Mechanics Roller Coaster

Q. Mechanics Roller Coaster

Question

At an amusement park, a child gently releases a toy roller coaster, Toy B, at  the top of a curved track. 

During the journey on the track, the toy roller coaster, Toy B, is subjected to the frictional force due to the toy's motion on the track. The frictional force between the toy and the curved track may be assumed to have an average value of 1.26 x 10^-2 N throughout the entire track under normal condition and 2.1 x 10^-3 N when lubricated. Mass of Toy B=50g, distance H =0.82m.

i) If Toy B moves along the track under normal condition, and just missed reaching point Z, find the energy loss to friction.

ii) Hence find the total distance travelled by Toy B on the curved track from the point of release to just before point Z.

iii) If the track is lubricated, find the speed of Toy B at point Z.

iv) If the track is lubricated, Toy B will make a head-on elastic collision with the identical stationary toy Toy Z at point Z. Find the horizontal distance d, that Toy Z will hit the horizontal surface, which is at 0.4m below point Z. (assume toy Z moves off with constant speed after collision, before free fall)

Answer 

i)

Kinetic energy gained during drop by height H is lost due to frction
=> Loss of energy due to friction
= mgH
= 50*10^-3*9.81*(0.82) J
= 0.402 J

ii)
Work done by frictional force 
= Force x displacement opposite to the direction of force = 0.402
=> Distance travelled 
= (0.402) / (1.26x10^-2) m
= 32 m.

iii)
If the track were lubricated, loss due to friction
= 32 * 2.1 x 10^-3 J
= 0.0672
KE = PE - Loss due to friction
=> (1/2) mv^2 = 0.402 - 0.0672 = 0.3348
=> v = √(2*0.3348)/(0.050) = 3.66 m/s.

iv)
In head-on elastic collision between two identical bodies, velocities get exchanged.
=> Toy Z will move with velocity = 3.66 m/s in the horizontal direction.
Time to move down by 0. 4 m with zero initial vertical velocity is given by
h = (1/2)*9.81t^2
=> t = √(0.8/9.81) = 0.286 s.
During this time toy Z moves with constant horizontal velocity of 3.66 m/s covering horizontal distance
= (0.286) * (3.66) m
= 1.047 m.