What is the percent of energy lost in an inelastic collision?

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In an inelastic collision, when block 1 collides with block 2 and they stick together, the final kinetic energy must be calculated using the combined mass of both blocks. The formula for percent energy lost is correct: [(K initial - K final)/K(initial)]*100. To find K final, include the mass of block 2 since they move as one mass after the collision. Understanding that the kinetic energy loss pertains to block 1's initial energy compared to the total energy after the collision is crucial. The discussion concludes with the clarification that using the combined mass for K final is essential for accurate calculations.
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percent of energy lost?

I'm confused... I have a case where block 1 moves with certain velocity, hits block 2 (that is at rest) and sticks to it - inelastic collision. Now, the problem is that I have to find the percent of energy (kinetic) that the first block has lost. I know that the formula is [(K initial - K final)/K(initial)]*100, but wha I'm wandering about is whether K(final) should include the mass of block 2 or not?
They are actually one mass after collision, but the question asks for the lost of energy of block 1. Can you somehow explain this to me? Thank you!
 
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Yeah, calculate the final kinetic energy with the combined mass.
 
Thanks, I got it right!
 
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