boxcarracer767 Posts: 1, Reputation: 1 New Member #1 Nov 30, 2004, 04:21 PM
Energy
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??
 drwls Posts: 14, Reputation: 2 New Member #2 Jan 8, 2005, 07:49 AM
<<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?? >>

You should have explained what portion of a circular arc makes up the track, the starting velocity (if any) and where the ball leaves the track. I will assume that the track is 1/4 of a circle, vertical where the ball starts with v=0, and horizontal where the ball leaves the track at the edge of the table.
In that case (1): V (while in the track) = sqrt (R -y), where y is the height above the top of the table. The velocity when the ball leaves the track at the edge of the table is
V = sqrt (2g R),
and is V horizontal at that point.
(2) Since (1/2) g t^2 = H during the falling period, with t = 0 being the time it leaves the table,
T = sqrt (2 H/g)
is when it hits the floor.
(3) X = V t = 2 R H
(4) Since the total distance it has fallen from the start of the track is R + H,
(Kinetic Energy)final = M g (R + H)

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