Thursday, March 21, 2013

Weird Person Who used to blog at me

So, there was this nutcase who used to blog at me. And one day she posted this:

If i have an asteroid which has a kinetic energy of 2.6x10^20 joules and with a density of 2500kg m-3 and earth which has a kinetic energy of 1.2x10^33 joules using the conservation of momentum how can i work out the earth and asteroid colliding inelastically as a single value?? the velocity of the asteroid is 20km s-1, the asteroid density is 2500kgm-3 and the m of the earth is 6x10^24kg the answer will be looking at the system before and after impact

Anyway, it seems like a fun question:

First, I think the word "inelastic" is important. This means that the two will collide, but neither will bounce, and the final momentum will simply be the sum of the energies with a new mass. Also, you need to calculate the mass of the asteroid

    a) kinetic energy = mv^2
        momentum = mv
        density = m/V
        V = volume
         v = velocity
         m = mass

        kineticEnergyAsteroid = 2.6E20 [massAsteroid/velocityAsteroid^2] [in Joules]

        velocityAsteroid = 20 [ km s-1]

        massAsteroid = 2.6E20 * velocityAsteroid^2 [ in kg]

        kineticEnergyEarth = 1.2E33 [massEarth/velocityEarth^2] [in Joules]

        massEarth = 6E24 [kg]
        velocityEarth = sqrt(1.2E33 / massEarth) [k s-1]

       These are all the variables you need to solve the final equation:
        massAsteroid(velocityAsteroid) + massEarth(velocityEarth) = (massAsteroid + massEarth)(finalVelocity)

        You are solving for the finalVelocity. Although, you will also get the final momentum.

Oh, but did I forget to tell you? Being able to do this is completely worthless. Go fuck some rich dude.

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