If I let a brick fall on the ground, does it create more heat or give more kinetic energy to the earth? Part of the kinetic energy of the falling brick will be converted to heat of the surrounding air and the earth. Another part of the kinetic energy will be transferred to the earth, influencing its speed a tiny bit. Which part is larger, and why?
I am interested in the case that the brick does not bounce at all.
 A: The amount of energy that goes into heat vs macroscopic kinetic energy depends on the elasticity of the collision: if the brick bounces almost as high as where it was dropped from, only a small amount of its kinetic energy right before impact goes to heat. If it just thuds into the ground, almost all of it goes into heat. In the former case, important to note that after the collision, kinetic energy remains with the brick. If you think of it as a two body system with the earth at rest immediately before impact, in a perfectly elastic collision, the earth ends up with about twice the momentum as the brick had. So after the collision, given a 3-kg brick and a 6-trillion-trillion-kg earth, the earth is going about one trillion-trillionth the speed of the brick. As kinetic energy is proportional to speed squared, you can see that the kinetic energy transferred to the earth is pretty negligible.
EDIT AFTER COMMENTS
Looking at the perfectly inelastic case in the same frame (earth at rest the instant before impact), the final speed of the earth is about half as large as in the elastic case above (so half a trillion-trillionth of the speed of the brick).  Thus the earth ends up with half a trillion-trillionth of the kinetic energy the brick had. The fraction of the kinetic energy of the brick that goes into heat is: $$\frac{9999999999999999999999995}{10000000000000000000000000}$$
