4 - test 01 – JACOBS AMBER – Due Feb 7 2008 6:00 pm 1...

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Unformatted text preview: test 01 – JACOBS, AMBER – Due: Feb 7 2008, 6:00 pm 1 Question 1, chap 4, sect 5. part 1 of 1 10 points Given: The battleship and enemy ships A and B lie along a straight line. Neglect air friction. A battleship simultaneously fires two shells (with the same muzzle velocity) at these two enemy ships. battleship A B If the shells follow the parabolic trajectories shown in the figure, which ship gets hit first? 1. both at the same time 2. B correct 3. A 4. need more information Explanation: The time interval for the entire projectile motion is given by t trip = t rise + t fall = 2 t rise , where t rise is the rising time from 0 to the maximum height, and t fall the falling time from h to 0. In the absence of air resistance t rise = t fall , h = 1 2 g t 2 fall , or t trip = 2 radicalBigg 2 h g . So the smaller is h , the smaller is t trip . In other words, enemy ship B will get hit first. Question 2, chap 4, sect 5. part 1 of 1 10 points The velocity of a projectile at launch has a horizontal component v h and a vertical com- ponent v v . Note: Air resistance is negligible. When the projectile is at the highest point of its trajectory, which of the following show the vertical and the horizontal components of its velocity and the vertical component of its acceleration in 3 columns? Vertical Horizontal Vertical Velocity Velocity Acceleration 1. v h 2. g 3. v v v h 4. v v 5. v h g correct Explanation: Basic concept: Newton’s second law of mo- tion vector F = mvectora The only force on the projectile is the gravita- tional force, which gives projectile a constant vertical acceleration of the magnitude g . There is no acceleration in the horizontal direction, which means at the highest point, the horizontal component of the velocity is the same as the initial value v h . One other obvious thing: the vertical com- ponent of the velocity is zero at the highest point, because from the trajectory, the projec- tile moves horizontally at the highest point. Question 3, chap 5, sect 6. part 1 of 1 10 points The net external force on a golf cart is 403 . 9 N north. If the cart has a total mass of 274 kg, what is the cart’s acceleration? 1. 1 . 34473 m / s 2 2. 1 . 38636 m / s 2 3. 1 . 42966 m / s 2 4. 1 . 47409 m / s 2 correct 5. 1 . 51974 m / s 2 6. 1 . 56695 m / s 2 7. 1 . 61587 m / s 2 Explanation: Basic Concept: vector F net = Σ vector F = mvectora test 01 – JACOBS, AMBER – Due: Feb 7 2008, 6:00 pm 2 Given: F net = 403 . 9 N north m = 274 kg Solution: a = F net m = 403 . 9 N 274 kg = 1 . 47409 m / s 2 to the north. Question 4, chap 2, sect 5. part 1 of 1 10 points Suppose the position equation for a moving object is given by s ( t ) = 3 t 2 + 2 t + 5 , where s is measured in meters and t is mea- sured in seconds....
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This note was uploaded on 04/28/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas.

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4 - test 01 – JACOBS AMBER – Due Feb 7 2008 6:00 pm 1...

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