What is the maximum speed and acceleration of a train on a big dipper?

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SUMMARY

The maximum speed and acceleration of a train on a big dipper can be calculated using principles of gravitational potential energy (GPE) and kinetic energy (KE). Given a radius of curvature of 20.0 m and a vertical drop of 20.0 m, the maximum speed at the bottom of the dip can be determined using the conservation of energy formula. Additionally, centripetal acceleration and the normal reaction force on a passenger can be calculated using the appropriate physics equations. The collision between two trains at the dip requires an analysis of momentum conservation and energy loss during the collision to determine how far the combined mass will rise afterward.

PREREQUISITES
  • Understanding of gravitational potential energy (GPE) and kinetic energy (KE)
  • Familiarity with conservation of energy principles
  • Knowledge of centripetal acceleration calculations
  • Basic concepts of momentum conservation in collisions
NEXT STEPS
  • Calculate maximum speed using the formula for GPE and KE
  • Determine centripetal acceleration using the formula a = v²/r
  • Analyze normal reaction force on a passenger using F = ma
  • Explore energy loss in collisions and its effect on vertical rise
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of roller coasters and the application of energy conservation principles in real-world scenarios.

lemon
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Hi:
I have a question to answer but I don't think I have enough information.
I know I have to consider that GPE goes to KE, with max.velocity at the bottom of the dip. I think that the amplitude is the radius of the dip. And I think it is important to think about Total energy is proportional to amplitude squared. But... I'm stuck how to start this without knowing the mass really.
any help appreciated. thank you.

On a fairground big dipper, a train free-wheels down a steep incline into a dip before rising out of the dip. The track at the dip has a radius of curvature of 20.0 m. Assuming the train is almost stationary at the top of the incline before descending, and given that the train drops 20.0m, calculate
(i) the maximum speed the train could have at the dip.
(ii) the centripetal acceleration of the train as it passes through the dip.
(iii) the normal reaction force on a passenger of mass 75kg at this point.
Unfortunately, another train is stationary at the bottom of the dip and a collision occurs in which the two trains move as one up the rise following the dip.
(iv) how far vertically will the combined two-part train go up the rise?



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The Attempt at a Solution

 
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Do you know the equations for gravitational potential energy and kinetic energy? Can you write out the conservation of energy formula?
 
Thank you:
I have done some calculations - if you could check please.
The last part of the question is...
Not sure. I started with conservation of momentum and collisions, but actually, I think it doesn't apply here. The question says 'how far vertically'. So again, is it a change back from KE to GPE? And what about any loss of energy in the collision?

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