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.

Asked by ayan1.chatterjee | 13th Jan, 2021, 08:47: AM

Expert Answer:

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

Answered by Thiyagarajan K | 13th Jan, 2021, 09:35: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.