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A tube of mass 2m and radius 'R' is kept on smooth horizontal surface. A small sphere of mass m kept inside tube is imparted velocity v_{0} along y axis as shown in figure. The sphere just fits into the tube. Motion of particle and tube take place in horizontal plane and there is no friction at any contact. Then, speed of particle when it reaches point B is

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Asked by salonirajpal123 15th March 2022, 7:19 AM
Answered by Expert
Answer:
Initially , when sphere is at A ,
 
x-component of velocity of centre of mass = Vcm_x = 0  .......................... (1)
 
y-component of velocity of centre of mass = Vcm_y =  (1/3) v......................(2)
 
( initially, tube is not moving )
 
When sphere reaches the point B, y-component velocity of sphere is zero . It has only x-component vx .
 
When sphere reaches the point B, Let Vx and Vy be the respective x-component and y-component velocities.
 
When sphere reaches the point B, x-component of velocity of centre of mass  is given as
 
Vcm_x = (1/3) vx + (2/3)Vx  = 0   ( eqn.(1) is used )
 
Hence we get ,  Vx = -(1/2) vx  ............................ (3)
 
When sphere reaches the point B, y-component of velocity of centre of mass  is given as
 
Vcm_y =  (2/3)Vy  = (1/3) vo    ( eqn.(2) is used )
 
Hence we get ,  Vy = (1/2) vo  ............................ (4)
 
By energy conservation , (1/2) m vo2 = (1/2) m vx2 + (1/2) ( 2m ) Vx2 + (1/2) ( 2m ) Vy2
 
Using eqn.(3) and (4) , above expression is simplified as
 
5 vx2  = 3 vo2
 
Hence , begin mathsize 14px style v subscript x space equals space square root of 3 over 5 end root space v subscript o end style
Answered by Expert 17th March 2022, 9:21 PM
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