In outer space rock 1, with mass 3 kg and velocity < 3900, -2700, 3600 > m/s, st

Question

In outer space rock 1, with mass 3 kg and velocity < 3900, -2700, 3600 > m/s, struck rock 2, which was at rest. After the collision, rock 1's velocity is < 3500, -2000, 4100 > m/s.

Suppose the collision was elastic (that is, no change in kinetic energy and therefore no change in thermal or other internal energy of the rocks). In that case, after the collision, what is the kinetic energy of rock 2?

On the other hand, suppose that in the collision some of the kinetic energy is converted into thermal energy of the two rocks, where Ethermal,1 + Ethermal,2 = 2.52106 J. What is the final kinetic energy of rock 2?

In this case (some of the kinetic energy being converted to thermal energy), what was the transfer of energy Q (microscopic work) from the surroundings into the two-rock system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.)

Explanation / Answer

This question really isn't as bad as it may look so rock1 has m = 3kg and v^2 = 3900^2 + 2700^2 + 3600^2 after the collision it has v^2 = 3500^2 + 2000^2 + 4100^2 It's Kinetic Energy (before) is 1/2 mv^2 = 1.5 * (3900^2 + 2700^2 + 3600^2) = 5.319E7 J It's Kinetic Energy (after) is 1.5* (3500^2 +2000^2 + 4100^2) = 4.959E7 J The difference in energy is therefore 5.319E7 - 4.959E7 = 3.6E6 J This is the Kinetic Energy of rock2 I'm not sure if that thermal energy is 2.52106 or 2.52E6 but either way, the Kinetic energy of rock1 after the collision is exactly the same. Simply subtract the thermal energy from the kinetic of rock2 and thats rock2's final kinetic(we don't have to worry about which rock the thermal energy goes to bc it doesn't contribute anything to kinetic energy.) For the last part I'm pretty sure the microscopic work just the negative thermal energy.

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