We would usually use the pythagorean theorem to nd

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Unformatted text preview: use the Pythagorean theorem to nd the magnitude of a vector, but in this case, our vectors have only one component each, so nding the magnitudes is easy. For ~ i , p j~ i j = j(0 :057 kg) h 50; 0; 0i m =sj p = jh2:85; 0; 0i kg m =sj = 2 :85 kg m =s : We are looking down on the orbit from above the north poles of the Sun and Mars, with + x to the right and + z down the page. Notice that there is no y component of momentum at either position, so the change in momentum vector will be of the form ~ = h ~ x; 0; ~ zi : p p p Find Correct answer: ~ f = j(0 :057 kg) h 42; 0; 0i m =sj p = jh 2:394; 0; 0i kg m =sj = 2 :394 kg m =s : 9: 6 1027 kg m =s. ~ x = ~B ~ A p px px = m~ x m ~ x vB vA = m ~ x ~x vB vA = m ( 15000 m =s 0) = (6 :4 1023 kg) ( 15000 m =s) = 9:6 1027 kg m =s : So the change in magnitude is j ~ij = p 0:456 kg m =s: The \red planet," Mars, has a mass of 6 :4 1023 kg and travels in a nearly circular orbit around the Sun, as shown in the following gure. ~B v C ~ x. Answer in kg m =s. p Explanation: To nd ~ x, we take the dierence between p the initial and nal momenta of the planet: And for ~ f , p ~f p 2 B Sun Find ~ z. Answer in kg m =s. p Correct answer: 9 :6 1027 kg m =s. Explanation: We nd ~ z by the method of part 1: p ~A v A +x D +z When Mars is at location A , its velocity is ~ z = ~B ~A p pz pz = m~ z m ~ z vB vA = m ~z ~z vB vA = m [0 ( 15000) m =s] = (6 :4 1023 kg) [ ( 15000) m =s] = 9 :6 1027 kg m =s : ~ A = h0; 0; 15000i m =s: v When it reaches location locity is B , the planet’s ve- ~ B = h 15000; 0; 0i m =s: v Consider the momentum of Mars at positions C and D on the diagram above. If you found the change in momentum vector between C and D , which direction would it point? cai (atc667) { homework 03 { turner { (57345) 3 A baseball has a mass of about 155 g. What is the magnitude of the momentum of a baseball thrown at a speed of 95 mph? Answer in kg m =s. (Note that you need to convert mass to kilograms and miles per hour to meters per second.) 1. Correct answer: 6 :58265 kg m =s. Explanation: Since the question asks for the magnitude of momentum, we don’t need to worry about direction or vectors. The magnitude of momentum is given by p...
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This note was uploaded on 02/11/2014 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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