homework 03-solutions

Find px in kg m s remember that any vector can be

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Unformatted text preview: es its unit vector; for example, ~ = j~ a . a aj ^ Correct answer: 2 :79205 10−21 kg m =s. Explanation: Since we already know the magnitude of the momentum from part 3, we just need to multiply by the x component of the unit vector pointing in the direction of the momentum. But since, relativistically, ~ = m~ p v; cai (atc667) { homework 03 { turner { (57345) meaning momentum and velocity point in the same direction, we can simply use the appropriate component of the given velocity unit vector: Here, punit;x refers to the x component of the unit momentum vector, and similarly for the unit velocity vector. Find py in kg m =s. 2:09723 10−21 kg m =s. Explanation: We follow the same procedure as in part 4: Correct answer: 4. d~ r dt d h 3 + 4 t; 3t 2; 2t = dt = h4; 6t; 2 21t 2i : ~= v 7t 3i Clearly, vx(0) = 4. Find vy(0) : Correct answer: 0. py = j~ punit;y pj = j~ vunit;y pj = (4 :26267 10−21 kg m =s)( 0:492) = 2:09723 10−21 kg m =s : Find pz in kg m =s. Correct answer: Find the instantaneous velocity at time t= 0. Start by nding the x component, vx(0). Explanation: To obtain the velocity from the position, we just take the derivative with respect to time. To take the derivative of a vector, we take the derivative of each component: px = j~ punit;x pj = j~ vunit;x pj = (4 :26267 10−21 kg m =s)(0 :655) = 2 :79205 10−21 kg m =s : Correct answer: 6 2:44251 vy(0) = 6(0) = 0 : Find vz(0) : 10−21 kg m =s. Explanation: Same procedure again: Correct answer: 2. Explanation: Taking the z component from above: pz = j~ punit;z pj = j~ vunit;z pj = (4 :26267 10−21 kg m =s)( 0:573) = 2:44251 10−21 kg m =s : Suppose the position of an object at time given by ~ = h3 + 4 t; 3t 2; 2t r Explanation: We just take the y component from above: 7t 3i : t is vz(0) = 2 21(0) 2 = 2 : Now nd the instantaneous acceleration at time 0. Start with a x(0). Correct answer: 0. Explanation: To nd the acceleration from the velocity, we take the derivative again: cai (atc667) { homework 03 { turner { (57345) d~ v dt d = h 4; 6t; 2 21t 2i dt = h0; 6; 42ti : 7 What is the magnitude of the correct relativistic momentum of the proton? Your answer must be within 5.0% −17 Correct answer: 1 :12144 10 kg m =s. ~= a Explanation: The...
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This note was uploaded on 02/11/2014 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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