Question

PLEASE SEE ATTACHED PICTURE.

Answer 1

Why did my teacher assume that vf does not change upon landing in the last sub-question? This makes no sense to me. Is there such a way that the skier can change direction of vf without changing the magnitude of it? Otherwise, the skier will collide in an inelastic collision with the ground (since he does not bounce off) and this means that all the vertical velocity is lost.

Answer 2

I would simply use the horizontal velocity to calculate work done by friction in that last-subquestion.

Answer 3

I presume she said vf didn't change and also change it's direction so that you wouldn't have to calculate the component of vf in the horizontal direction. That would require calculating the angle at which the skier lands. It can be done but it would certainly add more work.

As for the second half you are correct in your approach. The skier would come to a stop when the work done by friction equaled the skier's initial kinetic energy.

Answer 4

Psi, it is relatively easy to do to calculate the component of vf. What would be the most realistic approach here? Because I honestly cannot understand how an inelastic collision can simply mean change of direction of v. Are you an undergrad in physics?

Answer 5

By the way, how could we calculate this angle at which the skier lands?

Answer 6

The more realistic approach, especially if you're concerned with the integrity of the skier's skeleton would be to calculate the skier's landing angle and the components of velocity at landing, especially since 37.7 m/s is about 84mph.

Grad student here.

Answer 7

Oh a veteran of physics!

Answer 8

To calculate the angle at landing and the component velocities at landing, I'd start by calculating the height which the skier reaches, and that can be done by finding the component of the launch velocity in the vertical direction. Then use that velocity to calculate the kinetic energy of the skier's upward motion, which will in turn yield the skier's height gained via the equation:\[h=\frac{ v ^{2} }{ 2g }\]Then add that to the height of the ramp to calculate the total potential energy at that height above the ground. Equate that with kinetic energy to find the vertical component of velocity when the skier hits the ground. You'll be able to find the horizontal velocity just by finding the horizontal component of the launch velocity or by using the 37.7 resultant velocity and the vertical component velocity you calculated. You can then use those final component velocities to find the landing angle.

Without doing any calculations, it looks like the skier's vertical velocity at landing will be much bigger than his or her horizontal velocity.

Answer 9

Wow awesome! So, then I would use ONLY the horizontal velocity to calculate kinetic energy after the collision and the work done by friction? Should I tell my teacher that he might be mistaken?

Answer 10

I wouldn't do that. First, the launch angle isn't given, so in fact that necessitates using energy to find the landing velocity. Also that means that there is no way to find the components of the landing velocity. So in reality, your professor's only option was to state that the landing velocity was a horizontal velocity in order to give you the opportunity to use the Work-Energy Theorem.

Answer 11

I could set up an equation of the sort:
-(H+h)=10m/s sintheta t -1/2 9.8 t^2
cos35 *100=10m/s costheta t

Where theta is the initial launch angle.

Find theta and t. Use velocity equations to find final speed.

Is my method wrong? Then, I could plug in only the horizontal final speed to find work done.

Answer 12

The problem is that you have no way to determine what the launch angle is. When the problem is worked as your professor worked it, it's not necessary to know the launch angle. Without said launch angle you can't find the individual components of the final velocity because you won't know the trajectory angle at landing. You could try different θ to see what components you get for each θ, but the results will all be equally valid since there's not enough information in the problem to pick out the correct θ.

Answer 13

But, we have two equations with two unknowns! Isn't it just a matter of solving the quadratic?

Answer 14

@marianopolis, you have three unknowns:
1. h, the height above H to which the skier reaches
2. θ, the launch or landing angle (they'll be the same in a system with no drag)
3. t, the total time of flight.

So, you need a third equation.

Answer 15

@PsiSquared I do not want to enervate you x) but could you help me identify what is wrong with my two equations. In these two equations h and H refer to the height drawn on the schematic representation on the picture.
The equations seem to right to me that I do not see the point of using the max height the skier reaches!

I hope I'm not bugging you too much :S

Answer 16

You're not bugging me.

Sorry, I forgot about the h on the drawing. Your first equation is wrong. First, H-h is not really relevant. That's just the height above the top of the hill at which the skier launched. The equation thou would need would be as follows:\[y=H+vsin(\theta )t-\frac{ 1 }{ 2 }g t ^{2}\]You have to account for the starting height, and intuition should tell you that the skier will rise to some height, y-final-H, above H. While there are three unknowns in this equation--y, θ, and t--the good news is that with a decent amount of manipulation--which I won't do here--you can arrive at an equation that describes the time of flight of a projectile......or skier:\[t=vsin(\theta )+\frac{ \sqrt{\left( vsin(\theta) \right)^{2}+2gy _{0}} }{ g }\]where y0 is the starting height, in this case H. Now you can use your second equation to find either t or θ.

Answer 17

Hmm, then, I don't wish to adventure myself too much here. But are we left with two equations, two unknowns? :O

Answer 18

Yes, now we are.

Answer 19

Wow awesome.

Answer 20

Hmm... do you think I built a strong case to see my teacher about this?

Answer 21

Well, you could ask your teacher why he/she used the approach that he/she did, but unless your class had been introduced to the trajectory equations, it was likelier easier to go the route your teacher chose. More importantly, if you were learning about energy methods for solving problems, the teacher's approach was the correct one. Note that projectiles in physics problems don't have to necessarily survive in real life.

Definitely ask your teacher, though, about the problem and other methods of solution.

Note that the problem points out one fact of physics: that there is quite often more than one way to answer a problem.

Answer 22

Perfect, I'll ask him about the horizontal velocity calculation problem in the meantime.