Conservation of Energy - not expected result

[unr_]

I've been working over this relatively simple problem for quite a while, and I still cannot get the answer that I'm looking for.

Here is the problem:
A car of mass M traveling at speed V approaches a hill of height H. At the bottom of the hill the engine of the car is turned off.

Show that if V > SQRT(g(1+(2H/R)) the car would come off the hill at the top of the hill. R is the radius of curvature of the road at the top of the hill.



I have worked through this quite a few times, but I cannot get that answer.

Here's what I've done.
K_i = 1/2*mv^2
U_i = 0
U_f = mgH

To find K_f, I set g equal to the radial acceleration: g = v^2/R. Then, solving for that v, I get v = SQRT(gR).

So now my K_f is K_f = 1/2*mv^2 = 1/2*mgR.

Using the conservation of energy equation, I get:
1/2*mv^2 = 1/2*mgR + mgH.

After solving for V, I get V = SQRT(g(R + 2H)) which is not equal to the answer I am looking for.


If someone could please find where I've been doing something wrong or making a bad assumption, that would be awesome. A little guidance would be very helpful right now.

Thanks a lot,
J

 

[unr_]

Nevermind. Apparently an email was sent out correcting the answer in the worksheet that never made its way to my inbox.

 

[Motifs]

Sorry J, I was really late because I was really busy, I hope I wouldn't have stood you up! I will be in time next time on, Okay ?

You were right about the incorrectness of your first post's answer but the way you reasoned to find out the result, as what you posted in the first one, wasn't quite correct either, and I think if you hand that into your professor, he/she wouldn't give you a good grade.
One possible way I think to solve this problem is to start from energy conservation, just like what you have done. At the foothill, K=1/2MV^2. At the top of the hill, K_f=1/2Mv^2+MgH. Because of the EC you have K=K_f, but since the problem stated that the car flew off the hill, K must always be larger than K_f. Now that the car is drawing a curve when flying, you then also have g=v^2/R or v^2=gR. These means 1/2MV^2 > MgR/2 + MgH,...

The main point I think here is that you are not asked to go compute velos, or accels, (although you must write them down to substitute into other formula to get the desired results), so don't try to go deep into those or you will get 'hurt': the more you put them down, the more chances of being checked by your professors you would get, and of course you would have higher probabilities of making mistakes for computations. This is a small problem anyway, daijobu da to omoimasu.