rnstaphandbook2016.pc ? The Expert TA|Human × My Courses-Blackboardal t. ? on/Ta

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rnstaphandbook2016.pc ? The Expert TA|Human × My Courses-Blackboardal t. ? on/Take TutorialAssignment.aspo Class l Help Homework 5 Begin Date: 3/19/2018 11:5900 PM-Due Date: 49/2018 11:59:00 PM End Date: 492018 11.59 00 PMM (7%) Problem 5: Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1500 kg approaching at 9.5 m/s due south. The second car has a mass of 650 kg and is approaching at 15.5 m/'s due west 33% Part (a) Calculate the magnitude of the final velocity, i meters per second, of the cars Grade Summ Deductions Potential Submissions Attempts rema 4 5 6 cotan sn acos0 atanO acotan sinb cosh0 tan0 cotanh . Degrees Radians igive up SubrmitHint Feedback: 0 deduction per feedback Hints:?, deduction per hint Hints renaming : in degrees south of west, of the cars -& 33% Part (c) what is the change in kinetic energy, an joules, for the colision? (This energy g 33% Part (b) Calculate the direction ofthe final velocity oes into deformation of the cars)

Explanation / Answer

Given,

m1 = 1500 kg ; m2 = 650 kg ; u1 = 9.5 m/s ; u2 = 15.5 m/s

a)We know from conservation of momentum

Pi = Pf

m1u1(-j) + m2u2(-i) = (m1 + m2) v

v = m2 u2/(m2 + m2) (-j) + m1u1/(m1 + m2) (-i)

v = 650 x 15.5/(1500+650) (-j) + 1500 x 9.5/(1500+650) (-i)

v = -4.69 j - 6.63 i

lvl = sqrt (-4.69^2 + -6.63^2) = 8.12 m/s

Hence, v = 8.12 m/s

b)theta = tan^-1(-4.69/-6.63) = 35.28 deg

theta = 35.28 deg

c)KEloss = KEi - KEf

KEloss = 1/2 (m1u1^2 + m2u2^2) - 1/2 (m1+m2) v^2

KE(loss) = 0.5(1500 x 9.5^2 + 650 x 15.5^2) - 0.5 x (1500+650)8.12^2 = 7.5 x 10^4 J

Hence, KE(loss) = 7.5 x 10^4 J

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