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• O • S Power: Ep 203 - Like We're an XY EV WileyPLUS (95 Ahmad – a ca Secure | https: //edugen.wileyplus.com/edugen/studentmainfr.uni Y NEW a other Bookmarks I Apps R WileyPLUS O Differential calculus. LS University of South. D Apple & iCloud Yahoo B Bing w Wikipedia F Facebook y Twitter in LinkedIn D The Weather Channel 6 Yelp wleyPLUS: MywlePLUSI Help WileyPLUS haiday, Fundamentals of Physics10e Calculus-based Physics I & II (PH 201-202) Home Read, Study & Practice Assignment Gradebook ORION Downloadable eTextbook Assignment > Open Assignment PULL , SCREEN PRINTER VERSION BACK HDr ASSIGNMENT RESOURCES Chapter 9, Problem 050 2012 onnect 09, Montam A 6.50 g bullet moving at 647 m/s strikes a 970 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 453 m//ss. (a) What is the resulting speed of the block? ((b) What is the speed of the bullet-block center of mass? Chapter 09 Problem 008 09, Probl (a) Number Units 09 (b) Number Units Ch. 062 LINK TO TEXT LINK TO SAMPLE PROBLEMVIDEO VIDEO MINI-LECTURE Review Score Review Results by Study objective Question attempts: D of 6 used SAVE FOR LATER SUBMITT ANSWER licensa Agreement privacy Policy ID 2000-201Zohn Wiley & Sons, Inc. All Rights Reserved. A Division of ohn Wiley & Sons, Inc. Version 4.24.1.23Explanation / Answer
usling law of conservation of linear momentum
A 6.5 g bullety travelling at 647 m/s strikes a 970g wooden block at rest on a frictionaless surface
momentum of system before collision = momentum of the system after collision
(m*u) = (m*V)+(M*V1)
(6.5*10^-3*647) = (6.5*10^-3*453)+(0.97*V1)
V1 = 1.3 m/sec is speed of the block
b) speed of center of mass is Vcm = (m*u)/(m+M) = (6.5*647)/(6.5+970)
Vcm = 4.31 m/sec
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