How Much Kinetic Energy is Lost in a Completely Inelastic Collision?

[~angel~]

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.

 

[HallsofIvy]

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

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

Set up your coordinate system so that particle A is moving in the positive direction along the x-axis (so its momentum is <2p, o>) and particle 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 <v_x, v_y>. Since their total mass is now 3m we have <2p, p>= <3mv_x,3mv_y> so we must have v_x= 2p/(3m) and v_y= 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.

 

[~angel~]

Ok...thanks for your help, but that answer is incorrect.

 

[~angel~]

Does anyone else know. Thank you.

 

[~angel~]

I just keep on ending up with the same answer.

 

[dark_angel]

I just keep on ending up with the same answer.

hmmm why did i get 3p^2/2m

 

[~angel~]

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.

 

[jdstokes]

Okay, we have established that the initial kinetic energy is

$K_1 = \frac{9p^2}{4m}.$

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

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

Substituting into the kinetic energy equation you get

$K_2 = \frac{5p^2}{6m}$

and therefore

$K_\mathrm{lost} = \frac{17p^2}{12m}.$