame,(although it's messy),but how can I find V_cm?..should I just compute angular momentum before the hit and after the hit,in the CM frame,and equate them?

thanks for your help

Your initial idea for finding the angular velocity of the system after the collision is correct. Once you have found ω you have two ways to approach the solution. You can either find the center of mass of the whole system and use the distance from the pivot to the CM with ω to find the velocity, or you can find the linear momentum of the bullet using its distance from the pivot (a), ω, and m, and combine that with the linear momentum of the CM of the rod to get the total linear momentum of the system.
 
thank you very much..(I finally got it..spent like 3 hours on this)

thanks again :)
 

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