Completely Inelastic collision

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  • #1
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Please help.

Two particles move perpendicular to each other until they collide. View Figure 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.

a. 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, K_lost, is lost in the collision?

b.Consider an alternative situation: This time the particles collide completely inelastically. How much kinetic energy K_lost is lost in this case?

I've already found the answer to part a, which is 3p^2/4m, and as I was trying to work out the answer to b, I ended up with the asme answer, and I followed all the "instructions" in my textbook. Please help.
 

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  • #2
HallsofIvy
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Your answer for part (a) was correct (although I had to calculate it twice to get the right answer!).

In part (b), the particals collide "completely inelastically" which means that the "stick together" and have the same after-collision speed.

Set up your coordinate system so that partical A is moving in the positive direction along the x-axis (so its momentum is <2p, o>) and partical B is moving in the negative direction along the y-axis (so its momentum is <0, p>). Before the collision, their total momentum is <2p, p>. After the collision, the move off together: let's call their velocity vector <vx, vy>. Since their total mass is now 3m we have <2p, p>= <3mvx,3mvy> so we must have vx= 2p/(3m) and vy= p/(3m). v2 is now (3p/3m)2+ (p/3m)2= (10p2)/(9m2) and the kinetic energy is
(1/2)(3m)(10p2)/9m2)= (150/81)p2/m.

Of course, the total initial kinetic energy was still (9/4)p2/m so the energy lost is (9/4- 150/81)p2/m= ((720-600)/324)p2/m = (30/81)p2/m.
 
  • #3
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Ok...thanks for your help, but that answer is incorrect.
 
  • #4
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Does anyone else know. Thank you.
 
  • #5
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I just keep on ending up with the same answer.
 
  • #6
~angel~ said:
I just keep on ending up with the same answer.



hmmm why did i get 3p^2/2m
 
  • #7
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That is what I got as the final KE for the second question, but subtracting it from the initial KE to get the KE lost gave me the same answer as the first queston.
 
  • #8
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Okay, we have established that the initial kinetic energy is

[itex]K_1 = \frac{9p^2}{4m}.[/itex]

The initial x-component of momentum is [itex]P_{1x} = 2p[/itex] and the y-component is [itex]P_{1y} = p[/itex]. After collision [itex]P_{2x} = 3mv_x[/itex] and [itex]P_{2y} = 3mv_y[/itex]. Conservation of momentum states that

[itex]
\begin{align*}
\vec{P}_1 & = \vec{P}_2 \\
|\vec{P}_1| & = |\vec{P}_2| \\
4p^2 + p^2 & = 9m^2v_x^2 + 9m^2v_y^2 \\
5p^2 & = 9mv^2 \\
v ^2 = \frac{5p^2}{9m^2}.
\end{align*}
[/itex]

Substituting into the kinetic energy equation you get

[itex]K_2 = \frac{5p^2}{6m}[/itex]

and therefore

[itex]K_\mathrm{lost} = \frac{17p^2}{12m}.[/itex]
 

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