a metal track has a long inclined ramp followed by a 30 cm diameter loop-the-loo

Question

a metal track has a long inclined ramp followed by a 30 cm diameter loop-the-loop. A 20 g glass marble ( a solid sphere) is placed on the track at the bottom of the loop. A 60 g steel marble is placed at the top of the incline and allowd to roll down the track. At the bottom of the incline the steel marble has an elastic collision with the glass marble (which is initially at rest). How high must the steel marble start so that after the collision the glass marble will be able to just make it around the loop?

Hint: you should use the following concepts

centripetal acceleration

conservation of energy (twice!)

conservation of momentum (elastic collision)

Explanation / Answer

let v2' is the speed of glass marble at the top of the loop.

v2' = sqrt(g*R)

v2 is the speed of the marble glass after the collsion.

Apply Enrgy conservation,

0.5*m*v2^2 = 0.5*m*v2'^2 + m*g*d

0.5*m*v2^2 = 0.5*m*(g*R) + m*g*(2*R)

v2 = sqrt(5*g*R)

= sqrt(5*9.8*0.15)

= 2.71 m/s

let u1 is the speed of stell marble at the bottom

V2 = 2*m1*u1/(m1+m2)

u1 = (m1+m2)*v2/(2*m1)

= (60+20)*2.71/(2*60)

= 1.807 m/s

let h is the height from which steel marble rolls down,

Apply enrgy conservation

m*g*h = 0.5*m*u1^2 + 0.5*I*w^2

m*g*h = 0.5*m*u1^2 + 0.5*(2/5)*m*r^2*w^2

m*g*h = 0.5*m*u1^2 + 0.2*m*v^2

m*g*h = 0.7*m*u1^2

h = 0.7*u1^2/g

= 0.7*1.807^2/9.8

= 0.233 m <<<<<<<<<<<---------Answer

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