Translational Motion Passage & A few of the questions

[arc5005]

I'm really struggling with my Physics review. Physics is by far my weakest link in the pre-med reqs.

I took Physics I ~4-5 years ago, and took Physics II about 1.5 years ago. Getting a B and A respectively, but not truly understanding the concepts. I'm a little concerned, but trying to do my best and I'm currently studying from The Berkeley Review books and Khan Academy. I have a few shortcomings and a few concepts and algebra basics that I just don't remember. I probably took algebra like 7 years ago.

Important information from passage:

• studying the effect of throwing 1-kg stone horizontally from the top of a 20-m rocket
• For different trials: A, B, C, and D
• Stone A and B are thrown from the top of the rocket while on earth
• Stone C and D are thrown from the top of the rock while on another planet, with half the gravitational force of the earth & the same atmospheric composition.

Data:

Stone A, g= 9.8 m/s^2, Vi = 10.0 m/s
Stone B, g= 9.8 m/s^2, Vi = 5.0 m/s
Stone C, g= 4.9 m/s^2, Vi = 10.0 m/s
Stone D, g= 4.9 m/s^2, Vi = 5.0 m/s

This was a 7 part question, in which I got 4 of the questions wrong and 3 correct. Below are the three of the 4 I got incorrect that I am struggling with:

1: If Stone C were thrown instead from a rocket is four times as tall as Galileo's, the stone would remain in flight

A. Twice as long
B. √2 times as long
C. Just as long
D. Four times as long

I chose D. Correct answer is A. I erased my thought process, so I could copy their thought process into the answer, so I don't have the original reasoning of what I did, but I'm having some trouble understanding why this is the correct answer...

so, h = 1/2gt^2, I understand that this comes from y = Vi*t + 1/2gt^2, and then my boyfriend explained to be that h = 1/2gt^2 is equivalent to t=√2h/g. How would I know that this is the equation to use in this question though?

t=√2h/g
t=√2(4h)/g
t=2√2h/g

so, I can factor out a 2 and put it infront of a square root?

2. Next Galileo compares the range (i.e., horizontal distances traveled by the tossed stones) on each planet, he found that the stone in Trial D traveled:

A. 2 times as far as in Trial B
B. √2 times as far as in Trial B
3. 1/√2 times as far as in Trial B.
D. The same distance as in Trial B.

I chose A, correct answer is B.

My work:

Stone B:
R=Vi^2 Sin 2 θ
R= 5.0 m/s * Sin 90 degrees
R = 5/10 = .5

Stone D:
R=Vi^2 Sin 2 θ
R= 10.0 m/s * Sin 90 degrees
R = 10/10 = 1

For some reason, I thought he was throwing it at 45 degrees, which makes no sense, because he the passage states it is being thrown horizontally. -_- so that's my mistake; however, I don't fully understand the book's reasoning to why the correct answer is B.

Their reasoning:
need to determine the x-displacement, aka Range. d= rate * time. We are provided a x-direction velocity, but are lacking the time in flight. To determine flight, we use a kinematic eqn for free fall in the y-direction. Somewhere they state that the R = Vi_x * t, where Vi_x is a given and the time if found through h = q/2 gt^2. what?

I thought the range eqn was: R=Vi^2 Sin 2 θ.

Next:

R = Vi_x * t
h = 1/2 gt^2

t=√2h/g

R = Vi_x (√2h/g)

R_stoneb = Vi_x (√2h/g)


R_stoned = Vi_x (√2h/g)

They then state: ViB = ViD, and gB = 2gD, so we can write:

R_stoned = Vi_B (√2h/1/2gB)
= √2 ViB √(2h/gB) = √2R_stoneB, which is choice B... what????? This is the partof the question I probably need most help with. How do you factor out a √2 from Vi_B (√2h/1/2gB)??????

Then they say that this answer should be solved using process of elimination, and state that choices C and D are eliminated, because we know Stone D will travel further than Stone B due to reduced gravity, which I understand and that makes sense. They then also say "they start from the same height, which relates to flight time by a square root factor, so there must be a square root in the relationship somewhere" I don't understand this statement. Will this be true for most problems then???


3. How should the range of Stone A and Stone D compare?
A. Stone A"s range is longer.
B. Stone B's range is longer
C. Their ranges are equal.
D. It would vary with the temperature of the atmosphere.

I chose C, correct answer is A. Once again they use the equation:

R = Vi_x * t
h = 1/2 gt^2

t=√2h/g

R = Vi_x (√2h/g)

R_stoneb = Vi_x (√2h/g)

I'm still having trouble understanding how this range equation is dervied from the equation that I learned:
R=Vi^2 Sin 2 θ

Also confused by the factoring out and square root stuff. Please help!

They are basically saying

R_stoneA= 2 ViD (√2h/2gD) = √2 ViD (√(h/gD)) = √2 R_stone D..

what? how did they factor out and get that √2 in front of ViD?

---

Members don't see this ad.

 

[aldol16]

(1) You would need to know this is the equation to use because you're working with a kinematics problem and you're trying to find time while being given acceleration and height and only one of the kinematic equations has both of those in it. So the equation is, as you say, delta y = v*t + 0.5*g*t^2. Where did the v term go in your analysis?

(2) For this one, you have to recognize that the x- and y-components of velocity are separable. So the x displacement simply becomes delta x = v*t. You have to find t first and you can get that from using the equation above on the y-component of velocity and height. In both the x- and y-directions, the time must be the same because that's the time the object is in flight/moving. So once you find t there, you can use that to find delta X. (3) is related to this question.