TBR physics section 1 passage 9 Q# 59

[hmania]

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Experiment 1 : Parapharsed: Student wish to measure g. They drop balls out of the window that is 20 minutes above the ground.


Question: IN experiment 1, what effect does double the height have on the velocity of the ball when it hits the ground?

a. velocity increase by a factor of 2
b. velocity increase by a factor of sqrt (2)
c. the velocity increases by factor of 4
d. the velocity is independent of height


I alluded to the equation vf^2=vi^2+ 2ad. Solved for vf by plugging in 2d instead of d. I got A but the answer is B. I just do not see how it could be B.

Help

 

[ToldYouSo]

Just look at the equation; to find vf you have to take the square root of it. So when you double d, you end up with sqrt (2d).

 

[cloak25]

mgh = 1/2mv^2

h is proportional to v^2, so if you double h you get 2h = v^2. solve for v. answer is B.

there are many ways to solve this problem. you can go the translational motion route as you were doing or the energy route, which I did. Choose whatever is easier for you because it won't matter how you solve it on the MCAT as long as it'll help you eliminate wrong answers.

 

[hmania]

Just look at the equation; to find vf you have to take the square root of it. So when you double d, you end up with sqrt (2d).

vf= sqrt( 2a (2d))

so you would get sqrt 4= which is 2 giving me A. 😕

where am I wrong.

 

[cloak25]

vf = sqrt (2d) only. 2a is a constant so you dont need to include it in the proportion.

 

[V5RED]

vf= sqrt( 2a (2d))

so you would get sqrt 4= which is 2 giving me A. 😕

where am I wrong.

(pre doubling)vf1=sqrt(2ad)

(post doubling)vf2=sqrt(2a(2d))=sqrt(4ad)

vf1/vf2=sqrt(2ad)/sqrt(4ad)=1/sqrt(2)

therefore vf2=sqrt(2)*vf1

 

[ToldYouSo]

Cloak25 is right. Ignore everything else except d, since that is what's being varied. vi^2 + 2a is constant, so disregard it. You end up with:

vf^2 = d. The square root of vf is proportional to d.

 

[GeorgiadisMD]

Cloak25 is right. Ignore everything else except d, since that is what's being varied. vi^2 + 2a is constant, so disregard it. You end up with:

vf^2 = d. The square root of vf is proportional to d.

lol no offence but that explanation was actually a little confusing even for me.

1. I used the equation Vf^2 = Vi^2 + 2ax
2. Solved for Vfinal: Vf = sqrt(Vi^2 + 2ax)
3. Figured out that you are doubling the height which equals x
4. So basically if you pull it out of the equation, you end up with sqrt(2) so the same thing is implied by writing the final equation like Vf = sqrt(2) * sqrt(Vi^2 + 2ax) since all you are doing is pulling out a sqrt (2) since it is added onto the x (also the height)
5. ????
6. Profit

Good luck!

 

[sciencebooks]

vf = sqrt (2d) only. 2a is a constant so you dont need to include it in the proportion.

I always just use simply proportions this way as well and it works.

Vf^2 = vi^2 + 2ad = 0 + 2ad

Vf^2 is proportional to d

(Lol, whoops, ToldYouSo basically said just this... 😛)

 

[milski]

lol no offence but that explanation was actually a little confusing even for me.

1. I used the equation Vf^2 = Vi^2 + 2ax
2. Solved for Vfinal: Vf = sqrt(Vi^2 + 2ax)
3. Figured out that you are doubling the height which equals x
4. So basically if you pull it out of the equation, you end up with sqrt(2) so the same thing is implied by writing the final equation like Vf = sqrt(2) * sqrt(Vi^2 + 2ax) since all you are doing is pulling out a sqrt (2) since it is added onto the x (also the height)
5. ????
6. Profit

Good luck!

Be careful, this works only when Vi is 0.

 

[GeorgiadisMD]

Be careful, this works only when Vi is 0.

You are right. The answer would be completely different if there indeed was some number for Vi.