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A 1050-kg sports car is moving westbound at 15.0 m/s on a

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 36E Chapter 8

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 36E

A 1050-kg sports car is moving westbound at 15.0 m/s on a level road when it collides with a 6320-kg truck driving east on the same road at 10.0 m/s. The two vehicles remain locked together after the collision. (a) What is the velocity (magnitude and direction) of the two vehicles just after the collision? (b) At what speed should the truck have been moving so that both it and the car are stopped in the collision? (c) Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

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Solution 36E To solve this question, we shall have to apply the conservation of momentum principle before and after collision. Given, mass of the sports car m = 1050 sg Its speed v = s 15.0 m/s (negative sign taken for westbound motion) Mass of the truck m = 6320 kg t Its speed v = 10.0 m/s t They remain locked after the collision, hence they will have a common speed. Let this speed be V. (a) herefore, momentum before collision = Momentum after collision m vs+ s v = (m + m )V s t t t 1050 kg × ( 15.0 m/s) + 6320 kg × 10.0 m/s = (1050 + 6320) × V 47450 kg.m/s V = 7370 kg V = 6.44 m/s Since the speed value is positive, hence it will be eastbound. (b) Given that, both the car and the truck are stopped after the collision, therefore V = 0 Therefore, the initial momentum of the system will be zero. So, m v s msv = 0 t t 1050 kg × ( 15.0 m/s) + 6320 kg × v = 0 t v =t2.50 m/s Therefore, the speed of the truck should be 2.50 m/s for the system to come to a halt. (c) The change in kinetic energy = Final kinetic energy Initial kinetic energy For part a, Final kinetic energy K = 1 × (1050 + 6320) kg × (6.44) m /s 2 2 2 f 2 K =f152830 J Initial kinetic energy. 1 2 1 2 K =i 2 × 1050 × (15.0) J + 2 × 6320 × (10.0) J = 118125 J + 316000 J = 434125 J The change in kinetic energy = K K = f52830iJ 434125 J = 281295 J The change in kinetic energy K = 281295 J = 2.81 × 10 J 5 For part b, Final kinetic energy K f = 0 since the system is stopped. 2 2 Initial kinetic energy K = i 2 × 1050 × (15.0) J + 2 × 6320 kg × (2.50) J K = 118125 J + 19750 J = 137875 J i Therefore, the change in kinetic energy in this case 5 = 0 137875 J = 137875 J 1.38 × 10 J Therefore, for the part b, the approximate change in kinetic energy is 1.38 × 10 J . 5 Comparing the answers in part a and part b, it is concluded that the part a has the greater magnitude in the change in kinetic energy.

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Chapter 8, Problem 36E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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