This account cannot be access XPf Sign OnCourse HomeV engineering.pdfopenvellum.ecollege.com/course.html?courseld=17215962&OpenVellumHMAC=d17f81cd781bbdbc830fe3efc4e25812#10001My CoursesPart ECalculate the change in momentum for 10.0-g marble.Figure<) 1 of 1?Ap2 =kg · m/s0.200 m/s0.400 m/sRequest AnswerSubmit30.0 g10.0 gPart F Complete previous part(s)Part GCalculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before This account cannot be access XPf Sign OnCourse HomeV engineering.pdfopenvellum.ecollege.com/course.html?courseld=17215962&OpenVellumHMAC=d17f81cd781bbdbc830fe3efc4e25812#10001My CoursesSubmitRequest AnswerPart BFigure<) 1 of 1Find the magnitude of the velocity of 10.0-g marble after the collision.0.200 m/sv2 =m/s0.400 m/s30.0 gSubmitRequest Answer10.0 gPart CFind the direction of the velocity of each marble after the collision.

Answered: Image /qna-images/answer/30b4f516-c67c-45f1-bb5c-d19c4b36057c.jpg

onstants Part D A 10.0-g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0-g marble sliding to the right with a velocity of magnitude 0.200 m/s. Let +z be to the right. (Since the collision is head-on, all the motion is along a line.) (Eigure 1) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for 30.0-g marble. ? Api = kg - m/s Request Answer Submit Part E Calculate the change in momentum for 10.0-g marble. Figure < 1 of 1 ? Ap2 = kg - m/s 0.200 m/s 0.400 m/s Request Answer Submit 30.0 g 10.0 g Part F Complete previous part(s) Part G Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before

onstants Part D A 10.0-g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0-g marble sliding to the right with a velocity of magnitude 0.200 m/s. Let +z be to the right. (Since the collision is head-on, all the motion is along a line.) (Eigure 1) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for 30.0-g marble. ? Api = kg - m/s Request Answer Submit Part E Calculate the change in momentum for 10.0-g marble. Figure < 1 of 1 ? Ap2 = kg - m/s 0.200 m/s 0.400 m/s Request Answer Submit 30.0 g 10.0 g Part F Complete previous part(s) Part G Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A projectile has undergone an elastic one-dimensional collision with an initially stationary target, along an x axis. The figure below is a graph of position versus time for the projectile and target, before and after the collision. (Two line segments are parallel to the time axis.) Which is true about the masses of the projectile and target? 4 O the projectile's mass is greater O the target's mass is greater O the masses are equal
A 8.58-g bullet is moving horizontally with a velocity of +366 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1168 g, and its velocity is +0.625 m/s after the bullet passes through it. The mass of the second block is 1556 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision. +366m/s Block 1 (a) Before collision +0.625m/s mblock 1=1168 (b) After collision (a) Vblock2= Number (b) KE after/KEbefore = Number Block 2 17 block 2 = 1556 Mbullet =8.58E i block 2 K Units Units
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