Explanation Given info: The mass of the moving particle is 1.99 × 10 − 26 kg and the velocity of the particle in x-direction is 0.500 c i ∧ . collided with a particle of one third of its mass which was moving with a velocity of − 0.500 c i ∧ in the negative x-direction. Let m be the mass of the moving particle and m 3 be the mass of the smaller particle The velocity of the mass m particle be + u i → and the velocity of the smaller particle be − u i → . The resulting particle mass is M and its velocity is v f i ∧ For relativistic case, the momentum of a body is, m u 1 − ( u c ) 2 Here, m is the mass of the body. u is the velocity c is the speed of light. Apply conservation of relativistic momentum in X-direction Initial momentum = Final momentum Momentum of heavier particle + momentum of the lighter particle = momentum of the resultant particle. m u 1 − ( u c ) 2 + m 3 ( ( − u ) 1 − ( u c ) 2 ) = M v f 1 − ( v f c ) 2 2 m u 3 1 − ( u c ) 2 = M v f 1 − ( v f c ) 2 (1) The energy of body in relativistic case is m c 2 1 − ( u c ) 2 Apply conservation of energy Initial Energy of heavier particle + Initial energy of the lighter particle = final enrgy of the system of the resultant particle

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN: 9781305116399

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Chapter 39, Problem 39.63P

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The mass of the resulting particle after a collision.

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Chapter 39, Problem 39.62P

Chapter 39, Problem 39.64P

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Physics for Scientists and Engineers, Technology Update (No access codes included)

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The mass of the resulting particle after a collision.

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The mass of the resulting particle after a collision.

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