Speed of Coaster In Fig. 10-28, a frictionless roller coaster of mass m = 1000.0
ID: 2008790 • Letter: S
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
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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