- #1

Natalie456

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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.

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 Statement1. 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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