Sir Isaac Newton Contest Question (Energy)

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SUMMARY

The discussion revolves around solving a physics problem involving energy conservation and projectile motion for a space shuttle landing on a distant planet with a gravitational acceleration of 2.0. The shuttle initially travels at a speed of 6.0 and ascends a frictionless ramp of height 5.0 at a 30-degree angle. Participants utilize the equations for kinetic energy (KE) and potential energy (PE) to derive the final speed at the top of the ramp and subsequently apply kinematic equations to calculate the horizontal distance (S) traveled after the ramp. The final answer for S is confirmed to be 12 units.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) equations
  • Familiarity with conservation of energy principles
  • Knowledge of projectile motion equations
  • Basic skills in algebra and solving equations
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  • Study the conservation of mechanical energy in physics problems
  • Learn about kinematic equations for projectile motion
  • Explore the effects of gravitational acceleration on projectile trajectories
  • Practice solving energy conservation problems in different contexts
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jeran921
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Hey
This problem has been stumping me for awhile and I just don't seem able to get it


Homework Statement


A space shuttle lands ona distant planet where the gravitiational acceleration is 2.0 (We don't know the local units of length and timem but they are consistent throughout this problem.) The shuttle coasts alond a level, frictionless plane with a speed of 6.0. It then coasts up a frictionless ram of height 5.0 and angle 30 degrees. After a brief ballistic flight, it lands a distance S from the ramp. Solve for S in local units of length. Assume the shuttle is small compared to the local length unit and that all atmospheric effects are negligible.


Homework Equations


I used the equation
Eg=mgy
Ek=0.5mv^2



The Attempt at a Solution


first befor it hits the ramp
mv^2/2 = 18m (m is not meters)
on the ramp mgh would be introduced so
mgh+mv^2=?
And this is where I got lost

 
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I like to write out the whole thought before resorting to numbers!
I'd say energy before ramp = energy after ramp
KE = KE + PE
½mv² = ½mv² + mgh
Hope that is a start for you. Remember the v on the left is the 6 you know and the v on the right is the end of ramp speed you are looking for.
 
I'll try to do this but I'm not sure if I can get the right answer (I'm taking intro mechanics). So do you have the answer for me to check?
 
You go first, ahmadmz!
 
Ok I did it something like this :

Like you said Energy has to be same, so
Bottom of ramp, (call it point 1) K1 = (mv1)/2 and U1 = 0
Top most part of ramp , (point 2) K2 = (mv2)/2 + 5*m*2 (This 2 is the "g" here).
Equating K1 + U1 = K2 + U2 we can solve for v2 = 4
Then maybe we can use the kinematic equations for constant acceleration to find the S?

Or is there a way to solve this using energy methods only? We are doing this right now in class so I need more experience :)
 
The 4 looks good. The rest of it is a pretty standard 2D projectile motion question - just remember the end point is below the start point. You know the routine - write out your equations for both the horizontal and vertical parts.
 
Actually what it sys in the back of the book is 12 units. All that really needs to be done now is use a projectile motion equation and it should give me the answer.
 
Last edited:

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