
Answers and some guidance...
 1 m/s eastward
 400.6 m/s
 – 0.5 m/s
 Momentum is conserved, so depending on the ratio between the mass of the rowboat and your mass, your maximum velocity may be diminished.
 The water/hose system must have zero momentum if it is to remain in position to fight a fire (at rest). This is difficult due to the high mass and high velocity of the water leaving the hose having a large momentum in the direction of the flow of water. The fireman must provide an impulse (force applied for an amount of time) to change the water’s momentum to zero. b. 330.35 N
 Pointing the fan towards the rear of the boat provides a force (resulting from the air pushing against the blades) to the boat over a period of time – this impulse changes the momentum of the boat from zero to some value of v as a factor of the boat’s mass.
 The same force is experienced by the bug and the windshield. (Newton’s Third Law) From a conservation of momentum perspective, the bug and your car (the windshield is just part of this much larger mass object) both experience the same change in momentum. The car’s mass is so big, the resulting velocity change is negligible. For the bug, with tiny mass, the impulse is fatal!
 B hit A. Consider the location of the stop signs. If B was truthful, he would not have reached the other side of the intersection, having very low velocity such a short distance from the stop sign. If he had been stopped, A’s momentum would have carried them to the right, not towards the top of the intersection as it is drawn.
 A is the closest. Set up a coordinate system and assign a mass value of some unit (1 is convenient) for the shortest length of wire. All of the others have some mass value as a factor of this unit value. Each have an x,y coordinate associated with the center of mass of each piece. Use the center of mass equation for each coordinate, and you will have the location. I set the origin of my coordinate grid to be the equidistant point in the center of the four dots, where the wire bends to the right in the more or less geographic center. The resulting coordinate pair is negative in the x direction, and positive in the y direction, making the result in the upper left quadrant (I believe that is the 2^{nd} quadrant), which is the location of point A.
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Start with #110. I'll be adding more by this weekend.
Scratch #10. Here are the rest. 11, 17, 18, 20, 22, 23, 25, 27, 2934, 36, 39, 49, 52, 56, 67, 84
A total of 30 problems  due by Monday, November 6th (exam day). You are free to do more for practice  exam MC questions will be similar.
There will be one, multipart FRQ on the exam, as well. We will have plenty of time to work on these in class between now and then..
Answers as promised. Be sure to show your work for credit on the study guide problems. Just circling answers is just wrong...
1  C 2  E 3  C 4  E 5  B 6  A
7  B 8  E 9  D 11  B 17  B 18  B
20  A 22  C 23  E 25  A 27  D 29  D
30  B 31  D 32  E 33  C 34  B 36  E
39  E 49  D 52  E 56  C 67  C 84  A
PHYS Kinematics Retake Practice
Here are three example 2D kinematicstype problems where the projectile is launched at an angle:
51. A football is kicked with an initial velocity of 25.0 m/s at an angle of 35.0 degrees with the horizontal. Determine the time of flight.
52. A ball is launched into the air at an angle of 32.0 degrees to the horizontal with an initial speed of 18.0 m/s. Determine the horizontal distance.
53. A baseball was hit at 45 m/s at an angle of 55° above the horizontal. How long did it take to reach the top of its peak?
ANSWERS
51) 2.92 s 52) 29.8 m 53) 3.8 s
Here are the answers for 150 of the Unit A Study Guide. Show your work for full credit.
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 IGNORE
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