- #1
Natalie456
- 16
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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. Homework Statement
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?
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
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)
1. Homework Statement
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?
Homework 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
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)
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