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Unformatted text preview: ikram (ai2734) – hw1022 – Swinney – (58785) 1 This printout should have 8 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This is the last homework before the test on Monday Oct 26. 001 (part 1 of 2) 10.0 points The force of magnitude F x acting in the x direction on a 3 . 4 kg particle varies in time as shown in the figure. 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 F x (N) t (s) Calculate the impulse of this force. Correct answer: 20 N · s. Explanation: The impulse of a timedependent force F ( t ) is the integral I = integraldisplay F ( t ) dt. Graphically, this integral is the area between the F ( t ) plot and the t axis. Splitting this area into a triangle (for 0 < t < 4 s), a rectangle (for 4 s < t < 5 s) and another triangle (for 5 s < t < 9 s), we find I = (4 N) (4 s) 2 + (4 N) (1 s) + (4 N) (4 s) 2 = 20 N · s . 002 (part 2 of 2) 10.0 points Before the force is applied ( t < on the above graph ), the particle moves along the x axis with velocity v 1 = 4 . 6 m / s. Find the velocity v 2 of the particle after the force stops acting on it ( t > 9 s). Correct answer: 1 . 28235 m / s. Explanation: The momentum p = mv of the particle changes by the impulse of the force, Δ p = I mv 2 mv 1 = I v 2 v 1 = I m = 20 N · s 3 . 4 kg = 5 . 88235 m / s and therefore v 2 = 4 . 6 m / s + 5 . 88235 m / s = 1 . 28235 m / s . 003 10.0 points A 4 . 56 kg steel ball strikes a massive wall at 6 . 31 m / s at an angle of 86 . 5 ◦ with the plane of the wall. It bounces off with the same speed and angle....
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This note was uploaded on 02/02/2010 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas.
 Spring '07
 Swinney
 mechanics, Work

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