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Introductory Physics Homework Help
Find momentum transfer and force on the head with and without a helmet
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[QUOTE="jbriggs444, post: 6852509, member: 422467"] Note that the correct attribution would be to myself ([USER=422467]@jbriggs444[/USER]) as author of the quoted passage: When responding to short passage within a post, it is best if you can avoid quoting the whole thing and only quote the relevant passage, leaving that passage alone within the quote tags that are automatically supplied. One way to do that is to mouse over the relevant passage in the original post, selecting that passage and then clicking "quote" on the resulting pop-up. That will save the passage, making it available to use with the "insert quotes" button on the editting panel. One flaw in this technique is that ##\LaTeX## in the quotated passage will not be properly captured. Another way is to quote the whole thing (as you have been doing) and then erase all but the desired passage. This approach leaves ##\LaTeX## intact. Yet another variation on this latter technique is to turn off BB code processing on the editting panel. Then you can see how the quote is formatted behind the scenes. e.g. [PLAIN][QUOTE="jbriggs444, post: 6852126, member: 422467"][/PLAIN] [...]Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s[sup]2[/sup]. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. [B]This does not match the displacement given in the problem statement.[/B] [...] [PLAIN][/QUOTE][/PLAIN] You can see here what the undedrlying BB code looks like. You can chop up the original quoted text into pieces and insert your own BB code tags so that you can quote multiple passages and insert your own text in between. Do not forget to use the preview feature in case you butcher the result :-) Now, with all that said, let us look at your actual question: Your concern is that I used ##s=\frac{1}{2}at^2##. This equation is supposed to work only when ##v_i = 0##. The answer is that I am contemplating the collision in time reversed fashion, adopting the frame in which the head is at rest after the collision. In this frame the exterior of the crushed helmet is at rest and the head accelerates away. The amount by which the helmet (or head) is crushed is an "invariant". That means that we will get the same result no matter what frame we use to do the calculation. So I automatically chose a frame that makes the result easy to calculate. I apologize for not making that clear. Shifting frames like that is something that I do automatically without even thinking about it sometimes. [/QUOTE]
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Find momentum transfer and force on the head with and without a helmet
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