Incline question regarding heavy and light mass

[traitorman]

I have a question about this:

Lets say you have a heavy mass rolling down a frictionless incline. Then you have a lighter mass rolling down the same frictionless incline. The time it takes for both masses to reach the bottom is equal.

Why would it be equal? The heavier mass has a higher potential energy (mgh) meaning it will have a higher kinetic energy and thus higher velocity.

KE=1/2mv^2 ==> v=square-root (2KE)

If velocity is the change in distance over time, and the heavier mass has a higher velocity, the heavier mass is traveling a greater distance over time. So shouldn't the heavier mass roll down the incline faster?

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[traitorman]

nevermind, i figured it out. when the masses start to roll, the potential energy is converted to kinetic energy. the amount of potential energy that is converted is equal to the amount of kinetic energy.

so 1/2mv^2 = mgh meaning the masses cancel out. cant believe i did not realize this.

 

[Morsetlis]

The bigger mass will have more momentum, though.

 

[traitorman]

will that change the times it takes for the objects to reach the bottom then?

 

[BerkReviewTeach]

will that change the times it takes for the objects to reach the bottom then?

No, but it will increase the impulse of the collision at the bottom and open up different questions for them to ask if you were to see a passage on a ball-rolling experiment involving a ramp.