Relating Momentum and Kinetic energy in collision

1. Oct 20, 2007

gills

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

Two particles move perpendicular to each other until they collide. Particle 1 has mass m and momentum of magnitude 2p , and particle 2 has mass 2m and momentum of magnitude p . Note: Magnitudes are not drawn to scale in any of the figures.

Suppose that after the collision, the particles "trade" their momenta, as shown in the figure. That is, particle 1 now has magnitude of momentum p , and particle 2 has magnitude of momentum 2p ; furthermore, each particle is now moving in the direction in which the other had been moving. How much kinetic energy, Klost , is lost in the collision?

2. Relevant equations

$$\sum$$ m_i * v_i

?? K = Kcenter mass + Kinternal???

3. The attempt at a solution

I'm having trouble relating the momentums and kinetic energies. If someone can provide a little guidance, that would be great.

Since that collision seems to be elastic, i'm assuming that the kinetic energy of the system is being conserved.

Last edited: Oct 20, 2007
2. Oct 20, 2007

Staff: Mentor

Figure out the change in speed of each mass. For example: If the 1m mass has initial speed of v (and initial momentum mv), what's its final speed if its momentum is now 2mv?

Then just crank out the kinetic energies of each mass before and after the collision and find the difference.

(Be sure to express your final answers in terms of the given data: the momentum p and mass m.)

Alternatively, just express the KE of a particle in term of its momentum and mass. Then you can compare the KEs directly.

I don't know why you think it seems to be elastic, but it's not. (If it were elastic, there would be no change in KE. Calculate it and see for yourself!)

Last edited: Oct 20, 2007
3. Oct 20, 2007

Kushal

the Ek lost is equal to the Ek initially minus the Ek final.

get an equation from this, and substitute the speeds in this equation by m and p.

e.g. 2p = m*u1

u1 = 2p/m

where u1 is the initial speed of mass m.

4. Oct 21, 2007

gills

Ok if i relate the given information to the initial and final velocities and then use:

$$\Delta$$K = Kf-Ki or should i use:

Ki-Kf

The latter will give a negative sense i'm assuming.

Here's my processes, but i'm not getting the right answer.

Initial velocities:
1)ball of mass m ---> 2p = mv ----> v = 2p/m

2)ball of mass 2m ---> p = mv ----> v = 2m/p

after collision

Final velocites:
1) ball of mass m ---> p = mv ---> v = p/m

2) ball of mass 2m ---> 2p = 2mv ---> v = 2p/2m ---> v = p/m

Using conservation of Kinetic energy theorem:

(1/2)m*(2p/m)^2 + (1/2)2m*(2m/p)^2 = (1/2)m*(p/m)^2 + (1/2)2m(p/m)^2 --->

[(1/2)m*(2p/m)^2 + (1/2)2m*(2m/p)^2] - (1/2)m*(p/m)^2 + (1/2)2m(p/m)^2 = $$\Delta$$K

Simplified=

(2p^2)/m + (4m^3)/p^2 - (3p^2)/2m

....am i on the right track?

5. Oct 21, 2007

Staff: Mentor

OK.

You flipped this one around; it should be: p = (2m)v ---> v = p/2m

OK.

Yes, you're on the right track.

6. Oct 21, 2007

gills

DOH!! no wonder why it's wrong.

So with the correction i'm getting:

2p^2/m + p^2/4m - [(p^2/2m) + (p^2/m)] -->

9p^2/4m - 3p^2/2m -->

3p^2/4m

how's that look? I'm running out of attempts on my web assignment ;)

7. Oct 21, 2007

Staff: Mentor

Looks good to me.

8. Oct 21, 2007

gills

It was right, thank you!

9. Oct 21, 2007

gills

in the next question it asks how much K is lost if the collision were completely inelastic.

Just by messing around with it a little, i'm getting that it'll still be:

3p^2/4m

Is that correct?

10. Oct 21, 2007

gills

no, it's not correct. Damn, any advice?

11. Oct 21, 2007

Staff: Mentor

It better not be! A completely inelastic collision is one that loses the most KE.

What's the final speed after the collision and how did you figure it out?

12. Oct 21, 2007

gills

well a completely inelastic collision =

m1v1 + m2v2 = (m1 + m2)*Vf ---> in this case --->

m1v1 = 2p and m2v2 = p --> right or wrong? --> therefore -->

2p + p = (m + 2m)*Vf -->

3p = 3m*Vf ---> Vf = 3p/3m --> Vf = p/m --> which is the same as before. And plugged into the K_final part -->

(1/2)(m+2m)*(p/m)^2 ---> which turns out to be the same exact expression as before in K_f which =

(3p^2)/2m

What am i assuming incorrectly?

Last edited: Oct 21, 2007
13. Oct 21, 2007

Staff: Mentor

Your error is assuming that you can just add the momenta as if they were just numbers: 2p + p = 3p. Nope!

The momenta are vectors and you need to add them like vectors. Find the resultant total momentum and use that to figure out the final speed.

14. Oct 21, 2007

gills

i'm always forgetting little things like this!

2p_i + p_j --->therefore-->

|p| = $$\sqrt{(2p^2) + p^2}$$ = $$\sqrt{5}$$p --->

$$\sqrt{5}$$p = (3m)* Vf -->

Vf = $$\sqrt{5}$$p/(3m) ?

15. Oct 21, 2007

Staff: Mentor

Good!

16. Oct 21, 2007

gills

you rock! it was right, thanks again