Speed of Coaster In Fig. 10-28, a frictionless roller coaster of mass m = 1000.0

Question

Speed of Coaster In Fig. 10-28, a frictionless roller coaster of mass m = 1000.0 kg. tops the first hill with speed v1 = 12.00 m/s. Assume the first hill is h = 100.0 m tall.

Figure 10-28
(a) What is the speed of the coaster at point A?
m/s
(b) What is the speed of the coaster at point B?
m/s
(c) What is the speed of the coaster at point C?
m/s
(d) How high will it go on the last hill, which is too high for it to cross?
m

(e) If we substitute a second coaster with twice the mass, what is the speed of this coaster at point A?
m/s
What is the speed of this coaster at point B?
m/s
What is the speed of this coaster at point C?
m/s
How high will it go on the last hill?
m


DIAGRAM

http://img831.imageshack.us/i/diagram2.gif/

Explanation / Answer

a .

U + K = 0 ; [ since only conservative force act ] ;

U = 0 ; since height doesnt change ;

so K = 0 ;

so velocity will not change from point A ,

vA = 12 m /s ;

b.

U + K = 0 ;

mg ( h/2 - h ) + m / 2 [ v2^2 - v1^2 ] = 0 ;

or - gh / 2 + 0.5 ( vb^2 - 12^2 ) = 0 ;

or 0.5 ( vb^2 - 12^2 ) = 9.8 * 100 /2 ;

or vb = 33.5261 m /s ;

c .

U + K = 0 ;

mg ( 0 - h ) + m / 2 [ vC^2 - vA^2 ] = 0 ;

g ( 0 - h ) + 0.5 [ vC^2 - 12^2 ] = 0 ;

or vC = 45.8693 m /s ;

d . its velocity at bottom is vC = 45.8693 ;

it will go up till its velociy becomes 0

so

U + K = 0 ;

mg ( H - 0 ) + m /2 [ vf^2 - vc^2 ] = 0 ;

or g ( H - 0 ) + 0.5 [ 0 - 45.8693^2 ] = 0 ;

or h = 107.3469 m

e .

as we can see, in the above answewrs for calculation for velocities and height , they are independent of mass , so all the above answers remains unchanged , even if the mass is changed .

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