PHYS 303K HW 4 - kodeih (njk359) HW 4 Coker (56625) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: kodeih (njk359) HW 4 Coker (56625) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This assignment covers the material in Chapter 5, namely Newtons 2nd and 3rd laws and their application. 001 10.0 points Before the time of Galileo and Newton, some learned scholars thought that a stone dropped from the top of a tall mast of a moving ship would fall vertically and hit the deck behind the mast by a distance equal to how far the ship had moved forward while the stone was falling. In light of your understanding of Newtons first law, what is true? 1. The stone will fall in some trajectory depending on the speed of the ship. 2. All are wrong. 3. Everyone on the ship will see the stone fall vertically if released from rest. correct 4. If the ship speed is fast enough, the stone will drop into the sea. 5. The stone will have a horizontal motion; it will hit the deck in front of the mast. Explanation: A stone will fall vertically when released from rest. If the stone is dropped from the top of the mast of a moving ship, the hori- zontal motion is not changed while the stone drops providing the air drag on the stone is negligible and the ships motion is steady and straight. From the frame of reference of the moving ship, the stone falls in a vertical straight line path, landing at the base of the mast. 002 (part 1 of 2) 10.0 points The position of a toy locomotive moving on a straight track along the x-axis is given by the equation x = b t 3- ct 2 + dt, where b = 2 m / s 3 , c = 78 m / s 2 , and d = 1006 m / s, x is in meters and t is in seconds. Find the time t when the net force on the locomotive is equal to zero. Correct answer: 13 s. Explanation: From Newtons second law of motion the force is directly proportional to the accelera- tion, which means zero acceleration will give a zero force. The velocity is v = dx dt = 3 b t 2- 2 ct + d, and the acceleration is a = d 2 x dt 2 = dv dt = 6 b t- 2 c = 0 , so t = c 3 b = (78 m / s 2 ) 3 (2 m / s 3 ) = 13 s . 003 (part 2 of 2) 10.0 points At the time when the net force on the loco- motive is equal to zero, the velocity of the locomotive is 1. negative. correct 2. zero. 3. positive. Explanation: Since v =- 8 m / s, the velocity is negative, see Part 1. 004 10.0 points Consider a toy car on an inclined ramp. The car is pushed up the ramp and released, it then coasts up the ramp for a while, slows down to a complete stop, and eventually rolls back down the ramp. This problem focuses on the time period when the car is coasting up the ramp, after its released but before it stops. kodeih (njk359) HW 4 Coker (56625) 2 vector v Let us neglect the resistive forces such as friction and air drag and consider the net sum vector F net of all the other forces. Which of the following statements describes the behavior of this net force vector F net during the time the car coasts up the ramp...
View Full Document

This note was uploaded on 09/25/2010 for the course PHY 56640 taught by Professor Coker during the Fall '10 term at University of Texas at Austin.

Page1 / 11

PHYS 303K HW 4 - kodeih (njk359) HW 4 Coker (56625) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online