# Problem on conservation of energy/momentum

1. Apr 21, 2007

### siddharthmishra19

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Law of conservation of energy
Law of conservation of momentum

3. The attempt at a solution

Obviously as total kinetic energy is E,

$$m_1v_1^2/2 + m_2v_2^2/2 = E$$

I don't know if i'm allowed to use conservation of momentum because initial momentum is 0 and suddenly it is m1v1 + m2v2...

2. Apr 21, 2007

### interested_learner

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. Apr 21, 2007

### Pseudo Statistic

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

4. Apr 21, 2007

### siddharthmishra19

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

Hmm... thats possible but only is the momentum of the the split particles is equal ... can anyone verify this?

5. Apr 22, 2007

### siddharthmishra19

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. Apr 22, 2007

### Staff: Mentor

Good.

Sure you can. (You must!) Realize that momentum is a vector: one of those velocities will be positive, the other negative.

0 is a perfectly legitimate velocity!

7. Apr 22, 2007

### Staff: Mentor

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. Apr 22, 2007

### siddharthmishra19

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

$$m_1v_1^2/2 + m_2v_2^2/2 = E$$

Initial momentum = O (system at rest)
New momentum = $$m_1v_1 - m_2v_2$$

Assuming the above is correct and momentum is conserved,

$$m_1v_1 = m_2v_2$$

Two equations with two unknowns!!

Solving,

$$v_1 = sqrt(2Em_2/m_1(m_1+m2))$$

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. Apr 22, 2007

### siddharthmishra19

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. Apr 22, 2007

### Staff: Mentor

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.)

Good.

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.

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. Apr 22, 2007

### siddharthmishra19

Yes but the angle is not give. Could that be a problem?

12. Apr 22, 2007

### Staff: Mentor

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. Apr 22, 2007

### siddharthmishra19

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. Apr 22, 2007

### Staff: Mentor

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. Apr 23, 2007

### interested_learner

I see too. Thanks!

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

Last edited: Apr 23, 2007