Colliding balls: Conservation of momentum and changes in kinetic energy?

[cueballbullet]

I got curious about firearm ballistics and googled something similar to "bullet momentum vs kinetic energy".

IIRC, momentum P = mv (checked); and kE = (mv^2)/2 (also checked).

So I essentially wondered if it's worse to get hit by a bullet with greater kE than by one with lesser kE, presuming that P remains the same (same momentum (also same shape and size); yet different masses and velocities).

Quickly I learned that the faster, lighter bullet causes more damage and has (/because it has) more kE, as the greater amount of kE gets transferred to the bodily tissues.

Cool. Yet this led me to wonder about something else:

Posit that a rolling cue ball, B, of mass M, moving at velocity V, hits another cue ball, b, of mass M/2. If momentum is conserved, then the latter, lighter cue ball, b, will start rolling at velocity 2V... So, same momentum, and different velocities. This means that b has greater kinetic energy than B.

Everything makes sense in my non-physicist mind up until that last sentence. For the life of me I can't guess at all where that extra energy comes from. Same momentum, but twice the speed, because of half the weight. Cool. But again, if the momentum is indeed the same, but the speeds are different, then the kE should also be different, right? How does this work? I may have misunderstood something along the way and perhaps the energy is not greater in b than in B, afterall.

 

[Doc Al]

Posit that a rolling cue ball, B, of mass M, moving at velocity V, hits another cue ball, b, of mass M/2. If momentum is conserved, then the latter, lighter cue ball, b, will start rolling at velocity 2V... So, same momentum, and different velocities. This means that b has greater kinetic energy than B.

You are assuming that the rolling ball transfers all its momentum to the second ball, then stops dead. That's not how it works. To figure out the speeds of both after the collision, one must apply both conservation of momentum (total momentum of both) and conservation of energy. (If anything, in a real collision, some of the energy will be "lost" to heat and sound.)

Everything makes sense in my non-physicist mind up until that last sentence. For the life of me I can't guess at all where that extra energy comes from.

That's good instinct to sense something's not right. The answer: There is no extra energy!

 

[Doc Al]

Just for fun, here are the final speeds of each. (Assuming a perfectly elastic head-on collision, which is the simplest to analyze.)

Final speed of the first ball: V/3
Final speed of the second ball: 4V/3