homework 03-solutions

# The magnitude of momentum is given by p m v first

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Unformatted text preview: = m v: First, however, we need to convert our units. It’s easy to see that 155 g = 0 :155 kg : And if we look up the conversion factors, we have Following the trajectory of the planet as it goes along its orbit, it’s clear that at point C the momentum vector will point along the positive z direction (downward in the diagram), and that at point D the momentum vector will point along the positive x direction (to the right). Therefore the two vectors will be as follows: 95 mph If we subtract these two vectors ~ D p the resultant vector will be of the form ~D p ~ C; p ~ C = h+ pD; 0; pCi : p x z Since p ~ C = ~ D = 15000 kg m =s; p this vector will point at a 45 the upper right. ◦ 1 hr = 42 :4687 m =s: 3600 s Then we can write p = mv = (0 :155 kg) (42 :4687 m =s) = 6 :58265 kg m =s: ~ C = h0; 0; + pCi ; and p z ~ D = h+ pD; 0; 0i : p x 1609:34 m 1 mile The following gure shows a portion of the trajectory of a ball traveling through the air. Arrows indicate its momentum at several locations. ~C p ~D p ~E p D C ~B p E B angle toward is: At various locations, the ball’s momentum cai (atc667) { homework 03 { turner { (57345) ~B = p ~C = p ~D = p ~E = p h3:03; 2:83; 0i kg m =s h3:55; 0:97; 0i kg m =s h2:24; 0:57; 0i kg m =s h8:97; 1:93; 0i kg m =s You will calculate the change in the magnitude of momentum between each pair of adjacent locations. Begin by calculating j~ BCj, p the change between locations B and C . Answer in kg m =s. Correct answer: 0:465923 kg m =s. Explanation: To nd the magnitude of the change in momentum between points B and C , we rst nd the magnitudes of the momentum vectors corresponding to these two locations, and then we subtract them, always subtracting the earlier magnitude from the later. The magnitude of ~ B is given by p j~ Bj = p = q q ( pB) 2 + ( pB) 2 + ( pB) 2 x y z (3 :03) 2 + (2 :83) 2 + (0) 2 p = 17:1898 = 4 :14606 kg m =s : And for ~ C, p j~ Cj = p = q q p We use the same procedure as in part 1. We already know the length of ~ C, so we just p need the length of ~ D: p j~ Dj = p = (3 :55) 2 + (0 :97) 2 + (0) 2 = 13:5434 = 3 :68014 kg m =s : q q ( pD) 2 + ( pD) 2 + ( pD) 2 x y z (2 :24) 2 + ( 0:57) 2 + (0) 2 p = 5:3425 = 2 :31138 kg m =s : So the dierence is given by j~ Dj j ~ Cj = p p 1:36875 kg m =s: Now nd ~ DE, the change between locations p D and E . Answer in kg m =s. Correct answer: 6 :8639 kg m =s. Explanatio...
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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.

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