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Acceleration on a Ramp

  1. Feb 23, 2017 #1
    I'm new to this, so I'm sorry if I am doing this process incorrectly. I feel like the answer to this should be straightforward and should just be √(2Lgsinθ). For some reason, I thought there was an acceleration vector along the ramp made up of x and y components, but I don't think this is accurate. My reasoning for this was that the rock is moving in both the negative y and positive x directions, and, since it starts at rest, it will need to accelerate in both of those directions so that it can move in those directions.

    1. The problem statement, all variables and given/known data

    A rock is sliding down a frictionless ice ramp of length L and angle theta to the horizontal. It has an initial velocity of 0. You are not allowed to reorient the coordinate plane so that the x-axis is along the ramp. What are the components of the rock's velocity at the bottom of the ramp?

    2. Relevant equations
    We aren't allowed to use the work-energy equations or Newton's laws, only kinematics formulas.
    Formulas:
    1.v=v0+at

    2.Δx=v0t + .5at^2

    3.v^2=v0^2+2aΔx

    3. The attempt at a solution
    I think I overcomplicated the problem because of the different axis orientation.

    Δy=Lsinθ=.5(g)(t)^2 -- assuming ay=-g
    --> time to travel down ramp=√(2Lsinθ/g)
    Δx=Lcosθ=.5(ax)(2Lsinθ/g)
    ax=gcotθ
    vx=√(2Lgcotθcosθ)
    vy=√(2Lgsinθ)
     
    Last edited: Feb 23, 2017
  2. jcsd
  3. Feb 23, 2017 #2
    Welcome to Physics Forums.

    According to this equation, it looks like the smaller the angle, the less time it takes to travel down the ramp. Can that be?
     
  4. Feb 23, 2017 #3
    I hadn't even thought of that. Do you happen to know how to make sense of this problem using solely kinematics equations? Does the rock have x and y components of acceleration?
     
  5. Feb 23, 2017 #4
    I had a hard time making sense of it. Although I thought that the acceleration along the ramp due to the gravitational force might be equal to gsinθ. Then if you broke that down into x and y components, you would have the x component ax = gsinθcosθ, and the y component ay = gsinθsinθ.
    You might try working it that way. You can always check it using the normal way of solving the problem.
     
  6. Feb 23, 2017 #5
    Thanks. I'll give that a shot.
     
  7. Feb 23, 2017 #6

    haruspex

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    You mean, it will descend at the same vertical rate as if the ramp were not there?
    Quite so, but I am baffled as to how the student is expected to determine that without recourse to Newton's laws. And how else is one to determine the horizontal and vertical components of the acceleration?
     
  8. Feb 24, 2017 #7
    So I would need to use Newton's Laws? I will try using that knowledge, as well. Also, can I state that a=gsinθ along the ramp if the axis for this problem is not to be set along the ramp? Thank you for your time.
     
  9. Feb 24, 2017 #8

    haruspex

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    Whether you eventually work in terms of an axis parallel to the ramp or horizontal and vertical axes, at some point you need to establish that the acceleration parallel to the ramp is gsinθ. But I do not see how you can do that without using Newton's laws.
     
  10. Feb 24, 2017 #9
    Thank you so much! I guess the wording of the problem and my teacher's strictness with regards to the axis threw me off, initially, but now I understand. Thanks, again!
     
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