A single frictionless roller-coaster car of mass m = 825 kg tops the first hill

Question

A single frictionless roller-coaster car of mass m = 825 kg tops the first hill with speed v0 = 15.0 m/s at height 31.0 m.

A single frictionless roller-coaster car of mass m = 825 kg tops the first hill with speed vo = 15.0 m/s at height 31.0 m. First hill h/2 (a) What is the speed of the car at point A? (b) What is the speed of the car at point B? (c) What is the speed of the car at point C? rYnf: (d) How high will it go on the last hill, which is too high to cross? (e) If we substitute a second car with twice the mass, what is the speed of this car at point A? What is the speed of this car at point B? What is the speed of this car at point C? How high will it go on the last hll?

Explanation / Answer

A) by energy conservation,

mgh + 0.5 mv^2 = mgh + 0.5mv0^2

v = v0 = 15 m/s

B) again by energy conservation,

mgh/2 + 0.5 mv^2 = mgh + 0.5mv0^2

v ^2 = vo^2 +gh

v = sqrt (15^2 + 31*9.8)

= 23 m/s answer

C) again by energy conservation,

0.5 mv^2 = mgh + 0.5mv0^2

v ^2 = vo^2 +2gh

v = sqrt (15^2 + 2* 31*9.8)

= 28.5 m/s answer

D) at maximum height kinetic energy will be zero,

mgH = mgh + 0.5mv0^2

H = h + 0.5v0^2/g

= 31 + 0.5*15^2/9.8

= 42.5 m answer

E) mass will not impact anything here,

V=15 m/s

F) again v=23 m/s

G) again v=28.5 m/s

H) H = 42.5 m

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