I NEEEEED PART B!!! PLEASE!! ta The Ex TAIHuman-lik x https:// ust35nc.theexpert

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I NEEEEED PART B!!! PLEASE!!

ta The Ex TAIHuman-lik x https:// ust35nc.theexpertta.co /Common/TakeTutorialAssignment.aspx facebook YouTube Home nut Class Management I Help 2101 Final Exam Fall 2015 Extension Begin Date: 2015 9:00:00 AM Due Date: 12/11/2015 2:00:00 PM End Date: 12/11/2015 2:00:00 PM 10%) Problem 4: A car with mass m 1183 kg is traveling west through an NO) intersection at a magnitude of velocity of vc 14.5 m/s when a truck of mass m 1700 kg traveling south at vt 14.2 m/s fails to yield and collides with the car. The vehicles become stuck together and s e on the asphalt, which has a coefficient of friction of 0.5 Randomized Variables Assignment Status me 1183 kg 1700 kg Click here for EX) 14.5 m/s detailed view 14.2 m/s Otheexpertta.com Time Remaining 1:58:47 50% Part (a) Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j Problem Status Grade Summary i)) j)) mc mt Completed nic v mice mt mt v Deductions 92% Completed Partial a e 7 8 9 HOME Submissions Partial Attempts remaining: 1 (4% per detailed view END BACKSPACE CLEAR Submit Hint give up! 10 deduction per h Feedback: 5% deduction per feedback. Submission History Hints Feedback 80% 00% 00% A 50% Part b) How far, in meters will the vehicles slide after the collision? € 2015 Expert TA, LLC 11:08 AM rt TA 12/11/2015

Explanation / Answer

by conservation of momentum

initial horizontal momentum = final horizontal momentum

1183 * 14.5 = (1183 + 1700) * v_x

v_x = 5.949 m/s

initial vertical momentum = final vertical momentum

1700 * 14.2 = (1183 + 1700) * v_y

v_y = 8.373 m/s

net velocity = sqrt(v_x^2 + v_y^2)

net velocity = sqrt(5.949^2 + 8.373^2)

net velocity = 10.271 m/s

force = mass * acceleration

force = k * mass * g

mass * acceleration = k * mass * g

acceleration = k * g

acceleration = 0.5 * 9.8

acceleration = 4.9 m/s

v^2 = u^2 + 2as

0 = 10.271^2 - 2 * 4.9 * s

s = 10.764 m

distance covered = 10.764 m

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