Warning: Questions from AAMC 5

[Postictal Raiden]

Question #43 from the PS section:

The sled and the ride start from a location on the hill that is 10m lower than Point A [point A = 20m from ground]. How does the speed of the sled and the rider at point B [the bottom of the hill], starting from this location, compare to the speed of the sled and rider at point B when starting from the top of the hill?

a) It is lower by a factor of 4
b) It is lower by a factor of 2* sqrt(2)
c) It is lower by a factor of 2
d) It is lower by a factor of sqrt(2)



The answer is D. Why isn't A the correct answer? My reasoning is that V^2 = 2gh, so if h is reduced by a factor of 2, V should be reduced by a factor of 4. I don't understand their explanation.

I will post a couple more questions later.

Thanks

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[MedPR]

v=(2gh)^1/2. That's why.

 

[Postictal Raiden]

v=(2gh)^1/2. That's why.

But if h is halved, v will be reduced by (1/2)^1/2 = 0.707

 

[eric george delgado]

i solved it this way: E=U+K=0

for point 20m above:
mgh + 1/2mv^2 = 0

masses cancel out

gh+ 1/2v^2 = 0

1/2v^2 = -gh

1/2v^2 = - (-10m/s^2) x (20m)

[1/2v^2 = - (-10m/s^2) x (20m)] x 2

v ^2 =400

v=400^1/2

repeated for point 10 m above got and 200^1/2

and then divided first point by second point and got 2^1/2