Solving Energy Problem w/Radius, g, H & Mass

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The discussion revolves around solving a physics problem involving a ball on a frictionless curved track atop a table. Key points include the need for clarification on the ball's initial height above the table, as this affects calculations for velocity, time to hit the floor, and energy. The correct velocity of the ball on the table is derived as v = sqrt(2gR), while the time to hit the floor is t = sqrt(2H/g). The distance the ball lands from the table is calculated as D = 2sqrt(RH). The total energy upon impact is maintained as mg(R + H), assuming conservation of energy without friction or air resistance.
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I am stuck on this problem:

A ball with a mass of M is on a frictionless curved track with a radius of R. The track sits atop a table that has height of H. Solve the following in terms of R,g,H, and M.
a) the velocity of the ball
b) the time it takes to hit the floor
c) the distance D the ball lands from the base of the table
d)the total amount of energy the ball has when it strikes the floor

Here are my answers, are these correct.
a) v= sqrt(2gH) ?
b) t=g*sqrt(2gh) ?
c)?
d) would i add 1/2mv^2 + MgR+MgH ?
 
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Initial velocity??
How is the "track" positioned in relation to the table; in particular, would the ball leave the track&table with a strictly horizontal velocity?

There is too little information here..
 
As arildno said: not enough information.

The track sits ontop of the table so I visualize a curve that puts the ball coming out horizontally ON the table, rolls across the table and then off onto the floor.

The crucial information missing is "how high above the table does the ball start?"

That, I think, is NOT H though it might be R if the track is a quarter circle. You seem to be assuming that since you include potential energy mgR.

Assuming that, when the ball starts it has kinetic energy 0 and potential energy (relative to the floor) of mg(R+h) since R+ h is its height above the floor. When it rolls onto the table, its potential energy is reduced to mgh so it must have kinetic energy equal to mgR: its speed is given by v2= 2gR, and, of course, is horizontal.
IF (a) means velocity of the ball ON THE TABLE, then v= sqrt(2gR), not 2gH.

The time the ball hits the floor is exactly the same as if it were dropped off the table:
-(g/2)t2+ h= 0 or t= sqrt(2h/g)

The distance from the table the ball hits is vt= sqrt(2gR)(sqrt(2h/g)= 2sqrt(Rh).

Assuming no friction or air resistance so we have "conservation of energy", when the ball hits, it has, of course, exactly the energy it started with: mg(R+h). Of course, all that energy would be kinetic energy now.
 
If it's a frictionless curved track (as claimed) then it needs to be nailed to the table, rather than just 'sitting atop of it'. Either that or you need to know the mass of the track so that you can figure how much the track accelerates in the one direction, while the ball accelerates in the other. :-p :wink:
 
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