A marble is dropped into a frictionless U tube. If the tube is semi circular with  mean radius 5 cm and mass of the ball 2 gram, then find its velocity at the bottom of the tube.

Kinetic energy of marble is determined using conservation of energy .

 

Hence we equate the initial potential energy to kinetic energy of marble at bottom as given below.

 

( m g h ) = (1/2) m v² ........................(1)

 

where m is mass of marble, g is acceleration due to gravity , h is vertical height of marble

from bottom of semi-circular tube and v is speed of marble at bottom of semi-circular tube .

 

h = radius of  semi-circular tube = 5 cm

 

Hence from eqn.(1) , we get ,  v = { 2 g h }^1/2 = { 2 × 9.8 × 0.05 }^1/2 ≈ 1 m/s