Ps1 - Physics 1116 Fall 2011 Homework 1 Due Wednesday August 31st Reading Kleppner& Kolenkow Chapter 1.1-1.8 and Note 1.1 1 Vector Algebra

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Unformatted text preview: Physics 1116 Fall 2011 Homework 1 Due: Wednesday, August 31st Reading: Kleppner & Kolenkow Chapter 1.1-1.8 and Note 1.1 1. Vector Algebra. Consider two vectors A =- ˆ i + 2 ˆ j + 3 ˆ k and B = 4 ˆ i- 6 ˆ j + 3 ˆ k . Compute: a. A + B b. A- 2 B c. A · B d. A × B e. B × A f. A · ( A × B ) and B · ( A × B ) Solution: a. 3 ˆ i +- 4 ˆ j + 6 ˆ k b.- 9 ˆ i + 14 ˆ j +- 3 ˆ k c. -7 d. 24 ˆ i + 15 ˆ j + 8 ˆ k e.- 24 ˆ i- 15 ˆ j- 8 ˆ k f. Both are zero. The vector A × B is perpendicular to both A and B . 2. Distance of closest approach. Consider an arbitrary point in space given by a vector r , and an arbitrary direction given by a unit vector ˆ n . We can define a straight line in space, passing through the given point in the given direction, by the locus of points r = r + η ˆ n for all values of a real number η . Use vector techniques to answer the following questions. a. What is the distance of closest approach of this line to the origin? b. If a new line is drawn from the origin to the point of closest approach, at what angle do these two lines intersect one another? c. List two possible conditions that, if either is true, the distance of closest approach will be zero. Comment. In experimental particle physics it is often important to know the distance of closest approach of a particle track to the point where the primary beams collided. In many cases this distance helps to measure some other particle’s lifetime. Solution: a. Evaluate the length, r , of r , and minimize with respect to η : r 2 = r 2 + η 2 +2 η r · ˆ n . Minimum occurs for η =- r · ˆ n , so the point of closest approach is given by r min = r- ( r · ˆ n )ˆ n , and the distance (to the origin) of closest approach is r min = p r 2- ( r · ˆ n ) 2 . b. You can show r min · ˆ n = 0, so this implies that they intersect at right angles. c. r min = 0 if either r = 0 or ˆ n = ˆ r . 3. Kinematics in one dimension with contant acceleration. KK Problem 1.13 Solution: The elevator travels upward with (unknown) velocity v , starting from y = 0 at t = 0....
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This note was uploaded on 02/03/2012 for the course PHYS 1116 at Cornell University (Engineering School).

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Ps1 - Physics 1116 Fall 2011 Homework 1 Due Wednesday August 31st Reading Kleppner& Kolenkow Chapter 1.1-1.8 and Note 1.1 1 Vector Algebra

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