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Introductory Physics Homework Help
What is the magnitude of the final momentum?
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[QUOTE="jbriggs444, post: 5154979, member: 422467"] 200 in units of graph squares. [B]Always show your units when writing down numbers.[/B] [B]Avoid putting numbers into equations. Put variables into equations. Substitute the numbers in at the end.[/B] That way those of us looking at your work have a better chance of understanding what you are trying to do. As I reconstruct your work... You do not show your work computing the area under the curve but come up with a result of 200 graph square units. I'd expected some invocation of "1/2 base times height", "base times height" and "1/2 base times height" in your work product. You divide by the duration of the collision -- 4 graph width units. That gives you the average force in graph height units -- 50 graph height units. You convert this 50 graph height units to 50 Newtons and multiply by the elapsed time of .04 seconds. This gives you the average force multipled by the collision duration -- 2 Newton-seconds. [B]It would have been simpler if you had computed the area under the curve in Newton-seconds to start with. Then you would not have to uselessly divide by 4 and then multiply by 0.04 when all you are achieving is a unit conversion.[/B] You divide this impulse by the mass of the tennis ball, 0.057 kg to get a change in velocity -- 35.08 meters/second You assume a sign convention and subtract the initial velocity, 24 meters/second from the velocity change to get the final velocity -- 11.08 meters/second. You multiply by the mass of the tennis ball to get the resulting momentum -- 0.632 kg meters/second. [B]Rather than dividing by the mass of the tennis ball and then multiplying by the mass of the tennis ball, you could have worked with momenta directly.[/B] Yes, more complicated than it had to be and less symbolic, but it seems correct. [/QUOTE]
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Introductory Physics Homework Help
What is the magnitude of the final momentum?
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