An object of mass m is allowed to slide down a frictionless ramp of angle Θ, and its speed at the bottom is recorded as v. If this same process was followed on a planet with twice the gravitational acceleration as Earth, what would be its final speed?
Multiple Choice Options:
(D) v/ √2
I know that the right answer is A, but I do not understand why... I will explain what I attempted to do in my attempt at solution.
Newton's second law: F = ma, with the force of gravity parallel to the incline being Fg*sin(Θ).
This means that my acceleration will be g*sin(Θ)
The Attempt at a Solution
So let's say that we have a distance Δx, which represents the length of the ramp at the angle.
I attempt to use a kinematic equation to solve for the length of the ramp and then try to apply the calculated result in another kinematic equation using the doubled value of acceleration.
My attempt is attached in a photograph, and I got B. I do not understand why what I did is incorrect and why A is correct.
The solutions manual (which I am attaching the solutions to here) says something different, and while I understand their approach.. I believe that they are basing it on the faulty assumption that with the doubled acceleration, the amount of time that the object will take to reach the base of the ramp is the same.
Could anyone please explain why my approach is wrong and why, and what I could possibly do instead?
Thanks in advance
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