A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 30 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?

1 Answer
A08
Mar 16, 2018

It is assumed that there is no loss of energy on friction and while the roller coaster descends on its track no external force is applied.

When the roller coaster starts from rest at its highest point it has only potential energy #=mgh#
where #m# is mass of roller coaster, #g=9.81\ ms^-2# is acceleration due to gravity and #h# is height of highest point.

As it reaches ground level all its initial potential energy got converted into its kinetic energy

#1/2mv^2=1/2m(30)^2=mgh# ......(1)

Let its speed when its height was half that of its starting point be#=v_"half h"#

Using Law of conservation of energy at this half-height point

#mgh/2+1/2mv_"half h"^2=mgh#
#=>mv_"half h"^2=mgh#

Using (2) we get

#mv_"half h"^2=1/2m(30)^2#
#=>v_"half h"^2=(30)^2/2#
#=>v_"half h"=(30)/sqrt2#
#=>v_"half h"=21.2\ ms^-1#, rounded to one decimal place

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