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A 10.0-g marble slides to the left at a speed of 0.400 m/s

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

Solution for problem 48E 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 48E

A 10.0-g marble slides to the left at a speed of 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0-g marble sliding to the right at a speed of 0.200 m/s (?Fig. E8.48?). (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all motion is along a line.) (b) Calculate the ?change in momentum? (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the ?change in kinetic energy? (the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

Step-by-Step Solution:

Solution to 48E Step 1 Mass of first marble =10g=10x10 kg -3 Mass of the second marble =30g=30x10 kg -3 Initial velocity of first marble =0.4m/s Initial velocity of second marble =-0.2m/s Step 2 According to conservation of momentum, m 1 1 v2 2v’1m1’ 2 2 -3 -3 -3 -3 10x10 (0.4)+30x10 (-0.2)=10x10 v’ +30x10 v’ 1 2 3v’2v’ 10.2 ………………………………………………………..(1) Relative velocity equations for elastic collisions, v’ -v’ =-(-0.2-0.4) 2 1 v ’-v’ =0.6 ……………………………………………………………(2) 2 1 Solving equations (1) and (2) v ’=-0.5m/s 1 v’2=0.1m/s (b)

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Chapter 8, Problem 48E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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A 10.0-g marble slides to the left at a speed of 0.400 m/s

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