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Magnitude of Total Momentum
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[QUOTE="Steve4Physics, post: 6649062, member: 681522"] At the time of writing, this is a 17+ year old question. So I'm giving a fairly complete answer/explanation. You must account for the opposite directions. The total momentum for case-2 is therefore zero. That's because momenta are vectors; they must be added as vectors. (If the magnitude of the resultant is needed, this is found [U]after[/U] the vector-addition.) That means forcase-2, the carts (now stuck together) will end-up stationary. You can reach the same conclusion by thinking about symmetry. There is no preferred direction of motion for the combined carts in case-2. This is badly worded (or maybe a trick) question. Energy is [U]always[/U] conserved. So the total energy loss in both cases is zero. However, the question was probably intended to be about loss of mechanical energy (ME). Mechanical energy is potential energy + kinetic energy. In both cases, the starting ME is the same: - for case-1 it is mgh+ 0 = mgh - for case-2 it is mg(h/2) + mg(h/2) = mgh In case-2 [U]all[/U] the ME is lost, as the carts end-up stationary. (The ME gets entirely converted to heat (and maybe a little sound energy) as a result of the collision.) In case 1, the stuck-together cars are still moving. So not all of the ME has been converted to heat. So the loss of ME is greater in case 2. [/QUOTE]
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Magnitude of Total Momentum
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