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 45.0 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?
m/s

(d) How high will it go on the last hill, which is too high to cross?
m

(e) If we substitute a second car with twice the mass, what is the speed of this car at point A?
m/s

What is the speed of this car at point B?
m/s

What is the speed of this car at point C?
m/s

How high will it go on the last hill?
m

Explanation / Answer

a) same.

vA = vo = 15.0 m/s

b) use, vB^2 - vo^2 = 2*g*(h/2)

vB^2 = vo^2 + g*h

vB = sqrt(vo^2 + g*h)

= sqrt(15^2 + 9.8*45)

= 25.8 m/s

c)

use, vc^2 - vo^2 = 2*g*h

vc^2 = vo^2 + 2*g*h

vc = sqrt(vo^2 + 2*g*h)

= sqrt(15^2 + 2*9.8*45)

= 33.3 m/s

d) h = 45.0 + vo^2/(2*g)

= 45.0 + 15^2/(2*9.8)

= 56.5 m

e) same. speed of the cart does not depend on mass of the cart.

vA = 15.0 m/s

vB = 25.8 m/s

vC = 33.3 m/s

h = 56.5 m

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