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

In summary, after the bullet hits the rod, it has linear momentum and angular momentum. The angular momentum is the sum of the angular momenta of the bullet and the rod.
  • #1
Parallel
42
0

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
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
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?
 
  • #4
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]
 
  • #5
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
 
  • #6
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.
 
  • #7
thank you very much..(I finally got it..spent like 3 hours on this)

thanks again :)
 

1. What is a rigid body?

A rigid body is an object that does not deform or change shape when subjected to external forces. It is a simplified model used in physics to study the motion of objects.

2. How does a bullet and a rod behave as rigid bodies?

A bullet and a rod both behave as rigid bodies because they are solid and do not deform easily. This means that when they are subjected to external forces, they will maintain their shape and move as a single unit.

3. What is the difference between a bullet and a rod as rigid bodies?

The main difference between a bullet and a rod as rigid bodies is their shape. A bullet is a small, cylindrical object while a rod is a longer, thin object. This difference in shape can affect how they behave when subjected to external forces.

4. How do external forces affect the motion of a bullet and a rod as rigid bodies?

External forces such as gravity, air resistance, and contact forces can cause a bullet or a rod to change its direction or speed. However, as long as the forces are not strong enough to cause deformation, the bullet and rod will continue to move as rigid bodies.

5. What are some real-life examples of rigid bodies?

Some examples of rigid bodies in everyday life include cars, buildings, and furniture. These objects do not deform easily and can be modeled as rigid bodies in physics. Other examples include a spinning top, a spinning coin, and a spinning wheel.

Similar threads

  • Introductory Physics Homework Help
10
Replies
335
Views
7K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
821
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
821
  • Introductory Physics Homework Help
Replies
2
Views
415
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
2
Replies
62
Views
9K
Back
Top