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Referring to the famous example of a horizontally moving sticky ball that collides (and sticks) at the tip of a vertically floating rod, then the combination moves along the ball's incident course while rotating. For the above case, we assune that the final linear momentum is exactly the same as the ball's linear momentum, and even if the ball hits the rod at its center of mass we still apply the same conservation of linear momentum principle. However, if we look at the things from linear KE's perspective, then the first case implies a split of the ball's initial KE over the linear and rotatiinal KEs of the final combination, which means less final linear KE than the initial ball's KE, and the second case implies that the combination will have the same initial linear KE of the ball (since no rotation will occur).

My question is: How would the final combination in both cases have different linear KEs while having the same linear momentum (due the linear conservation)?

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  • $\begingroup$ In sticky (i.e. non-elastic) collisions, some of the kinetic energy is lost to heat. Even in your second case, the combination will not have the same KE as the initial ball. $\endgroup$
    – JF132
    Mar 3 '20 at 21:14
  • $\begingroup$ this is an inelastic collision . energy is not conserved. in the case of the collision with the end of the rod, a greater portion of the initial energy is conserved. $\endgroup$
    – trula
    Mar 3 '20 at 21:27
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    $\begingroup$ Energy as a whole is a conserved quantity as are both, linear and rotational momentum. A specific type of energy is not conserved. $\endgroup$ Mar 3 '20 at 21:37
  • $\begingroup$ Even if we count for the part of energy dissipated (thermal, optical, or any other form), this should happen in both cases (ball hitting the tip or the center of mass of the rod), we will still have two different values of final linear KE for each case, but the value of the final linear momentum is still to be considered the same. How is that possible? $\endgroup$ Mar 3 '20 at 22:33
  • $\begingroup$ To neutralise (as much as possible) the effect of energy loss due to conversion at collision time, let us imagine that the collision point (in both cases) has a latching mechanism to hold on to the ball (designed to consume relatively very low energy to be activated), not that both objects merge physically into each other and dissipate energy as a result. $\endgroup$ Mar 3 '20 at 23:04