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A roller coaster car rolls over the crest of 18 m high hill at aspeed of 3.2 m/s

ID: 1752147 • Letter: A

Question

A roller coaster car rolls over the crest of 18 m high hill at aspeed of 3.2 m/s. The car then rolls down the track to the bottomof the hill and continues up the next hill, which is 10 m high.(Neglect the size of the car and passengers and frictionaleffects.) a) How fast is the car moving at the bottom of the 18 mhill? b) How fast is the car moving at the top of the 10 mhill?
use the conservation of energy method a) How fast is the car moving at the bottom of the 18 mhill? b) How fast is the car moving at the top of the 10 mhill?
use the conservation of energy method

Explanation / Answer

      Given that the height of hillis H = 18 m       Speed at the top of hillis U = 3.2 m/s     ----------------------------------------------------------------- (a) Apply conservation of energy at top and bottomof the hill then we get                mgH + (1/2)mU2 = (1/2)mV2                          2gH + U2 = V2                                        V = ( 2gH + U2 )1/2                                            =--------- m/s (b) Apply conservation of energy at topand at h = 10m height of the hill then we get                mgH + (1/2)mU2 = (1/2)mv2 + mgh                          2gH + U2 = v2  + 2gh                                      V = ( 2gH + U2 - 2gh )1/2                                            =--------- m/s                                            =--------- m/s (b) Apply conservation of energy at topand at h = 10m height of the hill then we get                mgH + (1/2)mU2 = (1/2)mv2 + mgh                          2gH + U2 = v2  + 2gh                                      V = ( 2gH + U2 - 2gh )1/2                                            =--------- m/s                mgH + (1/2)mU2 = (1/2)mv2 + mgh                          2gH + U2 = v2  + 2gh                                      V = ( 2gH + U2 - 2gh )1/2                                            =--------- m/s                                            =--------- m/s     
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