Need help! I\'ve worked it out several times but I am not getting the corrrect a

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Need help! I've worked it out several times but I am not getting the corrrect answer:( I've updated the picture and the question i need help is number 7. It is the one with the blue circles. Thanks again!

Safari File Edit View History Bookmarks Window Help 96%-- Thu Oct 1 9:46 PM C E quest.cns.utexas.edu Assignment - Quest Learning & Assessment https://quest.cns.utexas.edu/student/images/problem_pdf?cour... Home | Chegg.com A disc of mass m moves horizontally to the zontally with speed vo strikes the block and right with speed v on a table with negligible embeds itself inside it. The system of block friction when it coides with a second disc of and bullet swings unil the string is precisely mass 6m. The second disc is moving horizon- horizontal, at which point the system is mo- tally to the right with speed at the moment mentarily at rest. of impact. before Mi+ m The two discs stick together upon impact. Find the string length in simple terms of m, after M, g and vp 1. None of these 2.1= What is the speed of the composite body immediately after the collision? -29(1+m) 3.1-29(-M) 2 1.vf- 4· 45 4 None of these is correct 6· 14 009 10.0 points A 7 g bullet is fired into a 321 g block that is initially at rest at the edge of a frictionless table of height 1.6 m. The bullet remains in the block, and after impact the block lands 2.5 m from the bottom of th table.

Explanation / Answer

masses m = m

             M = 6 m

Initial velocities u = v

                        U = v/6

From law of conservation of momentum,

mu+MU = (m+ M) V

mv +6m(v/6)=(m+6m) V

mv+mv=7mV

2mv=7mV

2v = 7V

From this velocity of the combined masses after collision V = (2/7)v

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