# The following problem deals with a movie of the Viper ride.You do not need  to view the movie to do this.The movie is scaled by assuming that the  support beams are 4 meters apart and the origin is set at the location of  center of mass of the first car of the roller coaster in frame 1. The following observations can be made:  (1) the initial speed of the first car is approximately 20.2 [m/s];  (2) In frame 12 the center of mass of the first car is at about y = 13.5 m  above the horizontal xaxis.  In frame 40 the entire set of cars ends up traveling on a level part of the track. In this problem let’s suppose we have an ideal roller coaster with no friction.  Assume that the mass of each car with four riders is 2.2x103 [kg].   Just using the data given in this problem (and not additional data extracted from a VideoPoint analysis of the movie),  show your equations and calculations as you find the following idealized  quantities: (a) The maximum kinetic energy of the first car when it is loaded with 4  riders. (b)The minimum potential energy of the first car relative to the location of  the first car in frame 1 when it is loaded with 4 riders. (c)The total mechanical energy of the first car relative to its location of the  first car in frame 1 when it is loaded with 4 riders. (d) The maximum potential energy of the first car relative to its location of  the first car in frame 1 when it is loaded with 4 riders. (e) The minimum kinetic energy of the first car when it is loaded with 4 riders (f) The velocity of the roller coaster in frame 40 if there is no friction.  Hint: Velocity is a vector! Specify its magnitude and direction! (g) Would you know how to find the final cart velocity in our friction-free roller coaster without using conservation of mechanical energy? Explain Assuming that the origin is located at the ground level this means that the first car is at zero height `h=0m` in first frame. Total energy of the first car is

`E = E_p+E_k`

and since for `h=0 rArr E_p =0` it means the maximum kinetic energy first car has...

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Assuming that the origin is located at the ground level this means that the first car is at zero height `h=0m` in first frame. Total energy of the first car is

`E = E_p+E_k`

and since for `h=0 rArr E_p =0` it means the maximum kinetic energy first car has is in the first frame (at the ground level).

Taking the mass of one person `m =80 kg` the maximum kinetic energy is

`E_k("max") =4*(mv^2)/2 =2*m*v^2 =2*80*20.2^2 =65286.4 J`

b) As said above the minimum potential energy relative to origin (location of the first car in frame 1) is zero.

`E_p =4*(m*g*h) =4*0=0 J`

c) Assuming again that there is no friction on tracks the total mechanical energy is the same along all the ride. Its value (constant along all the track) the can be found at origin (zero height) as:

`E = E_p +E_k =0 +65286.4 =65286.4 J`

d) Assuming that the maximum height of the first car is attained in frame 12 (and maintained until frame 40) at a height of `h =13.5 m` the maximum potential energy is

`E_p("max") =4*(m*g*h) =4*(80*9.81*13.5) =42379.2 J`

e) The minimum kinetic energy of the first car is at maximum height

`E_k("min") =E -E_p("max") =65286.4-42379.2 =22907.2 J`

f) As said, the first car attains the maximum height in frame 12 and maintain this height until frame 40. This means that its kinetic energy is maintained from frame 12 to frame 40.

`E_k("min") =4*(m*v^2)/2 =2*m*v^2`

` v=sqrt(E_k/(2m)) =sqrt(22907.2/(2*80)) =11.97 =12 m/s`

The direction of the velocity is horizontal (parallel to the x axis).

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