Question

part a
imagine that you replace the block in the video with a happy or sad ball identical to the one used as a pendulum, so that the sad ball strikes a sad ball and the happy ball strikes a happy ball. the target balls are free to move, and all the balls have the same mass. in the collision between the sad balls, how much of the balls’ kinetic energy is dissipated?
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○ all of it
○ half of it
○ none of it
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part b
now, consider the collision between two happy balls described in part a. how much of the balls’ kinetic energy is dissipated?
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○ all of it
○ none of it
○ half of it
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Explanation:

Part A

In a collision between two identical "sad balls" (assuming a perfectly inelastic or a collision where they stick together or transfer energy such that half the kinetic energy is dissipated, but more accurately, for identical masses in a one - dimensional collision where the first ball stops and the second moves, but if it's a case like a perfectly inelastic collision - no, wait, actually, when a ball of mass \(m\) moving with velocity \(v\) collides with an identical ball at rest (free to move), in a perfectly elastic collision, kinetic energy is conserved. But "sad balls" are likely to have a collision where they stick? No, wait, the key here: when two identical balls with the same mass collide, and the target is free to move. If we consider the case of a perfectly inelastic collision (they stick), but no, for identical masses, if the first ball has velocity \(v\), initial kinetic energy is \(\frac{1}{2}mv^{2}\). After collision, if they move together with velocity \(v/2\), kinetic energy is \(\frac{1}{2}(2m)(v/2)^{2}=\frac{1}{4}mv^{2}\), so the dissipated energy is \(\frac{1}{2}mv^{2}-\frac{1}{4}mv^{2}=\frac{1}{2}\) of the initial. But actually, the correct reasoning is that for two identical masses, when one is moving and collides with another at rest (free to move), in a perfectly inelastic collision (they stick), half the kinetic energy is dissipated. But the "sad ball" collision is such that they stick? Wait, no, the standard problem: when a ball hits an identical ball, if it's a perfectly elastic collision, kinetic energy is conserved (happy ball), but if it's a perfectly inelastic (sad ball), they stick, and half the KE is dissipated. So for sad balls, half of the kinetic energy is dissipated.

"Happy balls" undergo a perfectly elastic collision. In a perfectly elastic collision between two identical masses (where one is initially moving and the other is free to move), the kinetic energy is conserved. So no kinetic energy is dissipated.

Answer:

B. Half of it

Part B