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Home / Doubts and Solutions/JEE/Physics

Particle A moves with speed 10m/s in a frictionless circular fixed horizontal pipe of radius 5m and strikes with B of double mass that of A.Coefficient of restitution is 1/2 and particle A starts its journey at t=0.The time at which second collision occurs is 

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Asked by m.nilu 15th September 2018, 7:56 PM
Answered by Expert
Answer:
Initial momentum of both particles before collision = m×10 + (2m)×0 = m×10 ...........(1)
 
Let vA and vB are the speeds after collision of particle A and B respectively.
 
Then final momentum after collision = m×vA + (2m)×vB ................(2)
 
By conservation of momentum :-  m×10 = m×vA + (2m)×vB   or   vA + 2×vB = 10 ...............(3)
 
speeds before and after collision are related through coefficient of restitution e as,  vA-vB = -e×10
 
 vA = vB-10×0.5   or    VA = vB - 5 ...........(4)
 
solving (3) and (4), we get,  vA = 0,  vB = 5 m/s
 
Initial time taken by A to hit B, (π×5)/10  = π/2  seconds.....................(5)
 
After collision, A comes to rest. Time taken for b to hit A = (2π×5)/5  = 2π seconds....................(6)
 
total time = (π/2) + 2π  = 2.5×π seconds
Answered by Expert 19th September 2018, 1:34 PM
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