|QUIZ 4 CHAPTER 3: READ ALL BOOK EXAMPLES SINCE SOME ARE LIKE HINTS TO QUIZ PROBLEMS BELOW ! (REAL EXAM 2)|
https://www.physics.upenn.edu/sites/www.physics.upenn.edu/files/Error_Analysis.pdf ON STATISTICAL ANALYSIS OF PROJECTILE MOTION EXP.
|SAMPLE TEST 2|
|Discussions will be provided below as needed.|
DISCUSSION QUESTIONS: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 16
EXERCISES/PROBLEMS: 2*, 3*, 7*, 10*, 18*, 19*, 29*, 30*, 32*, 33, 35, 36,
56*, 62*, 65, 67(TRY), 78 (TRY), 81(TRY), 82,
|2. (NEW ED) READ EXAMPLE
3.1. See section 3.1. Use the definition: Average velocity
vector = change in r-vector/change in time. See eqn 3.2 .
(a) Remember each component of the average
velocity vector is simply given by the x or y displacement over
change in time: Vx average = (xb - xa)/(tb
- ta) and
(b) Compute the magnitude of the displacement
vector whose x and y components are given by
(xb - xa)*i + (yb - ya)*j , where i and j in bold represent our standard unit vectors. Review quiz 1.
ED) READ EXAMPLE 3.1 . Great
problem for you future or current web designers out there using
the same procedures as previous problem. The position vector = 2.5t2*i
+ 5.0*t*j = x(t)*i + y(t)* j, unit vectors in bold.
(a) Remember each component of the average velocity vector is simply given by the x or y displacement over change in time: Vx average = (xb - xa)/(tb - ta) and
Vy average = (yb - ya)/(tb - ta). Here is an example for the x-component of the problem:
x((t) = 2.5t2. We evaluate x(2) and x(0) by plugging in the time values into the formula for x(t). Compute x(2) - x(0), then compute [x(2) - x(0)]/(2 - 0). Also compute: [y(2) - y(0)]/(2 - 0). Once you get each component of the average velocity, find the magnitude using the Pythagorean Theorem. Find direction by determining the angle the vector makes with the x axis and the quadrant it points. See Ch. 1 and 4 problems and icq's/ quiz 1.
(b) For each formula x(t) and y(t), compute the derivative: for example dx/dt = 5.0*t. Differentiate and plug in the numbers for each component at the 3 different times.
(c) Sketch qualitatively the curve of motion on an x-y axes like we did with projectile motion and other examples. Hint: The curve is a parabola since x is proportional to the square of y.
|MORE DISCUSSIONS TO FOLLOW; WITH RESPECT TO PROJECTILE MOTION, THESE PROBLEMS CAN BE CONSIDERED "WARM UP "SINCE THEY ARE EASIER BUT ARE POWERFUL TEACHING AND LEARNING TOOLS.|
|10. (NEW ED) Think two dimensionally. Review the logic beyond the take home virtual lab: click here. Read book examples. In this case it is only matter of finding the time from y motion equations and plugging that time into an x-motion equation.|
|18. (NEW ED) SEE class notes and textbook examples on this asymmetric path with vertex at the highest point. Contrast this with previous exercise, where the particle returns to the same vertical level. In this problem, the particle lands at a lower level. General comments: Remember Vy = Voy - gt, where Voy = Vo*sin 51.0. You are given the time of flight. Vx is constant. The acceleration is constant and points downward.|
|19. (NEW ED) General comments. There is a nice derivation of the parabolic path on page 79. See equation 3.27. You could use this expression to find the height but please understand how to derive it. Of course you must use other means to find the y-component of velocity since equation 3.27 does not give that.|
|29. (NEW ED) SEE LECTURE CLASS EXAMPLES WHERE I SHOWED A PARTICLE IN UNIFORM CIRCULAR MOTION AT 4 OR SO POINTS ALONG THE PATH. SEE ALSO EXAMPLES 3.11 AND 3.12. Remember, the centripetal acceleration vector always points to the center of the circle whereas the velocity vector is always tangent to path.|
|For relative motion
problems, covered in the next 4 exercises, always
VPAx = VPBx + VBAx
VPAy = VPBy + VBAy,
where we have written the x and y components of the following equation
vector-VPA = vector-VPB + vector-VBA . This simply states that the velocity of a particle P relative to frame A equals the velocity of the particle relative to frame B plus the velocity of frame B relative to frame A. Remember the BART train examples. Let Frame B be the train. Let Frame A be the station, and let P be the person on a skateboard coasting along the floor inside the train. Recall in one example I said if VPBx = 10 mph and VBAx = 80 mph (Bart train's top speed), then VPAx = 90 mph as expected.
|32. (NEW ED) You are to find VPBx
in three cases when VBAx =13.0 m/s . Discussion of (a)
opens the door to the other parts:
(a) VPAx = VPBx + VBAx translates to 18 .0 m/s = VPBx + 13.0 m/s. Solve for VPBx .
(b) Same equation but now VPAx= -3.0 m/s.
(c) Use the same method to reach an obvious conclusion.
|33. USE (NEW
VPAx = VPBx + VBAx
VPAy = VPBy + VBAy, .
A = EARTH FRAME
B = RIVER FRAME
P = BOAT
ON YOUR PAPER , LET UP BE NORTH (N), DOWN BE SOUTH (S), RIGHT BE EAST (E) AND LEFT BE WEST (W). RIGHT IS THE POSITIVE X DIRECTION AND UP IS THE POSITIVE Y DIRECTION ON YOUR AXES.
VPAx = +0.40*cos45 (m/s)
Note: SOUTHEAST means 45 degrees south of
east, in the 4th quadrant; NORTHWEST means 45 degrees north of
west, in the second quadrant, etc.
USING THIS INFO, FIND THE MAGNITUDE OF
VECTOR-VPB BY FINDING BOTH OF ITS COMPONENTS AND
APPLYING THE PYTHAGOREAN THEOREM TO THEM.
|For problems 35 and 36, see in-class lecture notes and book examples 3.14 and 3.15 for reference. With regard to #35, see nearly identical lecture example I did with a river flowing to the right and a boat velocity relative to the river directed upward on the paper. Thus the boat velocity relative to the shore was not straight upward on the paper.|
|MORE DISCUSSIONS SHOULD FOLLOW BUT TRY THE REMAINING PROBLEMS USING THE ABOVE CONCEPTS. THEY REPRESENT AN IMPORTANT FOUNDATION, LIKE "WARM UPS" FOR THE MAIN RACE. FREE TO EMAIL ME !|
|56. (NEW ED)This problem is like the egg and the professor, #80, Ch. 2. The only difference is the non-vertical launch direction toward the front of the incoming boat. The horizontal displacement of the equipment from the launch point is x = Vo*cosao*t. Time t of course is obtained using Ch. 2 methods on the y- motion equations for projectile motion. Where V is the boat's speed, add V*t to the value of x to get the initial distance D of the boat from shore,|
|62. (NEW ED)
(a) Go to page 84, figure 3.26 to see why you must always aim above a target to hit it for any launch angle. If you aim at or below the target you will never hit target because of the downward pull of the gravitational force also known as the weight. Note: The difference between the straight line and the actual path is (1/2)*g*t2, simply seen by evaluating
Vo*sinao*t - (Vo*sinao*t
- (1/2)*gt2) ,