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A rollercoaster car passes the hill which is 5.5m above the ground at speed 9.3m/s, and rolls over the second hill which is 2.5m above the ground, and heads toward the third hill which is 4.0 m higher than the first one. If the track is frictionless,
a. What maximum height will the car climb on the third hill? [h max = 9.9m, so car will climb the entire 9.5m hill]
b. Will the speed of the car on top of the hill 3 be lower or higher than its speed on the top of the hill one? [lower]
c. Calculate the speed of the car when it is 1m lower than the top of the third hill. [5.3m/s]

Would somebody kindly go over the questions :D

Respuesta :

Answer:

Explanation:

Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.

Let m be the mass of the car and ground level is the origin.

on the 5.5 m hill, total energy is

E = PE + KE

E = mgh + ½mv²  

E = m(9.8)(5.5) + ½m(9.3)² = 97m J

a) The maximum height will occur when the total energy is all potential energy.

E = mgh

h = E/mg

h = 97m/m(9.8) = 9.9 m  

As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.

b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.

c) KE = E - PE

KE = 97m - m(9.8)(9.5 - 1.0)

KE = 97m = 83.3m

KE = 13.7m = ½mv²

v² = √(2(13.7)

v = 5.2345...

v = 5.2 m/s

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