TBR Physics Section 1 Questions- Have 2 Questions

[Gandyy]

Ok on Practice Passage 1 I dont understand the explanation to number 6. How do you resolve that expression you get at the end with 2v0x * sqrt (2h/2Gd)? I have no idea how to resolve that square root.

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[Gandyy]

Let me add that when I do try to resolve the root like I normally would and try to get the same expression in the answer I dont know how they got rid of the 2 in the 2v0x before the root.

 

[Gandyy]

Gotta Bump

 

[justadream]

@Gandy741

It would be helpful if you post the question + answer choices.

 

[Gandyy]

@Gandy741

It would be helpful if you post the question + answer choices.

Sure , here it is.

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

A. 2 times as far as in Trial B.

B. sqrt 2 times as far as in Trial B

C. 1/sqrt 2

D. The same distance as in Trial B.

Trials are as follows:

A: g= 9.8 ; Vo=10 m/s

D g=4.9 ; Vo 5.0 m/s

 

[justadream]

@Gandy741

Is the answer C?

 

[Gandyy]

@Gandy741

Is the answer C?

Thats what I got, but the answer is B "sqrt of 2"

 

[justadream]

@Gandy741

Is the V0 purely horizontal?

I can't seem to be able to get the correct answer.

 

[Gandyy]

Yes, sorry, the passage says it was "tossed horizontally off of a rocket."

 

[mehc012]

Horizontal distance is based solely off the time in flight, given identical initial velocities.
In half gravity, the stone will fall more slowly by a factor of sqrt(2), which means it is in the air for sqrt(2) times LONGER than when g=9.8, which means that the distance travelled will also be sqrt(2) longer. Or, in math terms:

h = gt^2 ________ (you use this equation because you know initial y velocity and height are constant - y velocity is zero- between trials, g changes, and you need to find t)
t1 = sqrt(h/g)
t2 = sqrt(2h/g)
t2 = t1*sqrt(2)

distance = v*t
d1 = v*t1
d2 = v*t2 = v*t1*sqrt(2)

 

[justadream]

@mehc012

But aren't the horizontal velocities different between the two scenarios?

 

[mehc012]

@mehc012

But aren't the horizontal velocities different between the two scenarios?

Missed that part of the question, sorry. From the math in the initial post though, I have a question...is the height identical both times? If so, the time works out as above, and it turns into
d1 = 2v2*t1
d2 = v2*t2 = v2*t1*sqrt(2) = 0.5d1*sqrt(2)

sqrt(2)/2 = 2/2sqrt(2) = 1/sqrt(2)

so, yeah, if the heights are the same, it should be D. But at this point, I'm not certain that the problem is complete as posted...there have been bits and pieces left out throughout, and of course, if OP is getting it wrong, the odds are they are overlooking a crucial bit of info (the most common reason to miss problems), so only providing what seems pertinent is kind of defeating the purpose of having another set of eyes on it.

Honestly, this whole thing would be easier if people would post the ENTIRE problem in the initial post. It's far harder to piece it together from 2 posts, neither of which has the whole story. People help out on these things as a favor, it is only hurting your chances of getting a good response if you do not make it easy on us by providing all of the info needed.

 

[justadream]

@mehc012

Do you mean "it should be c"?

I tried this question like 3 times (with the information the OP provided) and keep getting C.

 

[mehc012]

@mehc012

Do you mean "it should be c"?

I tried this question like 3 times (with the information the OP provided) and keep getting C.

Holy crap, yes. I am sorry, I am at work and it is really difficult to scroll all of the posts on this crappy tiny screen. If I were at home, I could get my book out and just look at the actual problem, but alas, I am limited to what is given above.

I have to reiterate, though, that OP's math (2v0x * sqrt (2h/2Gd)) makes me curious as to whether the heights are identical. Then again, why you would ever be multiplying the velocity by 1/d makes no sense to me, so...

I remember doing this problem and not having any issue with it, so I am really thinking there is some info missing.

 

[Cawolf]

We are comparing the stone in Trial D (g = 4.9 m/s^2 and v0 = 5 m/s) to Trial B (g = 9.8 m/s^2 and v0 = 5 m/s).

We want to know how the range of D compares to B.

Range = Velocity x time, we know the horizontal velocity, so we simply need the time it takes for each stone to fall.

To determine the time in flight, we use the kinematics formula that y = .5gt^2

This can be rearranged to solve for the time to fall as t = sqrt(2y/g) and we know both stones start at the top of the rocket where y = 20m.

We can either solve both equations or just make a ratio of the two and cancel out like terms.

Range D/Range B = Rd/Rb = [v0(sqrt(2y/4.9))]/[vo(sqrt(2y/9.8))]

= sqrt(1/4.9)/sqrt(1/9.8) = sqrt(9.8) / sqrt(4.9) = sqrt(9.8/4.9) = sqrt(2)

We can state that the range of D is sqrt(2) greater than B.

Edit: I looked back and saw you posted the data for trial A - this problem is comparing trials B and D as I posted in my first paragraph.

So @mehc012 had it a few posts back.

 

[mehc012]

We are comparing the stone in Trial D (g = 4.9 m/s^2 and v0 = 5 m/s) to Trial B (g = 9.8 m/s^2 and v0 = 5 m/s).

We want to know how the range of D compares to B.

Range = Velocity x time, we know the horizontal velocity, so we simply need the time it takes for each stone to fall.

To determine the time in flight, we use the kinematics formula that y = .5gt^2

This can be rearranged to solve for the time to fall as t = sqrt(2y/g) and we know both stones start at the top of the rocket where y = 20m.

We can either solve both equations or just make a ratio of the two and cancel out like terms.

Range D/Range B = Rd/Rb = [v0(sqrt(2y/4.9))]/[vo(sqrt(2y/9.8))]

= sqrt(1/4.9)/sqrt(1/9.8) = sqrt(9.8) / sqrt(4.9) = sqrt(9.8/4.9) = sqrt(2)

We can state that the range of D is sqrt(2) greater than B.

Edit: I looked back and saw you posted the data for trial A - this problem is comparing trials B and D as I posted in my first paragraph.

Ah, that makes sense...I was wondering about the v0 issue.

So, the original math which I posted was correct because the velocities are actually the same.

See, this is why I strongly believe that if people want a second perspective on a problem, they should post the full passage/question. Doing otherwise merely propagates the misinterpretation/misreading/overlooking of facts which most commonly are the cause of errors. Plus, it's always best to minimize the extra footwork the people helping you have to do - more likely to get help that way!

 

[Cawolf]

Yah I generally ignore posts w/o a problem but I am doing physics chapter 3 right now and had my book open!

 

[justadream]

This question almost shattered my kinematics physics confidence.

@Cawolf restored it.

 

[mehc012]

This question almost shattered my kinematics physics confidence.

@Cawolf restored it.

You've just got to trust in your calculations. We were all doing the problem correctly with the information given. That's all we can expect.