Does the angle of the slope matter in these cases? Why/not?

[grooveactiva]

Homework Statement

This is not a direct homework problem. Rather, I modified several that I saw that had different slopes but with the same initial velocity, and the same height answer: An icy, frictionless road slopes upward with angle Θ ° above the horizontal. A snowboarder of mass m approaching the road with a speed of v [m/s]. How high in the vertical direction does the snowboarder travel up the road before stopping? Assume no friction.

Homework Equations

½mv² = mgh, where h is the vertical height that we want.
The masses cancel, so: ½v² = gh.

The Attempt at a Solution

h = v²/(2g)
Or do we use v= v₀sinΘ because the sine of the angle is the opposite of that angle, which corresponds to height:
h = (v₀ sin Θ)²/(2g)
In other words, doesn't gravity affect how high up in the vertical direction that the snowboarder would travel?
I have seen similar problems with differing angles, 22°, 25°, 33° with the same velocity. All have same answer in terms of height. But wouldn't it take more energy to go up a steeper slope, even with no friction? If this problem were done with Newtonian kinematics, wouldn't the angle affect the result of how high in the vertical direction (and along the slope itself) the snowboarder would travel?

 

[jim mcnamara]

Three snowboarders, of identical mass m and identical velocity v, will have zero potential energy (max height) when they attain zero velocity. Is this true on a friction-less slope of any angle? Hmm.

So. Why does the angle matter?

Can you show mathematically how the angle of the slope changes the height required to achieve zero potential energy? I say "no". Remember
$M_1=M_2=M_3$

Consider applying the 'potential energy' concept to your question.

 

[haruspex]

All have same answer in terms of height. But wouldn't it take more energy to go up a steeper slope, even with no friction? If this problem were done with Newtonian kinematics, wouldn't the angle affect the result of how high in the vertical direction (and along the slope itself) the snowboarder would travel?

A steeper slope means you lose speed faster but also gain height faster. In the end, without friction, all KE is converted to GPE, so the height is the same.

 

[grooveactiva]

Thanks, but I have another question: When doing problems like this, how do I know when I can ignore the angle? I am working out another problem that says: A compressed spring with spring constant k is pushing an air cart of mass m up an air track of length L that is sloped at Θ° above the horizontal. Before moving up the air track, the air cart becomes disenaged with the spring. To what distance x was the spring originally compressed?

Conservation of energy: All PE of compressed spring becomes kinetic energy to move air cart up slope. Distance along slope depends on angle Θ?

½kx² = mgLsinΘ, ⇒ d = √(2mgLsinΘ/k)

or just
½kx² = mgL, ⇒ d = √(2mgL/k)

 

[haruspex]

Thanks, but I have another question: When doing problems like this, how do I know when I can ignore the angle? I am working out another problem that says: A compressed spring with spring constant k is pushing an air cart of mass m up an air track of length L that is sloped at Θ° above the horizontal. Before moving up the air track, the air cart becomes disenaged with the spring. To what distance x was the spring originally compressed?

Conservation of energy: All PE of compressed spring becomes kinetic energy to move air cart up slope. Distance along slope depends on angle Θ?

½kx² = mgLsinΘ, ⇒ d = √(2mgLsinΘ/k)

or just
½kx² = mgL, ⇒ d = √(2mgL/k)

The difference here is that you are given the distance along the slope. To figure out the energy required you need to find the height, and to get that from the distance you need the angle.
But in general you don’t need to determine in advance whether you need the angle. If you don’t then it will cancel out.