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tra The Ex TAIH x C fi https:// /Common/TakeTutorialAssignment.aspx ust35nc.theexpertta.co facebook YouTube Home My Account Log Out Class Management I Help 2101 Final Exam Fall 2015 Extension Begin Date: 2015 9:00:00 AM Due Date: 2015 2:00:00 PM End Date: 12/11/2015 2:00:00 PM (10%) Problem 6: A block of mass m J 15.5 kg slides along a horizontal surface (with friction, Aut 0.29) a distance d 2.45 m before striking a second block of mass m 5.25 kg. The first block has an initial velocity of v 775 ims Randomized Variables 15.5 kg 5.25 kg Assignment Status 0.29 2.45 Click here for m/s detailed view Otheexpertta.com A 50% Part (a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s Time Remaining Grade Summary 1:57:52 Deductions tial. 100% Problem Status S 9 HOME sino coso tano Submissions Completed cotan asino acoso Attempts remaining Completed per attempt sinho Partial atan0 acotana detailed view cosh 0 tanho cotanh Partial Degrees Radians deduction per h H Feedback: 5% deduction per feedback. 10 Submission History Hints Feedback E A 50% Part b) How far does block two travel, d2 in meters, before coming to rest after the collision? € 2015 Expert TA, LLC 11:09 AM 12/11/2015

Explanation / Answer

part a:

during just before the collision and just after the collision, there is no external force acting on the system

hence momentum just before the collision=momentum just after the collision

now, initial speed of m1=7.75 m/s

deceleration=-friction coefficient*acceleration due to gravity

=-0.29*9.8=-2.842 m/s^2

then using the formula:

final veloicty^2-initial velocity^2=2*acceleration*distance

==>final speed^2-7.75^2=-2*2.842*2.45

=>final speed^2=46.1367

==>final speed=6.7924 m/s

hence just before collision, total momentum of the system

=m1*speed of first block+m2*speed of second block

=15.5*6.7924+5.25*0

=105.2822 kg.m/s

as after collision, momentum of the system=momentum of the system before collision

==>m1*0+m2*speed of second block=105.2822

==>speed of second block just after collision=105.2822/5.25=20.05375 m/s

part b:

deceleration due to friction will remain constant at -2.842 m/s^2

then initial veloicty of the second blcok=20.05375 m/s

final veloicty=0

using the formula:

final veloicty^2-initial velocity^2=2*acceleration*distance

==>0^2-20.05375^2=2*(-2.842)*distance

==>distance =20.05375^2/(2*2.842)=70.7517 m

hence the second block will travel 70.7517 m before stopping.

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