# Using Momentum, KE and PE to solve this skier velocity problem

• OTSEngineer
In summary, the skier has a potential energy that is equal to m*g*h at the start position, and kinetic energy that is equal to that potential energy m*g*h at the top of the circular hill. However, the skier loses contact with the snow just as they reach the crest of the second hill, and their speed decreases as they move closer to the top. Centripetal force is responsible for the skiers height at the peak, which is 16.4 meters.f

#### OTSEngineer

Homework Statement
A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of 32.8m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill.
Relevant Equations
r=32.8m
Solve for height h

See a picture of the question above.
My thoughts are:
1. dp(y)/dy is negative such that when going up the slope, the momentum in the y direction is equal to 0 just as the skier reaches the top of the circular section.
2. Given that there is no friction on the slopes, the energy of the skier (potential+kinetic) is the same across the dashed line for all time.
3. The ground does not have a force acting on the skier when it reaches the top, because the skier had just enough momentum to reach the peak without interacting with the snow.
As such, my thoughts are that at the start position, the skier has a potential energy that is equal to m*g*h. At the top of the circular hill, the skier has kinetic energy that is equal to that potential energy m*g*h. However, the momentum p=m*v(x), exclusively.
I'm thinking that the potential energy gained by starting at height h is equal to the kinetic energies at the two points where the slopes and the dashed line intersect. As such, I could say that the kinetic energy at the first intersection point is equal to the kinetic energy at the second; 1/2*m*(mag(v))^2 at first intersection would then be equal to 1/2*m*(vy)^2 at the second. And, that the kinetic energy at the first point is equal to the potential energy at the starting point. I think that the angle between the velocity vectors at the first intersection point would determine height h, however, I'm not certain how to find that value with radius r. I think that the length r is the length of the hypotenuse of the two velocity vectors at the first intersection point, but I need 1 extra metric about the triangle (angle or the length of one side) to calculate h. Maybe I'm missing something? Or maybe my assumptions are incorrect?

Thanks,

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Maybe I'm missing something?
Calculating the skiers speed on the crest of the second hill should be the easy bit. Can you do that?

The more difficult question is why the skier would lose contact with the snow?

You will be better off if you set momentum in the y-direction aside. This is a two-step problem.
Step 1: Find the speed at which the skier just barely keeps contact with the surface at the top of the circle. Hint: Draw a free body diagram.

Step 2: Find the height ##h## which will result in the skier having that speed.
Hint: Mechanical energy is conserved.

Calculating the skiers speed on the crest of the second hill should be the easy bit. Can you do that?

The more difficult question is why the skier would lose contact with the snow?
##V=sqrt(2*g*h)##, where h is the height shown in the picture.

The skier would lose contact with the snow if there still is momentum in the y direction right when they reach the crest of the second hill. Or, if the momentum in the y direction reaches 0 just as the crest is reached.

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The skier would lose contact with the snow if there still is momentum in the y direction right when they reach the crest of the second hill.
True, but how will you work out what the vertical momentum is?
Easier to think in terms of forces and accelerations. What is the skier's acceleration while going over the arc?

True, but how will you work out what the vertical momentum is?
Easier to think in terms of forces and accelerations. What is the skier's acceleration while going over the arc?
Gravity is the only force acting on the skier, so the acceleration is ##-9.81m/s^2##.

Gravity is the only force acting on the skier, so the acceleration is ##-9.81m/s^2##.
Given that the skier is tracing out a circular path that matches the shape of the target hill, there is another expression that must also match the skier's acceleration as the peak is being passed.

Given that the skier is tracing out a circular path that matches the shape of the target hill, there is another expression that must also match the skier's acceleration as the peak is being passed.
Ah, centripetal force? That makes the height, ##h##, 16.4m. I did not think about this. I will review centripetal force. Thank you.

erobz, jbriggs444 and berkeman
Ah, centripetal force? That makes the height, , 16.4m.
Yes, can you show your work to get that new answer so we can check it please? Thanks

Ah, centripetal force? That makes the height, ##h##, 16.4m. I did not think about this. I will review centripetal force. Thank you.
So, ##h = \frac r 2##?

Given that the skier is tracing out a circular path that matches the shape of the target hill, there is another expression that must also match the skier's acceleration as the peak is being passed.
Unfortunately the question makes the invalid assumption that the skier can lose contact at the peak without having lost contact sooner. Perhaps it is a trick question, but more likely it was a blunder.

There is a geometric way to see this. Having lost contact at the top, the onward path would be a parabola until making landfall. The parabola would osculate the arc of the hill: tangential with the same radius of curvature at the top. Since that is the smallest radius of curvature for the parabola, the completed circle of the arc lies entirely inside it.
To arrive at the top with exactly the right speed, she must have leapt up along that parabola from a point before the arc section.

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OTSEngineer, jbriggs444, erobz and 1 other person
Yes, can you show your work to get that new answer so we can check it please? Thanks
m*g*h=(1/2)*m*v^2
v=sqrt(2gh)
Fg=Fc
g=(v^2)/r
g=(2gh)/r
r=2h
32.8=2h
h=16.4

PeroK
So, ##h = \frac r 2##?
That was my conclusion.