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 = 13.0 m/s at height 29.0 m.

(a) What is the speed of the car at point A?
m/s

(b) What is the speed of the car at point B?
m/s

(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

at A

total energy = Eo = 0.5*m*vo^2 + m*g*h

at B

total energy EA = 0.5*m*vA^2 + m*g*h

from energy conservation

Eo = EA

VA = vo = 13 m/s

(b)

EB = 0.5*m*vB^2 + m*g*hB = 0.5*m*vB^2 + m*g*h/2

EB = Eo

0.5*m*vB^2 + m*g*h/2 = 0.5*m*vo^2 + m*g*h

0.5*vB^2 + g*h/2 = 0.5*vo^2 + g*h

(0.5*vB^2) + (9.8*29/2) = (0.5*13^2)+ (9.8*29)

vB = 21.3 m/s

c)

at

Ec = 0.5*m*vc^2

Ec = Eo

0.5*m*vc^2 = 0.5*m*vo^2 + m*g*h

0.5*vc^2 = 0.5*vo^2 + g*h

(0.5*vc^2) = (0.5*13^2)+ (9.8*29)

vc = 27.15 m/s

d)

at the highest pointD, v = 0

total energy at the highest point = m*g*H

at D

ED = m*g*H

ED = Eo

m*g*H = 0.5*m*vo^2 + m*g*h

9.8*H = (0.5*13^2)+ (9.8*29)

H = 37.6 m

e)

the above equations does not depend on mass

vA = 13 m/s

vB = 21.3 m/s

vc = 27.15 m/s

H = 37.6 m

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