Answer check: A ball on a spring launched into the air. find speed/height

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The discussion focuses on calculating the speed and height of a ball launched from a spring, utilizing the conservation of mechanical energy principle. Initial calculations yielded a speed of 5.97 m/s and a height of 3.63 m, but participants pointed out errors in the approach, particularly regarding the point of maximum velocity and the final position of the ball. It was clarified that the maximum speed occurs immediately after release from the spring, not at the peak height. After correcting the final position to 0.15 m, the revised speed was calculated to be 8.3 m/s, with the height remaining at 3.63 m. The conversation emphasizes the importance of accurately determining the initial and final conditions in energy conservation problems.
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Homework Statement



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Homework Equations



Since energy is conserved, MEinital = MEfinal

The Attempt at a Solution


1/2 mvi2 + mgyi + 1/2kxi2 = 1/2 mvf2 + mgyf + 1/2kxf2

After finding what equals 0, I am left with:
1/2kxi2 = 1/2mvf2 + mgyf
v=5.97m/s

I found yf by: mgyf = 1/2kxi2
yf = 3.63m


Just checking to see if these answers are correct. THank you for the help!
 
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joe426 said:
After finding what equals 0, I am left with:
1/2kxi2 = 1/2mvf2 + mgyf
v=5.97m/s
I get a different result. Please post more details of your calculation.
 
haruspex said:
I get a different result. Please post more details of your calculation.

First I found yf, which is the answer to part b.
Since energy is being conserved, I came up with PEf = KEi. This seems weird because usually there is no potential energy at the final position but it was the only thing I could come up with.

mgyf = 1/2kxi2
y= (1/2kxi2) / g
y=3.63m

Then,

1/2kxi2 = 1/2mvf2 + mgyf

sqrt((kxi2 - mgyf) / m) = v
v = 5.97m/s
 
Last edited:
joe426 said:
mgyf = 1/2kxi2
...
1/2kxi2 = 1/2mvf2 + mgyf
Compare those two equations. Shouldn't you get vf=0? ...
sqrt((kxi2 - mgyf) / m) = v
.. but you didn't because you dropped a factor of 2.
For what point of the trajectory do you think you should be finding the velocity in part (a)? What is the value of y at that point.
 
haruspex said:
Compare those two equations. Shouldn't you get vf=0? ...

.. but you didn't because you dropped a factor of 2.
For what point of the trajectory do you think you should be finding the velocity in part (a)? What is the value of y at that point.

oh ok. i understand. the fastest the ball will be going is right after its released from the spring, not at its highest position in the air.

So its,
1/2mvi2 + 1/2kxi2 = mgyf
vi = sqrt( 2gyf - kxi2 )

But now I'm stuck not knowing what it's final position is.
 
Using yfinal = 3.63m. I got the velocity to be 7m/s
 
joe426 said:
Using yfinal = 3.63m. I got the velocity to be 7m/s

I am wrong. The inal position is .15m because that's how far the spring is compressed an that's how far the ball moves before its released from the spring

V= 8.3m/s for part a
Y= 3.6m for part b


Thanks for the help!
 
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