Explanation Given data: Two particles collide and bounce apart. Refer to the Figure P9.37 in the textbook for initial momentum of both the particles and final momentum of second particle. Formula used: Write the expression for law of conservation of momentum for an isolated system. p i = p f (1) Here, p i is the momentum of the particle before the collision, and p f is the momentum of the particle after the collision. Explanation: Modify the expression in equation (1) for conservation of momentum in the x -direction (horizontal direction) for the collision of two particles. ( p 1 x ) i + ( p 2 x ) i = ( p 1 x ) f + ( p 2 x ) f (2) Here, ( p 1 x ) i is the initial momentum of the particle 1 in the x -direction (before the collision), ( p 2 x ) i is the initial momentum of the particle 2 in the x -direction, ( p 1 x ) f is the final momentum of the particle 1 in the x -direction (after the collision), and ( p 2 x ) f is the final momentum of the particle 2 in the x -direction. Modify the expression in equation (1) for conservation of momentum in the y -direction (vertical direction) for the collision of two particles. ( p 1 y ) i + ( p 2 y ) i = ( p 1 y ) f + ( p 2 y ) f (3) Here, ( p 1 y ) i is the initial momentum of the particle 1 in the y -direction, ( p 2 y ) i is the initial momentum of the particle 2 in the y -direction, ( p 1 y ) f is the final momentum of the particle 1 in the y -direction, and ( p 2 y ) f is the final momentum of the particle 2 in the y -direction. From the Figure P9.37 in the textbook, the initial momentum of the particle 1 corresponds to the point ( 2 , 2 ) in the xy -coordinate system. Therefore, the initial momentum of the particle 1 in the x -direction ( p 1 x ) i is 2 kg ⋅ m s and the initial momentum of the particle 1 in the y -direction ( p 1 y ) i is 2 kg ⋅ m s . From the Figure, the initial momentum of the particle 2 corresponds to the point ( − 4 , 1 ) in the xy -coordinate system. Therefore, the initial momentum of the particle 2 in the x -direction ( p 2 x ) i is ( − 4 ) kg ⋅ m s and the initial momentum of the particle 2 in the y -direction ( p 2 y ) i is 1 kg ⋅ m s . From the Figure, the final momentum of the particle 2 corresponds to the point ( 0 , − 1 ) in the xy -coordinate system. Therefore, the final momentum of the particle 2 in the x -direction ( p 2 x ) f is 0 kg ⋅ m s and the final momentum of the particle 2 in the y -direction ( p 2 y ) f is ( − 1 ) kg ⋅ m s

4th Edition

ISBN: 9780134899978

Author: Knight

Publisher: PEARSON

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Question

Chapter 9, Problem 37P

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To show: The final momentum of particle 1 (p→1f) graphically in a coordinate system.

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Chapter 9, Problem 35P

Chapter 9, Problem 38P

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Chapter 9 Solutions

COLLEGE PHYSICS, VOL. 1-W/ACCESS

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The final momentum of particle 1 ( p → 1 f ) graphically in a coordinate system.

To show: The final momentum of particle 1 (p→1f) graphically in a coordinate system.

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Chapter 9 Solutions