A small particle starts moving from point P along the smooth inclined surface of circular cross section seen in figure (R=5m)

Options

A. The different of height between the highest point of the trajectory and the point P is `64/25 m `
B. The ratio of the kinetic energies at the highest point of the trajectory and the lowest point of the circular path is `64/125`.
C. The difference of height between the highest point of the trajectory and the point P is `36/25 m `
D. The ratio of the total mechanical energies at the highest point of the trajectory and the lowest point of the circular path is 1.

Answer

ABD

Explanation

`mg xx4=1/2 mv^(2)`
`v=sqrt(80)`
`h=(v^(2) sin^(2)37^(@))/(2g)=80/20xx9/25=36/25 m`
`DeltaH=4-36/25 =64/25 m`
`mgxx 5 =1/2 mv_(1)^(2)`
`v_("top")=v cos theta =sqrt(80) xx4/5`
`(K_("top"))/(K_("bottom")) =(1/2m xx80xx16/25)/(mgxx5)=64/125`