Conservation of Momentum of masses

In summary, a 4.0kg mess kit sliding on a frictionless surface explodes into two 2.0kg pieces, with one moving at 3.0m/s due north and the other at 5.0m/s, 30 degrees north of east. To find the original speed of the mess kit, the problem can be approached by considering the two 2.0kg pieces colliding and sticking together. By breaking the momentum of one piece into x and y components and finding the total momentum of the resulting mass, the original speed of the mess kit can be calculated to be 3.5m/s. The two pieces do not have the same momentum due to their different angles of emission relative to the initial motion
  • #1
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Homework Statement


A 4.0kg mess kit, ##m_1## sliding on a frictionless surface explodes in two 2.0kg parts, ##m_2## and ##m_3##: ##m_2##@3.0m/s due north and##m_3##@ 5.0m/s, 30 degrees north of east. What is the original speed of the mess kit?


Homework Equations


Momentum=mv


The Attempt at a Solution


After thinking about it for a bit, I thought that I could think about the problem in reverse, i.e. the two 2kg masses colliding in sticking. I broke the momentum of ##m_3## into x and y components and got ##P_{3x}=8.66N*s## and ##P_{3y}=5N*s## ##m_2## has only a y component of 6N*s. So the total momentum in x, after the collision, would be 8.66N*s, and in y would be 11N*s. Using these two values I found the magnitude of the momentum of the resulting mass,##m_1##, to be 13.99N*s, divided by its mass, 4kg, and found its velocity to be 3.5 m/s.

And to verify that this seemed right, it would have been moving at 51 degrees below the negative x axis.

It all seems within reason, but I'm not sure.

This was an even numbered problem and again I'm unsure if what I'm doing is right. Does this look right? If not, could someone steer me in the right direction.
 
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  • #2
Your approach looks fine, and the result looks good as well.
 
  • #3
Awesome, thank you. I have a follow up question though.

How is it possible for a moving object to break into two equal mass pieces, but those pieces have different magnitudes for their momentum? I'm having a hard time visualizing that in my head.
 
  • #4
The two pieces are not emitted in a symmetric way relative to the initial motion of the object. As extreme example, one object could be emitted in the original direction of motion (->quicker than the initial object), and the other object could be emitted in the opposite direction (can be slower than the original object, or even be at rest afterwards).
 
  • #5
Oh ok, I was picturinf a literal explosion, I guess it could be 2 springs directed at different directions. I think I keep getting hung up on the idea kinectic energy not alway being conserved.
 
  • #6
Kinetic energy is not conserved in this process, right. The mass needs some other type of energy to "explode".
 

1. What is the law of conservation of momentum?

The law of conservation of momentum states that the total momentum of a closed system remains constant in the absence of external forces.

2. How is momentum defined?

Momentum is defined as the product of an object's mass and velocity.

3. How does the conservation of momentum apply to collisions?

In a collision, the total momentum before the collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system.

4. What is the difference between elastic and inelastic collisions in terms of momentum?

In an elastic collision, both kinetic energy and momentum are conserved, while in an inelastic collision, only momentum is conserved.

5. How does the conservation of momentum affect the motion of objects in space?

In the vacuum of space, where there are no external forces, the law of conservation of momentum dictates that the total momentum of all objects in the system will remain constant.

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