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 = 11.0 m/s at height 30.0 m.

(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?


(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 hill?

Explanation / Answer

In part a,car will have only horizental speed. Also, there is no horizental acceleration, So speed will remain same throughout these point and will be 11m/s.

PART-B:-
v(y) at point B= sqrt(2gh/2)= 17.32m/s
So, overall velocity at point B= sqrt(17.32^2+ 11^2)= 20.51m/s

PART-C:-
V(y) at point c= sqrt(17.32^2 + gh)= 24.49m/s.
So, overall velocity, v(c)= sqrt(24.49^2+121)=26.85m/s

PART-D:-
Height,H= (v(y) at c)^2/2g= 30 meter.

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