Velocity after Free Fall and after Motion down an Incline

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  • #1
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I was doing some conceptual problems quickly before moving on to the numerical stuff when I glanced at this answer given in the solutions manual, and it surprised me:

Which object has greater speed at the bottom of its fall, object A of mass m dropped freely from height h or object B of mass m moving down an incline (no friction) from height h?

The manual says A. I googled around and found different people saying different things on this! I thought the answer was just that they have the same speed.

The manual gave this as its explanation: for A, v = sqrt(2gh), whereas for B, v = sqrt(2g(sinθ)h). Therefore, A is greater.

One of the online answers gave this: both A and B start with PE = mgh; so, at the end, all PE is now KE; so, KE = mgh = 1/2(mv^2); so both A and B have the same v, namely v = sqrt(2gh).

My analysis is that the manual mistakenly gives g(sinθ) for the acceleration on the incline. Maybe they were thinking of the force component?

I’d be grateful for help getting clear on this.
 

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  • #3
jbriggs444
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v = sqrt(2g(sinθ)h)
That formula would be accurate for a ramp of [diagonal] length h, not for a ramp of [vertical] height h.
 
  • #4
Dale
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I thought the answer was just that they have the same speed.
Yes, as you can see from a simple application of the conservation of energy. In both cases the same amount of gravitational PE was lost, in both cases there were no friction losses, so all the PE goes into KE meaning they have the same KE and therefore the same speed.
 
  • #5
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Thanks for the clarifications! I can see it more clearly now.
 

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