Problem on conservation of energy/momentum

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In summary: I think I see what you mean.I'm thinking you're trying to break the vector equation into componentsp1 + p2 = (m1v1 + m2v2) + (m2v2cosA)i + (m2v2sinA)jThe way the book is put it, it seems like they are making it two vectors with the same line of action i.e. the two vectors are parallel. So it should be:p = m1v1 + m2v2 and then I guess it should be something like this => p = sqrt ((m1v1)^2 + (m2v2)^2 + 2m1v1m2v2
  • #1
siddharthmishra19
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Homework Statement



A particle breaks into two pieces m1 and m2 traveling with velocities v1 and v2 respectively. The total kinetic energy of the particles is E. What is the velocity of each particle?

Homework Equations



Law of conservation of energy
Law of conservation of momentum

The Attempt at a Solution



Obviously as total kinetic energy is E,

[tex] m_1v_1^2/2 + m_2v_2^2/2 = E [/tex]

I don't know if I'm allowed to use conservation of momentum because initial momentum is 0 and suddenly it is m1v1 + m2v2...
 
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  • #2
The problem only makes sense if the original (non split) particle has a nonzero momentum. I think you are supposed to solve for the original velocity.
 
  • #3
The fact that the initial momentum is 0 is not a problem-- that tells you they travel in opposite directions, right?
 
  • #4
interested_learner said:
The problem only makes sense if the original (non split) particle has a nonzero momentum. I think you are supposed to solve for the original velocity.

There is no original velocity, the particle is at rest.

Pseudo Statistic said:
The fact that the initial momentum is 0 is not a problem-- that tells you they travel in opposite directions, right?

Hmm... that's possible but only is the momentum of the the split particles is equal ... can anyone verify this?
 
  • #5
This problem makes more sense:

A body (mass M) is moving with velocity Vo. It breaks into two pieces, one of which (mass m) is traveling with speed V1 in the direction perpendicular to Vo. What is the speed of the other piece.

Mass of other piece is (M-m). The problem I am finding is I don't know the direction of the second piece so I can project it into suitable axis. Or is it not needed?

I tried solving using conservation of energy (kinetic) but obviously ended up with wrong answer because the fact that the speed of first piece is perpendicular to original direction is not used...
 
  • #6
siddharthmishra19 said:
Obviously as total kinetic energy is E,

[tex] m_1v_1^2/2 + m_2v_2^2/2 = E [/tex]
Good.

I don't know if I'm allowed to use conservation of momentum because initial momentum is 0 and suddenly it is m1v1 + m2v2...
Sure you can. (You must!) Realize that momentum is a vector: one of those velocities will be positive, the other negative.

siddharthmishra19 said:
There is no original velocity, the particle is at rest.
0 is a perfectly legitimate velocity!
 
  • #7
siddharthmishra19 said:
I tried solving using conservation of energy (kinetic) but obviously ended up with wrong answer because the fact that the speed of first piece is perpendicular to original direction is not used...
This is a conservation of momentum problem. Remember that momentum is a vector. Hint: Call the original direction to be along +x. What's the y-component of the total momentum?
 
  • #8
Thanks, Doc! I think I've got it... please verify!
P.S. This is my first attempt to use LaTeX... forgive if something goes wrong.

According to problem, total kinetic energy of the particles is E

[tex] m_1v_1^2/2 + m_2v_2^2/2 = E [/tex]

Initial momentum = O (system at rest)
New momentum = [tex] m_1v_1 - m_2v_2 [/tex]

Assuming the above is correct and momentum is conserved,

[tex] m_1v_1 = m_2v_2 [/tex]

Two equations with two unknowns!

Solving,

[tex] v_1 = sqrt(2Em_2/m_1(m_1+m2)) [/tex]

V2 is found the same way, and it will also be +ve. I'm assuming that is the modulus of speed, because in the book the other is -ve.
 
