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Okay, so I am having a hard time wrapping my head around this. Before looking at the answer, I worked the example and got the answer only because I used dimensional analysis to get me to the right units for force. However, to be honest, I don't think I even understand the concept here. I know mass is changing and velocity is constant, but what is throwing me off is where it says "The water strikes the window at $32\, m/s$ so each kilogram of water loses $32\, kgm/s$ of momentum. Water strikes the window at the rate of $45\,kg/s$, so the rate at which it loses momentum to the window is $1400\, kgm/s^2$" How is it losing $32\, kgm/s$ of momentum and then at the same time also losing it at a rate of $1400\,kg m/s^2$?

Why does it even mention "The water strikes the window at $32\,m/s$ so each kilogram of water loses $32\,kgm/s$ of momentum" what is the purpose of that sentence. That is really confusing me.

EDIT: I added this simple diagram to show how I see this system. Where is $32\,kgm/s$ of momentum come in? (I am aware that the window is now showing)

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  • $\begingroup$ very good concept question and specific to boot +1 $\endgroup$ – Argus Oct 29 '13 at 6:25
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Let's take the picture you drew to visualize this and run with it; I think it's a pretty good way to think about the problem.

Every second, a $45\,\mathrm{kg}$ chunk of water strikes the window horizontally after which its horizontal velocity drops to zero instantly. Since the momentum of the water is its mass times its velocity, and since the velocity of the water before it strikes the window is $32\,\mathrm m/\mathrm s$, we find that before it strikes the window, each $45\,\mathrm{kg}$ chunk of water has a total momentum equal to $45\,\mathrm{kg}\cdot 32\,\mathrm m/\mathrm s = 1440 \,\mathrm{kg}\, \mathrm m/\mathrm s$.

On the other hand, right after the water strikes the window, its velocity is zero, so it has zero momentum. Therefore, the magnitude of the change in momentum of each such chunk of water is $1440 \,\mathrm{kg}\, \mathrm m/\mathrm s$, and this change in momentum occurs every second, namely a full chunk hits the window and stops every second.

It follows that the rate of change of the momentum of the stream of water has magnitude $1440 \,\mathrm{kg}\, \mathrm m/\mathrm s^2$; this is the rate at which the momentum of water stream decreases due to the successive chunks hitting the wall and stopping second after second.

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  • $\begingroup$ That is a great answer. I understand it better now, but why does it even mention "The water strikes the window at 32 m/s so each kilogram of water loses 32 kg.m/s of momentum." What is the meaning or purpose of that statement? Thanks a lot by the way. I appreciate the insight. $\endgroup$ – CrypticParadigm Oct 29 '13 at 7:19
  • $\begingroup$ Yes, but that was already given. I don't see the significance of letting me know the momentum for 1kg of water moving at 35m/s. Perhaps the author just wants to confuse students? $\endgroup$ – CrypticParadigm Oct 29 '13 at 7:29
  • $\begingroup$ @user31810 Since we are dealing with a continuous stream of water moving at constant velocity, you can really analyze the problem with any size chunk you want, and the author seems to want to use $1\,\mathrm {kg}$ chunks. If you pick $1\,\mathrm{kg}$ chunks, then the momentum of each such chunk is $1\,\mathrm{kg}\cdot 32\,\mathrm m/\mathrm s = 32 \,\mathrm{kg}\,\mathrm m/\mathrm s$. Then you simply note that one such chunk strikes the window each $1/45$ seconds (or equivalently $45$ such chunks each second), and you arrive at the same answer. $\endgroup$ – joshphysics Oct 29 '13 at 7:39
  • $\begingroup$ So then that is arbitrary. Okay, well I appreciate your help. Thanks, Joshphysics. $\endgroup$ – CrypticParadigm Oct 29 '13 at 7:56

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