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

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.
  • #1
OTSEngineer
10
5
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
Skiing.jpg

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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  • #2
OTSEngineer said:
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?
 
  • #3
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.
 
  • #4
PeroK said:
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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  • #5
OTSEngineer said:
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?
 
  • #6
haruspex said:
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##.
 
  • #7
OTSEngineer said:
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.
 
  • #8
jbriggs444 said:
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.
 
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  • #9
OTSEngineer said:
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 :smile:
 
  • #10
OTSEngineer said:
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##?
 
  • #11
jbriggs444 said:
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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  • #12
berkeman said:
Yes, can you show your work to get that new answer so we can check it please? Thanks :smile:
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
 
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  • #13
PeroK said:
So, ##h = \frac r 2##?
That was my conclusion.
 
  • #14
NECRORESPONSE:
haruspex said:
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.
You are discounting the capability of the skier change the distance of her CM from her feet (to keep that force minimal if possible thereby forcing a ballistic trajectory without inducing drag forcxe)
 

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

1. What is momentum and how is it related to velocity?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. In other words, it is the quantity of motion an object has. The greater the momentum, the harder it is to stop or change the direction of the object's motion. Velocity is directly related to momentum, as an increase in velocity will result in an increase in momentum.

2. How is kinetic energy (KE) calculated and what does it represent?

KE is calculated as 1/2 times the mass of an object times its velocity squared. It represents the energy an object possesses due to its motion. The greater the mass and velocity of an object, the greater its KE will be.

3. What is potential energy (PE) and how is it related to kinetic energy?

PE is the energy an object possesses due to its position or state. In the case of a skier, it can be the energy stored in their muscles or the energy stored in the position of the skis on the slope. PE is related to KE through the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. As the skier moves down the slope, their PE decreases while their KE increases.

4. How can momentum, KE, and PE be used to solve a skier velocity problem?

In a skier velocity problem, the initial and final states of the skier's motion are known, along with other relevant information such as the mass of the skier and the slope angle. By applying the principles of conservation of momentum and energy, equations can be set up to solve for the unknown variables, such as the skier's velocity. This involves using the equations for momentum, KE, and PE, along with the given information, to create a system of equations that can be solved simultaneously.

5. What are some real-life applications of using momentum, KE, and PE to solve problems?

The principles of momentum, KE, and PE are used in a variety of real-life applications, including sports, transportation, and engineering. For example, in sports like skiing, understanding these concepts can help athletes improve their performance and techniques. In transportation, these principles are used to design efficient and safe vehicles. In engineering, they are used in the design and construction of buildings and structures to ensure their stability and safety.

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