• What is the magnitude and direction (if any) of the acceleration of the ball at the instant it reaches the highest point in its trajectory? What is its acceleration just before it hits the ground upon its return journey? (0m/s2: 20; 9.8m/s2 down: 14; no answer: 16; other 3) Most got the second part correct.

• At the highest point the ball reaches, the ball stops, which means that the velocity of the ball is zero. So, the magnitude and direction of the accleration of the all would be zero, also.

• the magnitude is 9.8 m/s^2. i would say the direction is positive; going up would be negative. the acceleration is still 9.8 m/s^2. What does positive or negative mean?? (direction is downward, +/- only has meaning with a defined frame of reference).

• 1) The acceleration would be 9.8m/s^2. There wouldn't be a direction because the velocity is 0. 2) The acceleration would be -9.8m/s^2. Why does the velocity have anything to do with it?

• From the moment the ball is released upward to the time it returns to its initial position, the ball is in free fall flight and has a constant acceleration (a = -g = -9.8m/s^2) with a magnitude of 9.8m/s^2 in the negative (downward) direction. ESSENTIALLY PERFECT! Why is it constant? Only one force acts, that of gravity: F=mg (down) =ma

• Estimate the time it takes for a free fall drop from 10 meters height. Also, estimate the time a 10 m platform diver would be in the air if he takes off straight up with a vertical speed of 2 m/s (and clears the platform of course!). Please provide a brief description of how you arrived at your estimates. (A: 23 correct B 17 numerically correct, but many used slightly faulty thinking in getting there. Correct answer: 1.43 sec for A, 1.65 sec for B)

• A free fall drop from 10 meters would take 1.02 seconds because the force of gravity is 9.8 m/s^2. Dividing 10 by this number gives a quotient of 1.02 ?Units??? About 9 people answered this way

• For the first part of the problem, I figured an equation to look like; 10=1/2(9.8)t^2 therefor it took roughly 1.4s from a 10m height. For the second part, I got that it takes about .2s for the diver to reach it's maximum height. This I can use to add to the first part of the question and come to the conclusion that it took about 1.6s for the diver to reach the water. First part is correct, second part makes a slight mistake, and makes the problem a little more complicated than necessary; most who got about 1.6 sec answered something like this.

• Part B: Just use K1 and solve it as a quadratic equation; but be careful to take the correct root (of the two that the math gives you).

Examples– Chpt. 2

Since we didn’t go through this question in detail in class, here is the Solution:

a). Use K3 and note the final position is below the initial position (i.e. change in x is –h): vf

2= vo2-2g(-h) =>, vf =(vo

2+ 2gh)1/2

b). Use K1 and the result from a): -vf=-vo-gt => t = [(vo2+2gh)1/2 –vo]/g (note,

you could also just use K2 directly).

c) As discussed in class for 2-112, the speed will be the same as in a, since the sign of vo matters not at all in the result given for a).

d). The difference between this case and b). Is just that the sign of vo changes, so we have: t = [(vo

2+2gh)1/2 +vo]/g

Examples– Chpt. 2

We know t (2.5 sec) and vf (0.0m/s) for the flight up, SO: 0=vo -9.8m/s2(2.5s), or vo=+24.5 m/s. With this info, we can use K2 to get the height (H):

H= 0.m + 24.5 m/s*1.5 s -4.9m/s2(1.5s)2 = 25.7 m = 26 m. (2 sig. figs. justified)

Examples– Chpt. 2

When you drive somewhere in your car, is the change in the reading on your odometer equal to, greater than, or less than the magnitude of the displacement you experienced on your journey? (GE:29; EQ. 7; LE: 3; no answer 14)

• I think it is equal to the magnitude of the displacement. The odometer measures the number of miles the car has travelled, but it does not inform about the direction. The sum of magnitudes is not the magnitude of the sum!!!

• The reading on the odometer is equal to the displacement only if the car moved in a straight line. The odometer is greater than the magnitude if any turn is made. That’s the idea!

Please give a concise, simple description of the physical significance of the dot product of two vectors.

(projection: 13; definition: 9; other 16; none 16) • The main significance is that it allows for turning a vector into

a scaler quantity. Also, its main use is if A and B are vectors, their dot product = |A||B|Cos(theta). This is the definition; is it the physical significance??

• the dot product is the product of the magnitude of one vector and the scalar component of another vector, so basically its a way to combine vectors and see how they look and how they affect each other. This captures the book’s expression

• The physical significance of the dot product of two vectors would be that it is a projection of vector A onto the direction of vector B multiplied by the length of vector B. That captures the essence; and we’ll find this way of looking at it useful as we start using dot products in chapter 7.

Examples– Chpt. 3

(d) What is the angle between the two vectors?

I didn’t solve part c in class, so here it is: the vector c must add to the answer from b to give all components equal to zero, hence:

c =-5.0 i + 4.0 j +3.0 k (all numbers in m, and i, j, k are unit vectors (with out the carat, which I don’t know how to make Powerpoint do).