Solving Homework: Speed of Mass on Wagon

AI Thread Summary
The problem involves a wagon with a vertical pole and a pendulum mass, requiring the calculation of the speed of the mass as it passes the pole. The initial potential energy of the pendulum mass is calculated to be 50J, with no initial kinetic energy. The equations for energy conservation are set up, but the user initially struggles with the momentum conservation aspect. After correcting an error in summing the masses, the final speed of the mass relative to the wagon is determined to be 3.87 m/s. The discussion highlights the importance of correctly applying energy conservation principles in physics problems.
Alex.malh
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Homework Statement


There is a wagon (W) on wheels with a vertical pole of 2m. (mass 20kg)
On the top of that pole there is a rope pendulum with an attached mass (M) of 10kg and length 1m.
The starting position is a pendulum angle of 60°.

No friction taken into account.
What is the speed of the mass compared to the wagon the first time it passes the pole?

Homework Equations


Ep= m * g * h
Ek= m*v²/2

The Attempt at a Solution


1.
EM0= 10kg * 10 m/s² * (1m *cos60) = 50J
EW0= 0
2.
EM1= 10kg*VM²/2
EW1= (10kg+20kg) *VW²/2

50J=EM1+EW1

And now i don't know how to continue.
 
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Alex.malh said:

Homework Statement


There is a wagon (W) on wheels with a vertical pole of 2m. (mass 20kg)
On the top of that pole there is a rope pendulum with an attached mass (M) of 10kg and length 1m.
The starting position is a pendulum angle of 60°.

No friction taken into account.
What is the speed of the mass compared to the wagon the first time it passes the pole?

Homework Equations


Ep= m * g * h
Ek= m*v²/2

The Attempt at a Solution


1.
EM0= 10kg * 10 m/s² * (1m *cos60) = 50J
EW0= 0
2.
EM1= 10kg*VM²/2
EW1= (10kg+20kg) *VW²/2

50J=EM1+EW1

And now i don't know how to continue.
Conservation of linear momentum?
 
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Likes Qwertywerty
Ok, thanks!
I made an error by summing the masses for EW1.
Solution is 3.87m/s for the interested.
 
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