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I'm usually pretty hesitant to post about homework things on StackExchange, but I've got a problem at the moment that I can't quite wrap my head around.

The problem is stated as follows:

An 8.00kg object moving 4.00 m/s in the positive x direction has a one-dimensional collision with a 2.00kg object moving 3.00 m/s in the opposite direction. The final velocity of the 8.00kg object is 2.00 m/s in the positive x direction. What is the total kinetic energy of the two-mass system after the collision? (Not necessarily a perfectly elastic collision!)

The part in parentheses is obviously the point of the question, but I'm still not quite sure how to tackle it. I assume that since it's not necessarily (though, at the same time, it never explicitly states it's not) a perfectly elastic collision, I can't use the conservation of momentum or kinetic energy in my process, but I'm confused about what assumptions I can make in this question in order to solve it without being given the velocity of the 2.00kg object. I might be missing something really simple here, but I would really appreciate a little nudge. :)

Thanks!

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    $\begingroup$ You can use conservation of momentum in non-elastic collisions. $\endgroup$
    – stafusa
    Oct 29 '17 at 22:43
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Whenever you are dealing with such a collision, remember that you can always resort to the assumption that momentum will be conserved (Conservation of Momentum):

$p_f=p_i$

Since you have all of the masses, and all of the velocities except for the final velocity of the $2.00\text{kg}$ mass, you can solve for the final velocity using:

$p=m\cdot v$

After which finding the kinetic energy is trivial.

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  • $\begingroup$ I think that for some reason I had lumped kinetic energy and momentum into the same general box in my head—that they aren't conserved in imperfect collisions, but I hadn't fully thought through that simply the sum of each object's initial momentum has to equal the sum of each object's final momentum (since momentum is a vector whereas kinetic energy is a positive scalar, and thus energies can't both get smaller and still equal the original energy). I guess I was totally overthinking it. Thanks! (I would vote up your answer, but I don't have the necessary reputation) $\endgroup$ Oct 29 '17 at 22:06

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