  • #9
About the other problem... original velocity is along x-axis so y component is 0.

Since after splitting momentum is conseved then along the y-axis the momentum of the second particle will cancel out the one of the first particle and will also be along the y-axis in the opposite dir. (perpendicular to original velocity). m1v1=m2v2.

Is that what you mean? If so then why can't it be that the velocity of the second particle is at some angle "A" from the vertical and m1v1 = m2v2cosA ?
 
  • #10
siddharthmishra19 said:
Solving,

[tex] v_1 = sqrt(2Em_2/m_1(m_1+m2)) [/tex]

V2 is found the same way, and it will also be +ve. I'm assuming that is the modulus of speed, because in the book the other is -ve.
Looks good to me! (I see you took v_1 to be the magnitude of the speed of mass 1 and v_2 to be the magnitude of the speed of mass 2; no problem.)

siddharthmishra19 said:
About the other problem... original velocity is along x-axis so y component is 0.
Good.

Since after splitting momentum is conseved then along the y-axis the momentum of the second particle will cancel out the one of the first particle and will also be along the y-axis in the opposite dir. (perpendicular to original velocity). m1v1=m2v2.
The y-component of the second particle's momentum must be equal and opposite to the y-component of of the first particle's momentum.

Now figure out the x-component of its momentum.

Is that what you mean? If so then why can't it be that the velocity of the second particle is at some angle "A" from the vertical and m1v1 = m2v2cosA ?
The momentum (and thus velocity) of the second particle will surely be at some angle. Figure it out by finding the components of its momentum.
 
  • #11
Yes but the angle is not give. Could that be a problem?
 
  • #12
Why is that a problem? If you solve for the x & y components of momentum, you can figure out the angle if you wanted to. (But why bother? All you really need to find is the magnitude.)
 
  • #13
Thanks doc, you're a lifesaver... another question I can't seem to grasp...

Two particles (mass m1 and m2) are moving on a horizontal plane with speeds v1 and v2. Angle between them is A. What is the total momentum.

Won't it just be m1v1 + m2v2cosA? In the book its give as sqrt((m1v1)^2 + (m2v2)^2 + 2m1v1m2v2cosA)... (or something of this form)... I just can't seem to grasp the concept.
 
  • #14
siddharthmishra19 said:
Won't it just be m1v1 + m2v2cosA?
Looks like you are taking the momentum of the first particle (say it's moving in the +x direction) and then adding just the x-component of the second particle's momentum. That just gives you the x-component of total momentum. What about the y-component of the second particle's momentum?
 
  • #15
I see too. Thanks!

I was really scratching my head about where the momentum came from.
 
Last edited:

1. What is the law of conservation of energy/momentum?

The law of conservation of energy/momentum states that energy/momentum cannot be created or destroyed, only transferred or transformed from one form to another. This means that the total amount of energy/momentum in a closed system remains constant over time.

2. Why is the conservation of energy/momentum important in science?

The conservation of energy/momentum is important because it helps us understand and predict the behavior of systems. It allows us to make accurate calculations and predictions about the motion and interactions of objects in our world.

3. Can the law of conservation of energy/momentum be violated?

No, the law of conservation of energy/momentum is a fundamental principle in physics and has been proven to hold true in countless experiments. While energy/momentum may appear to be lost or gained in certain situations, it is simply being transferred or transformed into a different form.

4. How does the conservation of energy/momentum apply to everyday life?

The conservation of energy/momentum applies to everyday life in many ways. For example, when you kick a soccer ball, the energy/momentum you put into the ball is transferred to make it move. In a car accident, the total momentum of the vehicles involved before and after the collision will remain the same.

5. What are some real-world applications of the conservation of energy/momentum?

The conservation of energy/momentum has countless real-world applications, including in the fields of engineering, mechanics, and thermodynamics. It is also essential in understanding the behavior of celestial bodies and the formation of the universe.

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